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VISTAS OF
SPECIAL FUNCTIONS
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VISTAS OF
SPECIAL FUNCTIONS
Shigeru Kanemitsu & Haruo Tsukada
Kinki University, Japan
World Scientific
N E W J E R S E Y ? L O N D O N ? S I N G A P O R E ? B E I J I N G ? S H A N G H A I ? H O N G K O N G ? TA I P E I ? C H E N N A I
Published by
World Scientific Publishing Co. Pte. Ltd.
5 Toh Tuck Link, Singapore 596224
USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601
UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE
British Library Cataloguing-in-Publication Data
A catalogue record for this book is available from the British Library.
VISTAS OF SPECIAL FUNCTIONS
Copyright Е 2007 by World Scientific Publishing Co. Pte. Ltd.
All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means,
electronic or mechanical, including photocopying, recording or any information storage and retrieval
system now known or to be invented, without written permission from the Publisher.
For photocopying of material in this volume, please pay a copying fee through the Copyright
Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission to
photocopy is not required from the publisher.
ISBN-13 978-981-270-774-1
ISBN-10 981-270-774-3
Printed in Singapore.
ZhangJi - Vistas of Special.pmd
1
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To Professor Michel Waldschmidt with deep respect
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Preface
This book is intended for aspirant readers who are eager to have basic
knowledge of special functions in an organic way. We have kept paying attention to make an order in various equivalent statements on special functions. A unique feature is that the reader can gain a grasp of (almost) all
existing (and scattered around) formulas in the theory of gamma functions
etc. in a clear perspective through the theory of zeta-functions. Thus,
this is a book of special functions in terms of the zeta-functions. Reading
through this book, the reader can master both fields efficiently. Here a
hunter looking for two rabbits gets two.
Here are some descriptions of the contents.
In Chapter 1, we present a unified theory of Bernoulli polynomials with
all equivalent conditions properly located. We have revealed that the difference equation (DE) satisfied by the Bernoulli polynomial corresponds
to differentiation while the Kubert identity (K) corresponds to integration
(the Riemann sum into equal division). This new view point makes the
whole theory very lucid.
In Chapter 2 we shall present rather classical and standard theory of the
gamma and related functions. Classical as it looks, we shall provide some
very unique features of the Euler digamma function from which we may
deduce the corresponding properties of the gamma function. Especially, we
shall give three proofs of the remarkable formula of Gauss on the values
of the digamma function at rational arguments. One is classical and is
presented in Chapter 2. Other two proofs are more original given in Chapter 8, one is the limiting case (Theorem 8.2) of the Eisenstein formula in
its genuine form (a theorem due to H.-L. Li, L.-P. Ding and M. Hashimoto,
describing a basis element in terms of another basis of the space of periodic
Dirichlet series), the other is the theorem of M. Hashimoto, S. Kanemitsu
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Vistas of Special Functions
and M. Toda about the equivalence between the finite form of the value of
the Dirichlet L-function at 1 and the formula of Gauss.
In Chapter 3, we shall present the theory of the Hurwitz zeta-function.
The main ingredient is the integral representation for its partial sum. This
is to the effect that once we have an integral representation as the one we
have, we may immediately draw information for the derivatives, i.e. we
have an inheritance of the information. The integral representation for
the partial sum is so informative that it contains all information we need
(Theorem 3.1). The versatility of this result will be developed in Chapter 5,
where through Lerch?s formula, we transfer the results on the Hurwitz zetafunction to those on the gamma and related functions. Especially, the
asymptotic results established in Chapter 3 will immediately transfer to the
Stirling formula and other asymptotic formulas for relatives of the gamma
function.
In Chapter 4, we shall present the theory of Bernoulli polynomials
through the negative integer values ?(?n, z) of the Hurwitz zeta-function.
Here we shall establish only three statements, i.e. the Fourier series (H),
the difference equation (DE) and the Kubert identity (K) from any of which
we may complete the theory following the logical scheme in Chapter 1.
In Chapter 5, first we shall reveal the power of theorems in Chapter 3
to exhibit what the Dufresnoy-Pisot type uniqueness theorem means. Then
we shall go on to presenting the first circle (krug p?iervyi) which connects
various identities between gamma and trigonometric functions to the functional equations (zeta-symmetry) of the zeta-functions . Thus we shall show
that everything comes from the functional equation. A remarkable notice
is that such trigonometric identities like the infinite product for the sine
function or the partial fraction expansion for the cotangent function are
equivalent to the functional equation, thus revealing why Euler succeeded
in solving the Basler problem.
In Chapter 6, we shall further pursue this zeta-symmetry in relation to
the crystal symmetry through the Epstein zeta-function. We surpass the
preceding results by introducing the signs and giving the Chowla-Selberg
type formula (based on the Mellin-Barnes integrals) and provide a quick
means for computation of the Madelung constants.
In Chapter 7, we shall provide rudiments of the theory of Fourier series
and integrals to such an extent that is sufficient for applications and reading
through this book, for the sake of the reader who wants to learn it quickly.
Chapter 8 is, so to say, a discrete version of Chapter 7, i.e. the finite
Fourier series (transforms). Through this we make clear the orthogonality
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ix
of characters and other bases of the space of Dirichlet series with periodic
coefficients, giving rise to the theorem mentioned above. We can naturally
extend our method to develop the similar theory for higher derivatives of
the Dirichlet L-function, including Kronecker?s limit formula. But because
of limitation of time, we cannot go further.
Appendix A gives the very basics of the theory of complex functions.
We present mostly results only, and the interested reader should consult a
standard book for their proofs. We shall give, however, some details on the
use of residue theorem.
Appendix B assembles summation formulas and convergence theorems
used in the book. Especially, the Fourier series for the first periodic
Bernoulli polynomial is so essential and important, we give two proofs,
one depending on ordinary Fourier theory (Chapter 7) and the other on
the polylogarithm function of degree 1, where we apply the theorem of
Abel and Dirichlet in place of Fourier theory.
As is explained above, Chapters 1 and 4 are parallel, so are Chapters 2
and 5. To understand Chapters 4 and 5, one should read Chapter 3 first.
If one finds some difficulties, then one is referred to Appindices A and B.
Chapters 7 and 8 can be read independently, but it will be more instructive
to read both in parallel. Chapter 6 can be read separately which requires
more knowledge of Bessel functions. Because of lack of time, we could not
state much about them.
This publication was supported by Kinki University Grant for Publication, No. GK04 in the academic year 2006. The authors are thankful to
Kinki University for their generosity of this support. They also would like
to thank Ms. Chiew Ying Oi who helped them all through the process with
her efficient editorial skills. And toward the end of the process Ms. Zhang
Ji supported us and we would like to express our heartily thanks to her.
The authors would like to express their hearty thanks to their close
friend Professor Y. Tanigawa for his constant support, encouragement, and
stimulating discussions. The first author would like to thank his close friend
Professor Heng Huat Chan for his enlightening remark on the equivalent
statements to the functional equation, thanks to which he got motivated
enough to start writing this book. The second author was naturally got
infected the passion of the first. Thanks are also due to Ms. L.-P. Ding and
Mr. M. Toda for their devoted endeavor, without their enthusiastic help,
the book would have not been risen out.
the authors
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Contents
Preface
vii
1. The theory of Bernoulli and allied polynomials
1
2. The theory of the gamma and related functions
29
2.1 Gamma function . . . . . . . . . . . . . . . . . . . . . . . .
2.2 The Euler digamma function . . . . . . . . . . . . . . . . .
3. The theory of the Hurwitz-Lerch zeta-functions
3.1
3.2
3.3
3.4
3.5
Introduction . . . . . . .
Integral representations
A formula of Ramanujan
Some definite integrals .
The functional equation
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4. The theory of Bernoulli polynomilas via zeta-functions
51
54
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70
77
5. The theory of the gamma and related functions via zeta-functions 81
5.1 Derivatives of the Hurwitz zeta-function . . . . . . . . . . .
5.2 Asymptotic formulas for the Hurwitz and related zetafunctions in the second variable . . . . . . . . . . . . . . . .
5.3 An application of the Euler digamma function . . . . . . . .
5.4 The first circle . . . . . . . . . . . . . . . . . . . . . . . . .
6. The theory of Bessel functions and the Epstein zeta-functions
81
91
93
97
105
6.1 Introduction and the theory of Bessel functions . . . . . . . 105
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6.2 The theory of Epstein zeta-functions . . . . . . . . . . . . . 109
6.3 Lattice zeta-functions . . . . . . . . . . . . . . . . . . . . . 115
6.4 Bessel series expansions for Epstein zeta-functions . . . . . 125
7. Fourier series and Fourier transforms
7.1
7.2
7.3
7.4
Fourier series . . .
Integral transforms
Fourier transform .
Mellin transform .
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8. Around Dirichlet?s L-functions
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8.1 The theory of periodic Dirichlet series . . . . . . . . . . . . 165
8.2 The Dirichlet class number formula . . . . . . . . . . . . . . 174
8.3 Proof of the theorems . . . . . . . . . . . . . . . . . . . . . 176
Appendix A Complex functions
183
A.1 Function series . . . . . . . . . . . . . . . . . . . . . . . . . 183
A.2 Residue theorem and its applications . . . . . . . . . . . . . 193
Appendix B Summation formulas and convergence theorems
197
B.1 Summation formula and its applications . . . . . . . . . . . 197
B.2 Application to the Riemann zeta-function . . . . . . . . . . 202
Bibliography
207
Index
213
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Chapter 1
The theory of Bernoulli and allied
polynomials
Abstract
In this chapter we shall develop the theory of Bernoulli
polynomials in a way different from many existing books in
that we shall reveal the relationships between the Propositions (D?)-(H) and that any one of them (or a combination
thereof) can be adopted as a definition of Bernoulli polynomials (cf. Fig. 1. 5 for loose equivalence). Our intension is
not to provide a proof of the exact equivalence but equivalence in a loose sense (e.g. up to the initial condition
or the normalization) so that the reader can have a better grasp of the formulas scattered around the literature
([Bo?h], [Ca], [Erd], [Ni]). We shall also state some facts
about the cyclotomic polynomials (used in Chapter 2.)
We adopt Lehmer?s terminology [Leh2].
Definition 1.1 The Bernoulli polynomial Bn (x) of degree n (n =
0, 1, 2, . . .) can be defined by either of the following defining conditions.
(D0 ) (Appell sequence 1832)
Bn0 (x) = n Bn?1 (x)
(1.1)
with initial value B0 (x) = 1 and with normalization
Z 1
Bn (x) dx = 0 (n ? N).
0
If we know the differentiation formula (D 0 ), it is immediate to calculate the
k-th derivative:
1
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Fig. 1.1
Bn(k) (x) =
Jacob Bernoulli
n!
Bn?k (x).
(n ? k)!
(1.2)
By (1.2) we have the Taylor expansion
n
Bn (x) = (B + x) =
n X
n
k=0
k
Bn?k (0) xk ,
and the normalization condition amounts to the recurrence
n?1
X
k=0
n
Bk (0) = 0,
k
n ? 2.
(1.3)
We denote the value Bn (0) by Bn and refer to it as the n-th Bernoulli
number. Throughout in what follows we understand they are defined
by (1.3) once and for all. Another definition by (1.6) leads to the same
recurrence as (1.3) and these two definitions are consistent (cf. Remark 1.1).
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The theory of Bernoulli and allied polynomials
(A) Addition formula
Bn (x + y) =
n X
n
k=0
=
k
n X
n
k=0
k
Bn?k (x) y k
Bn?k (y) xk .
(U ) Umbral calculus formula (Lucas 1891)
The n-th Bernoulli polynomial can be expressed as (B + x)n :
n X
n
Bn (x) = (B + x)n =
Bn?k xk ,
k
(1.4)
k=0
i.e. (1.7), where, by umbral calculus, we mean that after expanding the
binomial, the exponent of B is to be degraded to subscript.
Theorem 1.1 The defining conditions in Definition 1.1 are equivalent to
conditions (DE)-(H).
(DE) {Bn (x)} are (principal) solutions of the difference equation
?Bn (x) = Bn (x + 1) ? Bn (x) = n xn?1 ,
where ? signifies the difference operator ?u(x) = u(x + 1) ? u(x).
(G) Generating functionology (Euler 1738).
Bernoulli polynomials Bn (x) are defined as the Taylor coefficients of the
generating function
?
X
Bn (x) n
z exz
=
z
z
e ? 1 n=0 n!
(|z| < 2?),
(1.5)
whence, in particular, Bn = Bn (0) are defined by the generating function
?
X
Bn n
z
=
z
z
e ? 1 n=0 n!
(|z| < 2?).
(1.6)
As will be indicated in Remark 1.1, it follows from (1.5) and (1.6) that
Bn (x) is a polynomial of degree n given by
n X
n
Bn (x) =
Bn?k xk .
(1.7)
k
k=0
(K) Kubert identity (Raabe 1851)
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Bn (x) is a monic polynomial of degree n satisfying
Bn (x) = mn?1
m?1
X
Bn
k=0
x+k
m
,
(1.8)
for each m ? N and x ? R. This identity is often referred to as the distribution property or the multiplication formula.
(H) Fourier series (Hurwitz 1890)
B n (x) = ?
?
X
n!
e2?ikx
,
n
(2?i)
kn
(1.9)
k=??
k6=0
where B n (x) = Bn (x ? [x]) , [x] being the integral part of x, for n ? N (in
the case n = 1, we should have (7.9)).
(S) Sums of powers (J. Bernoulli 1705?) As in Comtet [Com] let
Z(n, r) =
n
X
kr .
(1.10)
k=1
Then
Z(n, r) =
1
(Br+1 (n + 1) ? Br+1 ) ,
r+1
(r ? N)
(1.11)
(r ? N).
(1.12)
and
r 1 X r+1
Bk и (n + 1)r+1?k ,
Z(n, r) =
r+1
k
k=0
This was known to Jacob Bernoulli in his Ars Conjectandi, 1713
(posthumously published; J. Bernoulli died in 1705, and so (S) may be
proved in 1705? cf. Lehmer [Leh2]).
Although Formula (1.9) has been known since 1713, numerous papers
are still appearing which claim new closed formulas for Z(n, r) up to a
certain r, say 1,000.
z
Remark 1.1 We note that the function z
is analytic in |z| < 2?
e ?1
(including the origin cf. Theorem A.8), so that it has the Taylor expansion (1.6), which implies the recurrence (left-hand side of (1.3))= 0, n ? 2.
z
exz has its
= 1, n = 1, in conformity with (1.3). On the other hand, z
e ?1
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The theory of Bernoulli and allied polynomials
Fig. 1.2
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5
Ars Conjectandi
Taylor expansion in |z| < 2?, given by the Cauchy product (or sometimes
called Abel convolution)
! ?
!
?
X
X xl
B
z
k
exz =
zk
zl ,
ez ? 1
k!
l!
k=0
l=0
which establishes (1.7) on comparing the coefficients, i.e. (G)?(U ). The
special case x = 1 gives
n X
n
Bn (1) =
Bk ,
k
k=0
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Fig. 1.3
which reduces, by (1.3), to
Bn (1) = Bn (0) = Bn ,
n?2
(1.13)
and B1 (1) = ?B1 .
Example 1.1 We may calculate Bernoulli numbers from (1.3) or (G):
1
, B2k+1 = 0 (k ? N). The first few
B0 = 1, B1 = ? 21 , B2 = 16 , B4 = ? 30
Bernoulli polynomials are:
1
B0 (x) = 1, B1 (x) = B0 x + B1 = x ? ,
2
1
B2 (x) = B0 x2 + 2B1 x + B2 = x2 ? x + ,
6
3
1
B3 (x) = B0 x3 + 3B1 x2 + 3B2 x + B3 = x3 ? x2 + x,
2
2
B4 (x) = B0 x4 + 4B1 x3 + 6B2 x2 + 4B3 x + B4
1
1
= x4 ? 2x3 + x2 ? , B4 = ? ,
30
30
5 4 5 3 1
5
B5 (x) = x ? x + x ? x,
2
3
6
5 4 1 2
1
1
6
5
B6 (x) = x ? 3x + x ? x + , B6 =
.
2
2
42
42
We shall prove the equivalence of some of the conditions (D 0 ) ? (H),
some being left unproved.
(D0 )?(A)
Indeed, (A) is nothing but the Taylor expansion of Bn (x + y) in y, and
n!
(k)
the Taylor coefficient is Bn (x), which is
Bn?k (x) by (1.2), whence
(n ? k)!
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The theory of Bernoulli and allied polynomials
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we have (A).
(A)?(D0 )
For y 6= 0 we have
n Bn (x + y) ? Bn (x) X n
=
Bn?k (x) y k?1 ,
y
k
k=1
whence we deduce Bn0 (x) = n Bn?1 (x).
(A)?(U )
We note that the umbral calculus formula (U ) is the special case of (A)
with y = 0.
(U )?(A)
We have by (1.4)
Bn (x + y) =
n X
n
k
k=0
=
n X
n
k
k=0
=
Bn?k (x + y)k
Bn?k
k X
k
l=0
n X
n X
n k
k
l=0 k=l
l
l
y k?l xl
Bn?k y k?l xl ,
where we have changed the order of summation. Applying the formula
n k
n n?l
=
,
(1.14)
k
l
l
k?l
we obtain
Bn (x + y) =
n X
n
l=0
=
l
n X
n
l=0
l
y
l
n X
n?l
k=l
yl
k?l
Bn?k xk?l
n?l X
n?l
k=0
k
Bn?l?k xk ,
by the change of variable. Now the inner sum is Bn?l (x) by (1.7).
(U )?(DE) is proved in the following
(1.15)
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Exercise 1.1
Deduce (DE) and the reciprocal relation
Bn (1 ? x) = (?1)n Bn (x)
(1.16)
from umbral calculus (U ).
Solution
By (A), on using (1.15),
Bn (1 + y) =
n X
n
l
l=0
y l Bn?l (1),
(1.17)
which is Bn (1 + y) = Bn (y) + n y n?1 , by (1.3) and (1.4), and (DE) follows.
By (1.13) and (1.17)
Bn (1 ? x) =
n X
n
l=0
l
(?x)l (?1)n?l Bn?l
which is (?1)n Bn (x) by (1.4).
(U )?(D0 )
0
Bk (x) =
k X
k
r=1
r
Bk?r r xr?1
k
(s + 1)Bk?1?s xs
s
+
1
s=0
k?1
X k ? 1 =k
Bk?1?s xs
s
s=0
=
k?1
X
= kBk?1 (x).
Proof of
Z
1
0
Bn (x) dx = 0 for n ? N.
Z
1
Bn (x) dx =
0
n X
n
k=0
k
Bn?k
Z
1
xk dx
0
n 1 X n+1
Bn?k
n+1
k+1
k=0
n+1 1 X n+1
=
Bn+1?k .
n+1
k
=
k=1
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9
The theory of Bernoulli and allied polynomials
Pn+1 n+1
Now the sum is
k=0
k Bn+1?k ? Bn+1 , which is Bn+1 (1) ? Bn+1
by
R 1 (1.4); this is in turn Bn+1 (0) ? Bn+1 by (1.16) and is 0. Note that
0 B0 (x) dx = 1.
(A)?(DE)
By (A)
Bn (x + 1) ? Bn (x) =
=
n X
n
k=0
k
Bn?k (1) xk ? Bn?k xk
n
(B1 (1) ? B1 ) xn?1
n?1
= n xn?1
on using (1.13).
(G)??(U )
(G)?(U ) is proved in Remark 1.1.
(U )?(G): We form the generating function
(1.4) to get
?
?
X
Bn (x) n X 1
z =
n!
n!
n=0
n=0
?
X
=
k=0
=
ez
X
k+l=n
Bk k
z
k!
!
z
exz ,
?1
?
X
Bn (x) n
z and substitute
n!
n=0
n!
B k xl
k! l!
?
X
xl
l=0
l!
z
!
l
zn
!
whence (1.5) follows.
(H)?(K)
Substituting (1.9) into the right-hand side of (1.8), we obtain
m
n?1
m?1
X
k=0
Bn
x+k
m
=?
x
m?1
?
X
n!
e2?i m r n 1 X 2?i r k
e m .
m
(2?i)n r=?? rn
m
r6=0
k=0
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The inner sum is 0 except when r is a multiple of m by (8.5). Hence the
?
X
n!
e2?ixr
right-hand side becomes ?
, which is B n (x).
(2?i)n r=?? rn
r6=0
To extend the range (0, 1) to R we may refer to Milnor?s argument [Mi,
Lemma 7].
(G)?(K)
Consider the generating function
? m?1
X
X
Bn
n=0 k=0
=
x+k
m
m?1
X mz emz x+k
m
(mz)n
=
mz
n!
e ?1
k=0
?
X
ze
zn
e ?1
mz e
=
m
=
m Bn (x) ,
mz
z
z
e ?1 e ?1
e ? 1 n=0
n!
xz
xz
mz
whence (1.8) follows.
(DE)?(S)
This follows on summing (DE) with x = 0, 1, и и и , n.
Although the above implication is the most natural, we may also apply
(G) to deduce (S).
(G)?(1.11)
Consider the generating function
fn (z) =
?
X
Z(n, r)
z r+1 .
r!
r=0
On one hand we have
fn (z) =
?
X
(r + 1) Z(n, r)
r=0
(r + 1)!
z r+1 ,
and on the other, substituting (1.10), we have
fn (z) = z
?
n
n X
?
X
X
1
1 X r r
k z =z
(kz)r ,
r!
r!
r=0
r=0
k=1
k=1
whence
fn (z) = z
n
X
k=1
ekz .
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Rewriting the sum of the geometric series appearing above as
z
e(n+1)z ? ez
ze(n+1)z
z
= z
? z
? z,
z
e ?1
e ?1
e ?1
we have
fn (z) =
?
X
zr
r=0
r!
Br (n + 1) ?
?
X
zr
r=0
r!
Br ? z
by (G), whence
fn (z) =
?
X
r=1
(Br+1 (n + 1) ? Br+1 )
z r+1
+ nz;
(r + 1)!
but for r = 0:
Z(n, 0) = B1 (n + 1) ? B1 ? 1 = n,
and so
fn (z) =
?
X
r=0
(Br+1 (n + 1) ? Br+1 )
z r+1
.
(r + 1)!
Hence (G)?(1.11) by comparison of the coefficients. (1.12) follows from
(1.11) by (U ).
We proceed to give another explicit expression for Br :
Br =
r
X
(?1)n r + 1
Z(n, r).
n+1 n+1
n=0
(1.18)
We shall deduce (1.18) from (1.12) with the help of (1.19) which gives the
closed form for the Stirling number of the second kind (cf. (1.20))
k
1 X
j k
(?1)
(k ? j)n
S(n, k) =
k! j=0
j
k
1 X
k?i k
(?1)
=
in
k! i=1
i
=
1 k n
? 0 ,
k!
1 ? k ? n,
(1.19)
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where ? is defined in (DE). These are defined as the coefficients in the
fundamental relation
x
,
? 0
S(k, j)(x)j = x =
j
j=1
j=1
k
X
k
k
X
j k
(1.20)
where (x)j = x(x ? 1) и и и (x ? j + 1) indicates the falling factorial. The
last equality of (1.19) can be proved as follows:
k
X
j=0
(?1)j
k
X
k
k
E k?j 0n
(?1)j
(k ? j)n =
j
j
j=0
= (E ? 1)k 0n
= ?k 0n ,
with E = ? + 1 the shift operator.
(1.20)?(1.19)
Applying the shift operator, Eu(n) = u(n + 1), n-times, we obtain
nn = E n 0 n .
The left-hand side is the same as that of (1.20), while on the RHS, we apply
E = ? + 1 formally n-times to deduce that
E n 0n = (? + 1)n 0n =
n X
n
j=0
j
?j 0 n =
n
X
1 j n
? 0 (n)j ,
j!
j=1
whence (1.19) follows.
Substituting (1.12) into the right-hand side of (1.18), we have
r r
X
(?1)n r + 1
1 X r+1
the RHS =
Bk (n + 1)r+1?k
n
+
1
r
+
1
n
+
1
k
n=0
k=0
X
r
r X
r+1
r+1
1
(n + 1)r?k
(?1)n
Bk
=
n
+
1
r+1
k
n=0
k=0
X
r
r X
1
r+1
r+j r + 1
=
(?1)
Bk
(r + 1 ? j)r?k
r+1
k
j
j=0
k=0
on changing the order of summation and then writing r ? n = j. Divide
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The theory of Bernoulli and allied polynomials
the sum over k into two: 0 ? k ? r ? 1 and k = r.
?
r?1 r+1
X
1 ?X r + 1
r
j r+1
RHS =
Bk (?1)
(?1)
(r + 1 ? j)r?k
r+1
k
j
j=0
k=0
?
??
r+1
X
r
+
1
+Br ? (?1)j
? (?1)r+1 ??
j
j=0
=
r+1 1 X r+1
Bk (?1)r (r + 1)!S(r ? k, r + 1) + (?1)r Br ,
r+1
k
k=0
r
= (?1) Br = Br ,
where we used the fact that
S(n, r) = 0 for 1 ? n < r
and
r+1
X
(?1)
j=0
j
r+1
= (1 ? 1)r+1 = 0,
j
completing the proof of (1.18).
We shall state an example of (S).
Example 1.2
n
X
1
(B3 (n + 1) ? B3 )
3
k=1
1
3
1
= (n + 1) (n + 1)2 ? (n + 1) +
3
2
2
1
= (n + 1)(2n2 + 4n + 2 ? 3n ? 3 + 1)
6
1
= n(n + 1)(2n + 1).
6
Z(n, 2) =
k2 =
Compared with (S), the following is less well-known ([Com, p.155]):
Proposition 1.1
r+1
X
n
Z(n, r) =
(j ? 1)! S(r + 1, j)
.
j
j=1
(1.21)
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Proof.
By induction, for n + 1, the RHS is
r+1
X
1
j=1
we rewrite (n + 1)j as
r+1
X
1
j=1
j
j
S(r + 1, j)(n + 1)j ,
j
n+1?j +j
=1+
to obtain
n+1?j
n+j?1
S(r + 1, j)(n)j +
r+1
X
j=1
S(r + 1, j)n и и и (n ? j + 2),
the first term is Z(n, r) by hypothesis and the second can be written as
r+1
1 X
S(r + 1, j)(n + 1)j ,
n + 1 j=1
which is
1
(n + 1)r+1 = (n + 1)r , on applying (1.20).
n+1
The second proof. We first prepare auxiliary results. First,
n X
j
n+1
=
, 0 ? k ? n.
k
k+1
(1.22)
j=k
This may be proved by writing (x + 1)n+1 ? 1 in two ways: First, it is
n
X
(x + 1)n+1 ? 1
x=
(x + 1)j x,
x
j=0
which becomes
j n X
X
j
j=0 k=0
k
xk+1 =
n X
n X
j
k=0 j=k
k
xk+1
by changing the order of summation. Since (x + 1)n+1 ? 1 =
n
P
k=0
n+1
k+1
xk+1 ,
we obtain (1.22) by comparing the coefficients of xk+1 .
Secondly, we also need the triangular recurrence formula for S(r, j):
S(r + 1, j) = S(r, j ? 1) + j S(r, j),
1 ? j ? r + 1.
This may be proved by writing xr+1 in (1.20) in two ways:
(1.23)
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On one hand, it is
r+1
P
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15
S(r + 1, j)(x)j , and on the other, it is
j=0
x и xr = x
r
X
S(r, j)(x)j .
j=0
Since
x (x)j = (x ? j + j)x(x ? 1) и и и (x ? j + 1)
= (x)j+1 + j(x)j ,
we have
xr+1 =
r
X
j=1
(S(r, j ? 1) + j S(r, j)) (x)j ,
whence (1.23) follows by comparing the coefficients.
We may now prove (1.21). Substituting (1.20), we obtain
Z(n, r) =
r
X
j=1
S(r, j) j!
n X
k
k=1
j
after inverting the order of summation. We rewrite the innermost sum as
n?1
X k n n n
+
=
+
j
j
j
j+1
k=1
by (1.22). Hence
Z(n, r) =
r+1 X
j=1
n
n
S(r, j) j!
+ S(r, j ? 1) (j ? 1)!
j
j
n
=
(j S(r, j) + S(r, j ? 1)) (j ? 1)!
,
j
j=1
r+1
X
which is the RHS of (1.21) in view of (1.23).
Example 1.3
We take up the special case of (1.21).
n
Z(n, 2) =
(j ? 1)!
S(3, j).
j
j=1
3
X
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Since from (1.19) we have
S(3, 1) = 1, S(3, 2) = 3, S(3, 3) = 1,
it follows that
n
n
S(3, 2) + 2!
S(3, 3)
2
3
1
3
= n + n(n ? 1) + n(n ? 1)(n ? 2)
2
3
1
2
= n(2n + 3n + 1)
6
1
= n(n + 1)(2n + 1).
6
Z(n, 2) = S(3, 1)n +
Exercise 1.2
Prove that (H) implies (U ) under Euler?s identity (5.66).
Solution Since B1 (x) = x ? 12 for 0 < x < 1, the unique polynomial that
coincides with B 1 (x) is B1 (x) = x ? 21 = x + B1 . Denoting the right-hand
side of (1.9) by bn (x), we obtain
1 d
bn (x) = bn?1 (x),
n dx
whence
1 dn
bn (x) = B1 (x),
n! dxn
Integrating
0 < x < 1.
1 d
2 dx b2 (x)
= B1 (x), we deduce that
Z
1
1
b2 (x) = B1 (x) dx = x2 + B1 x + C,
2
2
where
?
1
1 2 X 1
1
1
C = b2 (0) =
= 2 ?(2) = = B2
2
2 ? 2 n=1 n2
?
6
by (5.66).
Repeating this procedure, each time using (5.66), we arrive at (U ).
Now we shall follow Lehmer [Leh2] to deduce some of the above defining
conditions from (K). First we state a lemma.
Lemma 1.1 (Lehmer) For a given n ? N there is a unique monic
polynomial of degree n satisfying (1.8).
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17
Proof. That Bn (x) satisfies (1.8) is a consequence of (3.69) and (4.1).
To prove uniqueness, suppose there are two polynomials Pn (x) and
Qn (x) of degree n satisfying the conditions. Then
Rr (x) := Pn (x) ? Qn (x) = a0 xr + a1 xr?1 + и и и ,
where Rr (x) is a polynomial of degree r < n satisfying (1.8):
m
X
x+k
Rr
mn?1
= Rr (x).
m
k=0
Identifying the coefficients of xr on both sides, we obtain
r
1
n
m
a0 = a 0 ,
m
which contradicts the fact that r < n and a0 6= 0.
This completes the proof.
Theorem 1.2 (Lehmer) For n, m ? N, there exists a unique monic
polynomial of degree n satisfying the functional equation
m?1 x+k
1 X
f
= m?n f (x).
(1.24)
m
m
k=0
Proof. Since for m = 1, (1.24) reduces to a trivial equality, we may
assume that m > 1.
We substitute a candidate polynomial
Pn (x) = b0 xn + b1 xn?1 + и и и + bn ,
b0 6= 0
with bk as indeterminates, into (1.24). If f satisfies (1.24), then so does
any multiple of f , so that we may assume b0 = 1.
The left-hand side of (1.24) becomes
X
m?1
n
n??
X
1 X
x+k
n??
Pn
=
b?
xn???х m?n+??1
Z(m ? 1, х)
m
m
х
?=0
х=0
k=0
n
r
X
X
n??
= m?n
xn?r
b?
m??1 Z(m ? 1, r ? ?).
r
?
?
r=0
?=0
This must be equal to the right-hand side of (1.24):
m?n Pn (x) =
n
X
r=0
xn?r br m?n .
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Identifying the coefficients of xn?r , we conclude that
m
?n
r
(m ? 1)br = ?m
Now, from Z(n, 1) =
?n
r?1
X
bk
k=0
n?k
mk?1 Z(m ? 1, r ? k).
r?k
(1.25)
n(n+1)
,
2
it follows that
n
b1 =
B1 .
1
Suppose inductively that b1 , и и и , br?1 (r > 1) are determined. Then
(1.25) determines br , completing the induction.
We note that by elaborating the above proof, we may actually prove
Proposition 1.2 The only polynomial satisfying (1.24) must be Bn (x),
on the ground of (U ), (A) and (1.16).
Proof. Indeed, suppose inductively that bk = nk Bk , k < r. Then the
right-hand side of (1.25) is
?m
?n
r?1 X
n n?k
k=0
r?k
k
Bk mk?1 Z(m ? 1, r ? k).
The product of binomial coefficients is
RHS = ?m
?n
n
r
r
k
, and so
X
r?k
r?1 X r ? k + 1 n
r
1
k?1
Bl mr?k+1?l .
Bk m
r?k+1
l
r
k
l=0
k=0
Now the innermost sum is
mr?k+1 +
r?k
X
l=1
r?k+1 r?k
Bl mr?k?(l?1) .
l
l?1
Hence
RHS = ?m?n+r
n
n
S1 ? m?m
S2 ,
r
r
where
S1 =
r?1 X
r
k=0
k
Bk
1
r?k+1
(1.26)
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r?1 X
r
S2 =
r?k X
1 r?k
Bk
Bl mr?l .
l l?1
k
k=0
r
k
Rewriting the product
r?l+1 r
l?1 r?k?l+1 , we obtain
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19
(1.27)
l=1
r?k
l?1
of binomial coefficients in S2 by
r?l r
X
X
r
1
r?l+1
r?l
Bl m
S2 =
Bk
l l?1
r?k?l+1
k=0
l=1
whose innermost sum is
r?l r?l+1
X
X r ? l + 1 r?l+1
Bk =
Bk ? Br?l+1
k
k
k=0
k=0
= Br?l+1 (1) ? Br?l+1
(
1, l = r
=
0, l 6= r
by (U ) and (1.16) successively. Since the last member is 0 except when
l = r, we infer that
r
1
Br = B r .
(1.28)
S2 =
r r?1
On the other hand,
S1 =
=
Z
r?1 1X
0 k=0
Z 1
r
r?k
(Br (x) ? Br ) dx
Bk x
dx =
k
0
1
(Br+1 (1) ? Br+1 (0)) ? Br = ?Br .
r+1
Hence
LHS = ?m
or
m
whence
?n
r
?n+r
n
?n n
(?Br ) ? m
Br
r
r
(m ? 1)br = ?(m
?n
?m
n
br =
Br ,
r
?n+r
n
)
Br ,
r
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completing the proof.
Exercise 1.3
Solution
With Bn (x) as defined by Theorem 1.2, prove (1.5).
Let
F (x, t) =
t ext
et ? 1
and expand it into the Taylor series in t:
F (x, t) =
?
X
1
Gn (x)tn ,
n!
n=0
|t| < 2?
(1.29)
with Gn (x) a polynomial of degree n,
(n)
Gn (x) = a0 xn + и и и + a(n)
n ,
say.
(n)
We may determine a0 as follows: Replacing x by
(1.29), we get
1
y
and t by ty in
?
X
1 n
1 n
yt
t
e
=
y
G
t ,
n
eyt ? 1
n!
y
n=0
which leads, as y ? 0, to
?
X
1 (n) n
a0 t
e =
n!
n=0
t
(since y n Gn
(n)
Hence a0
Now,
1
y
(n)
? a0 ):
= 1 and Gn is monic.
m?1 ?
m?1 1 XX 1
1 X
x+k n
x+k
t =
F
Gn
,t
m
n!
m
m
m
n=0
k=0
k=0
=
m?1
X x+k
t
1
e m t
m et ? 1
k=0
x
1 t emt
,
=
m e mt ? 1
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The theory of Bernoulli and allied polynomials
which is equal to
t
F x,
m
=
?
X
1
1
Gn (x) n tn .
n!
m
n=0
1 n
n! t ,
Hence, comparing the coefficients of
we-conclude that
m?1
x+k
1 X
Gn
= m?n Gn (x),
m
m
k=0
whence, by Theorem 1.2, that Gn (x) = Bn (x), completing the proof.
Lemma 1.2
DE
For every m ? N there is a unique polynomial satisfying the
f (z + 1) ? f (z) =
m X
m
k=1
k
bk z m?k ,
b1 6= 0
(1.30)
with the initial condition f (1) = 0.
Proof.
form
From (1.30), f (z) must be of degree m and may be put in the
f (z) =
m X
m
j=0
j
aj z m?j ,
a0 6= 0.
(1.31)
Forming the difference f (z + 1) ? f (z), thereby using the expansion
m?j
aj z m?j?r
r
r=1
m X
m?j
=
aj z m?k ,
k?j
(z + 1)m?j ? z m?j =
m?j
X
k=j+1
we find that
f (z + 1) ? f (z) =
m
X
j=0
aj
m X
m m ? j m?k
z
.
j
k?j
k=j+1
Using (1.14), we get
m m?j
m
m?j
m
k
m k
=
=
=
.
j
k?j
m?j
m?k
m?k
k?j
k
j
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Hence
f (z + 1) ? f (z) =
m X
m
k=1
k
?
?
m?1
X
j=0
?
k
aj ? z m?k .
j
(1.32)
Comparing (1.30) and (1.32), we see that it is enough to show that the
following system of m+1 linear equations in m+1 unknowns a0 , a1 , и и и , am
and f (z) has a unique solution:
k?1
X
j=0
k
aj = b k ,
j
k = 1, 2, и и и , m
(1.33)
m
aj z m?j + f (z) = am .
j
(1.34)
and
?
m?1
X
j=0
This is indeed the case because the coefficient matrix is lower triangular,
its determinant is the product of all diagonal components
so that
m
1 2
и
и
и
m?1 = m! 6= 0. Hence f (z) is determined by b1 , и и и , bm and am .
0 1
Now, comparing (1.33) with k = m and (1.34) with z = 1, we find that
am = bm by the condition f (1) = 0. Hence f (z) is determined uniquely
by b1 , и и и , bm .
Theorem 1.3 For each m ? N there exists a unique polynomial fm (z)
satisfying the conditions
fm (z + 1) ? fm (z) = mz m?1
(1.35)
fm (1) = 0.
(1.36)
And the m-th Bernoulli polynomial Bm (z) is defined by
Bm (z) = fm (z) + Bm (1).
Proof.
0.
(1.37)
This is a special case of Lemma 1.2 with b1 = 1, b2 = и и и = bm =
Remark 1.2 As is remarked in [Mi, p.284] , the r + 1-st Bernoulli polynomial can be characterized as the unique polynomial satisfying (1.11) for
every natural number r. Hence (S) can be also used as the definition as (K)
and (DE).
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Lemma 1.3
Under (G), Formula (1.37) implies
0
fm
(z) = mBm?1 (z) = m(fm?1 (z) + Bm?1 ),
m ? N ? {0}.
(1.38)
If we differentiate with respect to z
Proof.
?
X
fm (z) m
x
x ex
x = x
exz ? x
,
m!
e ?1
e ?1
m=0
we obtain
?
?
?
0
X
X
X
x2
Bm (z) m+1
Bm?1 (z) m
fm
(z) m
x = x
exz =
x
=
x ,
m!
e
?
1
m!
(m ? 1)!
m=0
m=1
m=0
whence we have
1 0
f (z) = Bm?1 (z),
m m
m ? N,
and
f00 (z) = 0,
which amounts to (1.38).
Proposition 1.3
(DE) and (G) together imply (K).
Proof. We shall prove that Theorem 1.3 and Lemma 1.3 imply the Kubert identity for fm (z):
fm (qz) = q m?1
q?1
X
k=0
k
+ (q m?1 ? 1)Bm (1)
fm z +
q
(1.39)
for any q ? N.
We follow Bo?hmer [Bo?h] to add (1.35) in the form
fm+1 (z + 1) = fm (z) + (m + 1)z m
to the trivial identity
q
m
q?2
X
fm+1
k=0
k+1
z+
q
=q
m
q?1
X
fm+1
k=1
k
z+
q
to obtain
q
m
q?1
X
k=0
fm+1
k+1
z+
q
=q
m
q?1
X
k=0
fm+1
k
z+
q
+ (m + 1)q m z m . (1.40)
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24
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Now subtract (1.40) from
fm+1 (qz + 1) = fm+1 (qz) + (m + 1)q m z m
to deduce that
q?1
X
k+1
fm+1 z +
fm+1 (qz + 1) ? q
q
k=0
q?1
X
k
fm+1 z +
= fm+1 (qz) ? q m
q
m
(1.41)
k=0
= F (z),
say, whence
1
= F (z)
F z+
q
i.e. F (z) is periodic of period 1q . But, F (z) being a polynomial, we must
have F (z) = constant.
Hence differentiating (1.41), we infer that
0
qfm+1
(qz)
?q
or
q(fm (qz) + Bm (1)) ? q
m
m
q?1
X
0
fm+1
k=0
q?1 X
k=0
fm+1
k
z+
q
k
z+
q
= 0,
+ Bm (1)
= 0,
by Lemma 1.3. This amounts to (1.39), thereby completing the proof.
Remark 1.3 Since (DE) and (K) correspond to differentiation and integration, respectively, we need an extra information (i.e. (G)) in Proposition 1.3 to deduce a result on integration from that on differentiation. We also remark that Pan and Sun?s result [Su2] implies differentiation from differencing for polynomials, which enables us to prove
(D?)+(DE)?(K). Indeed, differencing the right-hand side of (1.8), we
Pm?1 d
d
Bn x+k
= dx
Bn (x), or
get ?(Bn (x)), which implies mn?1 k=0 dx
m
P
m?1 n
x+k
n?1
m
= n Bn?1 (x) by (1.1), whence (1.8) ensues.
k=0 m Bn?1 m
Finally, we add the following as a memorial of the work of L. Lagrange,
who was a contemporary of Euler, which is often used in number-theoretic
settings.
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The theory of Bernoulli and allied polynomials
(D ? )
?
( A)
?
?
??
(G )
?
(S )
?
(DE )
Fig. 1.4
Fig. 1.5
(U )
?
?
(H )
?
?
March 27, 2007
? (K )
?
Logical scheme
L. Lagrange
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25
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The Lagrange Interpolation Method says that:
Given n values g(?i ), 1 ? i ? n of a function g(x) at n distinct arguments ?i , 1 ? i ? n, the Lagrange interpolation polynomial is defined
by
L(x) =
n
X
i=1
f (x)
g(?i )
(x ? ?i ) f 0 (?i )
(1.42)
interpolates g(x) (L(?i ) = g(?i )), where
f (x) =
n
Y
i=1
(x ? xi ).
Example 1.4 (Carlitz [Car]) Let k1 , k2 , и и и , kn be pairwise relatively
prime positive integers and let
gi (x) =
n?1
Y
xk i ? 1
x ? ?il ,
=
x?1
l=1
1 ? i ? n,
where ?i signifies a primitive ki -th root of 1. Further put
fi (x) = (x ? 1) gi (x) = xki ? 1 =
n?1
Y
l=1
x ? ?il
and
Gi (x) =
Y
k6=i
gk (x),
1 ? i ? n.
(1.43)
Then the polynomial
Li (x) =
n?1
1 X fi (x)
1
?ik
+ ?i (1) gi (x)
k
k
ki
x ? ?i Gi (?i ) ki
k=1
(1.44)
interpolates the polynomial ?i (x) of degree < ki ? 1 such that
n
X
Gi (x) ?i (x) = 1.
i=1
Proof. We note the following facts. Since fi0 (x) = ki xki ?1 , we have, for
a primitive ki -th root of 1 and 1 ? l ? n ? 1,
l(k ?1)
fi0 ?il = ki ?i i
= ki ?i?l .
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27
Also Gi ?il = 0 for k 6= i, 6= 0 for k = i. Hence by (1.43),
n
X
Gi ?il ?i ?il = 1.
Gi ?il ?i ?il =
k=1
Now note that the polynomial
Li (x) =
n?1
X
k=1
fi (x) ?i ?ik
1
+ ?i (1) gi (x)
k
k
0
ki
x ? ? i f i ?i
(1.45)
is the Lagrange interpolation
of gi (1) = ki .
polynomial for ?i (x) in view
1
,
so that (1.45)
From the above data, fi0 ?ik = ki ?i?k and ?i (?ik ) = Gi (?
k
i )
transforms into (1.44).
This identity is essentially used in the proof of the three-term relation of
the Dedekind sum, which is a finite sum of the product of two first periodic
Bernoulli polynomials.
Qn
Example 1.5 Let f (x) = i=1 (x ? ?i ) be an irreducible polynomial
over Q (?i being necessarily distinct and called conjugates). Then
f (x) ?r
= xr , 0 ? r ? n ? 1,
Tr
x ? ? f 0 (?)
where ? = ?1 , say, and Tr means the sum of conjugate elements.
This is because, by the Lagrange interpolation
xr =
n
X
?ri
f (x)
0
x ? ?i f (?i )
i=1
and all summands on the right are conjugate one another, so that the righthand side is the trace.
This is essentially used to find the dual basis in a finite extension of Q.
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Chapter 2
The theory of the gamma and related
functions
Abstract
In this chapter we develop the standard theory of the
gamma and related functions in a classical fashion starting
from the Eulerian integral of the second kind. We follow
partly Bo?hmer [Bo?h] and Hata [Hata] (cf. also Erde?lyi et
al [Erd]). Most of the results in this chapter are restated
from the zeta-function theoretic point of view in Chapter 5,
which can be read parallel to the present chapter.
2.1
Gamma function
First, we develop the theory of the gamma function defined by (2.1), one
of equivalent conditions to be discussed in Chapter 5. The gamma function,
being the Mellin transform of e?x to be mentioned in Д7.4, is defined by
the Eulerian integral of the second kind
Z ?
?(s) =
e?x xs?1 dx
(2.1)
0
for ? > 0. This improper integral is absolutely and uniformly convergent
in the wide sense in ? > 0, whence it follows that ?(s) is analytic in the
right half-plane ? > 0.
The gamma function may be continued meromorphically over the whole
plane by the difference relation
?(s) =
? (s + n + 1)
,
s (s + 1) и и и (s + n)
29
n ? N ? {0}.
(2.2)
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Fig. 2.1
Euler
Indeed, for ? > 0, we have by integration by parts,
?(s) =
1 ?x s
e x
s
?
0
?
1
s
Z
?
e?x xs dx =
0
1
?(s + 1),
s
(2.3)
whose right-hand side is analytic for ? > ?1. From (2.2) we see that
s = ?n, n ? N ? {0} are simple poles and that the residues at these poles
are
Res ?(s) =
s=?n
(?1)n
.
n!
(2.4)
Also if we put s = n ? N in
?(s + 1) = s ?(s),
(2.5)
then we have ?(n + 1) = n!, which means that the gamma function is a
function which interpolates the factorial n!.
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The theory of the gamma and related functions
Exercise 2.1
The generalized binomial coefficient is defined by
?(z + 1)
z
(z)r
=
=
r!
?(r + 1) ?(z + 1 ? r)
r
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31
(2.6)
for r ? N ? {0}, where (z)r = z(z ? 1) и и и (z ? r + 1) indicates the falling
factorial. Show that for n ? N ? {0}
z
r n+r?1
lim
= (?1)
(2.7)
z??n r
r
and confirm that the usual definition
?n
r n+r?1
= (?1)
r
r
is realized as the limit (2.7).
Solution
Writing by (2.2)
?(z + n + 1) (z + 1 ? r) и и и (z + n)
?(z + 1)
=
?(z + 1 ? r)
(z + 1) и и и (z + n)
?(z + n + 1)
(z + 1 ? r) и и и (z + n ? 1)
=
,
(z + 1) и и и (z + n ? 1)
we see that
(?n + 1 ? r) и и и (?1)
?(z + 1)
?
?(z + 1 ? r)
(?n + 1) и и и (?1)
(?1)n+r?1 (n + r ? 1)!
=
(?1)n?1 (n ? 1)!
as z ? ?n. By (2.5), this is
(?1)
+ r)
r n+r?1
= (?1)
r!,
?(n)
r
r ?(n
and (2.7) follows.
Remark 2.1 In contrast to (?)r , h?ir indicates the Pochhammer symbol
(or the shifted factorial, since h1ir = r!) defined by
(
1
(r = 0)
?(? + r)
=
h?ir :=
?(?)
? (? + 1) и и и (? + r ? 1) (r ? N) .
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The following formula (Prym?s decomposition) extracts all poles of
the gamma function and renders visible its behavior at the poles (cf. (2.4)):
Z ?
?
X
(?1)n 1
?(s) =
+
e?x xs dx.
n! s + n
1
n=0
(2.8)
The improper integral is absolutely and uniformly convergent in any disc
|s| < R and represents an integral function. The proof follows on dividing
the interval of integration and noting that
Z
1
x
s?1
0
Z
?
?
?
X
X
X
(?1)n 1 n+s?1
(?1)n 1
(?1)n xn
dx =
.
x
dx =
n!
n!
n! s + n
0
n=0
n=0
n=0
Let the beta function B(?, ?) be defined by the Eulerian integral
of the first kind
B(?, ?) =
Exercise 2.2
Z
1
0
t??1 (1 ? t)??1 dt,
Re ? > 0, Re ? > 0.
(2.9)
Prove the formula
?(?) ?(?) = ?(? + ?) B(?, ?),
(2.10)
whence in particular
?
?
1
= ?,
2
or the value of the probability integral
Solution
R?
0
(2.11)
2
e?x dx =
?
?
2 .
First, in (2.9), put t = sin2 ? to obtain
B(?, ?) = 2
Z
?
2
sin2??1 ? cos2??1 ? d?.
0
If in (2.1), we put t = x2 , then
?(?) = 2
Z
0
?
2
x2??1 e?x dx,
(2.12)
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33
The theory of the gamma and related functions
whence for Re ? > 0, Re ? > 0,
?(?) ?(?) = 4
Z
?
2
x2??1 e?x dx
0
Z
= 4 lim
X??
= 4 lim
X??
= 4 lim
X??
X
?
2
y 2??1 e?y dy
0
2
x2??1 e?x dx
0
Z XZ
Z0Z
Z
X
Z
X
2
y 2??1 e?y dy
0
x2??1 y 2??1 e?(x
2
+y 2 )
!
dx dy
0
x2??1 y 2??1 e?(x
2
+y 2 )
dx dy,
(2.13)
D
n o
p
x where D =
x2 + y 2 ? X . By the change of variable x =
y 0 ?
r cos ?, y = r sin ?, we have the correspondence
r ?
D ? D? =
,
0
?
r
?
X,
0
?
?
?
2
? where the absolute value of the Jacobian of this transformation is ?(x,y)
?(r,?) =
r. Hence
ZZ
2
r2?+2??2 e?r sin2??1 ? cos2??1 ? r drd?
?(?) ?(?) = 4 lim
X??
=2
Z
D?
?
0
2
r2?+2??1 e?r dr и 2
= ?(? + ?) B(?, ?)
Z
?
2
sin2??1 ? cos2??1 ? d?
0
by (2.12) above. This completes the proof of (2.10).
Putting now ? = ? = 21 , we obtain
?
2
Z ?2
1
= ?(1) 2
d? = ?,
2
0
whence (2.11) follows. It follows that
Z
?
e?
x2
2
dx =
?
2?,
??
which is used to normalize the distribution function of the Gaussian (or
normal) distribution.
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Vistas of Special Functions
Remark 2.2
for ? = ? = 21
Z
vista
?
e
?x2
0
dx
An ordinary procedure for proving (2.11) is to use (2.13)
2
= 4 lim
R??
Z
R
re
?r 2
dr
0
Z
?
2
d?
0
!
?
?
1 ?r2
= ?.
=4 ? e
2
0 2
Thus we see that if we generalize the problem by introducing the parameters
? and ?, we get a wider perspective.
Exercise 2.3
Use the formula in Corollary A.4 to prove that
?
B(s, 1 ? s) =
sin ?s
(2.14)
or
?(s) ?(1 ? s) =
?
,
sin ?s
(2.15)
the reciprocity relation for the gamma function, where 0 < s < 1 in the first
place and then for all s except for integer values by analytic continuation.
Solution
Make the substitution t =
Z
B(s, 1 ? s) =
1
x+1
?
0
in (2.9) to deduce that
x?s
dx,
1+x
which proves (2.14) in view of Corollary A.4.
Determine the value of the probability integral ? 12
Z x
2
2
and g(x) =
by considering two functions f (x) =
e?t dt
0
Z 1
2 2
1
e?x (t +1) dt.
2+1
t
0
Exercise 2.4
Solution
Recalling the fundamental theorem of calculus, we obtain
Z x
2
0
?x2
e?t dt.
f (x) = 2 e
0
We may differentiate g(x) under the integral sign to get
Z 1
1
d ?x2 (t2 +1)
0
g (x) =
e
dt
2
0 t + 1 dx
Z x
Z 1
2
2 2
2
e?u du
= ?2x
e?x (t +1) dt = ?2 e?x
0
by the change of variable u = xt.
0
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The theory of the gamma and related functions
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35
Hence we conclude that
f 0 (x) + g 0 (x) = 0,
whence by the Newton-Leibniz principle (cf. Lemma B.1) that
f (x) + g(x) = f (0) + g(0)
Z 1
1
?
1
dt = arctan(t) 0 = .
=
2
t
+
1
4
0
Now letting x ? ?, thereby noting that
Z 1
2 2
1
lim g(x) =
lim e?x (t +1) dt = 0,
2
x??
x??
0 t +1
we conclude that
lim f (x) =
x??
Z
?
2
e?t dt
0
2
=
?
,
4
i.e.
Z
Exercise 2.5
?
e
?t2
dt =
0
?
?
.
2
Use Formula (2.10) in the form
?(?) ?(?)
(2.16)
?(? + ?)
?
and the value of the probability integral ? 12 = ? to deduce Wallis?
formula
2
2 и 4 и и и (2n)
1
?
= lim
.
(2.17)
n?? 1 и 3 и и и (2n ? 1)
2
2n + 1
B(?, ?) =
Solution
First, by (2.16) and (2.12)
Z ?2
? 12 ? n + 12
1
1
=2
cos2n ? d? = S2n ,
= B ,n+
?(n + 1)
2
2
0
say, and
?
Z ?2
? (n + 1)
1
,
n
+
1
=
2
=
B
cos2n+1 ? d? = S2n+1 .
2
?(n + 32 )
0
1
2
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Clearly, 0 < S2n+1 < S2n < S2n?1 , so that
1<
n+
S2n?1
S2n
<
=
S2n+1
S2n+1
n
1
2
=1+
1
?1
2n
as n ? ?. Hence
lim
n??
S2n
= 1.
S2n+1
(2.18)
Since
? n + 12 ? n + 32
S2n
=
S2n+1
?(n + 1)2
!2 n ? 12 n ? 32 и и и 12 ? 21
1
n+
=
n и (n ? 1) и и и 1
2
2
(2n ? 1)(2n ? 3) и и и 1
?
=
(2n + 1) .
(2n)(2n ? 1) и и и 2
2
Formula (2.18) is nothing but Wallis? formula.
Exercise 2.6
lim
N ??
Use Wallis? formula (2.17) to prove that
N
X
1
log n ? N +
2
n=1
log N + N
!
= log
?
2?,
(2.19)
whence via (5.23) that
? 0 (0) = ? log
Solution
?
2?.
(2.20)
We write from Corollary 5.1
log ?(x) =
x?
1
2
log x ? x + c + o(1)
and determine the value of c through the asymptotic formula for
log
2n
?(2n + 1)
.
= log
n
?(n + 1)2
(2.21)
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The theory of the gamma and related functions
First, from (2.21),
log ?(2n + 1) ? 2 log ?(n + 1)
1
= 2n +
log(2n + 1) ? (2n + 1) + c
2
? (2n + 1) log(n + 1) + 2(n + 1) ? 2c + o(1)
2n + 1 1
? log(2n + 1) + 1 ? c + o(1).
= (2n + 1) log
n+1
2
Secondly, since the first term on the right is
(2n + 1) log
2n + 1
2n + 2
= (2n + 1) log 2 ? (2n + 1) log
n+1
2n + 1
1
= (2n + 1) log 2 ? (2n + 1) log 1 +
2n + 1
1
= log 22n + log 2 ? 1 + O
,
n
we obtain
log
?
?(2n + 1)
= log 22n ? log 2n + 1 + log 2 ? c + o(1),
2
?(n + 1)
(2.22)
whence
log
2n
n
?
2n + 1
= log 2 ? c + o(1).
22n
(2.23)
Since
1 и 3 и и и (2n ? 1)
2 и 4 и и и 2n и 1 и 3 и и и (2n ? 1)
1 2n
=
,
=
2n
n
2
2
n
(2 n!)
2 и 4 и и и 2n
q
?
2n+1
2
the LHS of (2.23) is log 1и3иии(2n?1)
,
whose
limit
is
log
2и4иии2n
? by Wallis?
formula. Hence
r
?
2
= log 2?.
c = log 2 ? log
?
Recalling the more exact form of (2.21), we may deduce Stirling?s formula from the above (cf. Corollary 5.1):
?
1
1
log ?(x) = x ?
log x ? x + log 2? + O
.
(2.24)
2
x
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Exercise 2.7 Prove that for x ? 2, the highest power of a prime p ? x
that divides [x]! is given by
[log x/ log p] X
l=0
x
,
pl
where [y] indicates the greatest integer function introduced in Chapter 1.
h i
Solution First note that pxl is the exact number of multiples of pl
i
h
x
numbers that are mulbetween 1 and x. Among them there are pl+1
h i
tiples of pl+1 . Hence pxl indicates the exponent of p that appears in
h i
h
i
h i
x
pl (2pl ) и и и pxl pl but not in pl+1 (2pl+1 ) и и и pl+1
pl+1 . Hence adding pxl
i
h
x
over all l up to r such that pr ? x < pr+1 , i.e. r = log
log p , then we obtain
the exact exponent ? of p such that p? k [x]! (p? divides [x]! but p?+1 does
not).
Exercise 2.8 Let ?(x) denote the number of primes ? x. Then using a
special case of Stirling?s formula (Exercise 2.6, (2.24))
r
1
2n
2
22n
+O
log
= log ?
,
(2.25)
+ log
n
?
n
2n + 1
deduce Chebyshe?v?s inequalities
x
x
?(x) ,
log x
log x
(2.26)
where is Vinogradov?s symbolism, meaning that there are constants c1 >
0, c2 > 0 such that
x
x
< ?(x) < c2
.
c1
log x
log x
Solution
First we note the inequalities
hxi
hxi
2
? [x] ? 2
+ 1.
2
2
(2.27)
x
x
The first
one
is
trivial
while
the
second
one
follows
from
?
1
<
2
2 or
x
x < 2 2 +2. We consider the highest power of a prime p ? 2n that divides
2n
2n
?
n : p k n . From Exercise 2.7, it is given by
r X
2n
n
log 2n
?=
?2 l
, r=
.
(2.28)
pl
p
log p
l=1
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The theory of the gamma and related functions
Bearing (2.27) in mind, we see that each summand is 0 or 1, and so ? ?
2n
r ? log
log p . Hence
Y
2n
=
pr < (2n)?(2n)r log p < (2n)?(2n) log 2n.
n
p?2n
Now by (2.25),
2n
n
2n , so that
2n
n
log 2 ? log
? ?(2n) log 2n.
n
Hence
n
?(2n),
log 2n
so that for x ? 2, using (2.27), we obtain
x
h x i
x
2 .
?(x) ? ? 2
2
log x
log 2 x2
To prove the other inequality in (2.26), we recall the remark after (2.28)
to the effect that those prime p, n < p ? 2n do not divide the denominator,
Q
and therefore 2n
p. Hence
n must be divisible by
n<p?2n
n?(2n)??(n) ?
Y
n<p?2n
p?
2n
.
n
We deduce from this, on using (2.25) in the form log
(?(2n) ? ?(n)) log n n,
2n
n
n, that
whence that
?(2n) log 2n ? ?(n) log n = (?(2n) ? ?(n)) log n + ?(2n) log 2 n
in view of ?(x) < x.
Hence we have
x
x
x.
2
i
h
x
(1 ?
Replacing x in (2.29) by 2xl , l = 0, и и и , r, r = log
log 2
obtain
x
x x
x
1
? l log l ? ? l+1 log l+1 l x.
2
2
2
2
2
?(x) log x ? ?
2
log
(2.29)
x
2r
< 2), we
(2.30)
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Summing (2.30) for l = 0, 1, и и и , r, we conclude that
?(x) log x ? ?
r
x
X
1
x
x
log
r
r
2
2
2l
l=0
so that ?(x) log x x, completing the proof.
Theorem 2.1 The integral representation (2.1) and Euler?s interpolation formula (telescoping series)
?(n + z)
= 1,
?(n) nz
lim
n??
0 < Re z < 1
(2.31)
are equivalent.
Proof. ?(s) being meromorphic over the whole plane, it suffices to consider the case 0 < z = x < 1.
First we show that (2.1) implies (2.31). In (2.1) with s = n ? N put
t = nu to obtain
Z ?
e?nu un?1 du = ?(n) n?n .
(2.32)
0
Similarly, by (2.1) with s = n + 1 and (2.5), we have
Z
?
e?nu un du = ?(n) n?n .
(2.33)
Z
(2.34)
0
Note that
Z
1
0
e?nt tn?1 dt ?
Z
1
e?nt tn dt =
0
1
n
1
0
d ?nt n
1
(e t ) dt = e?n .
dt
n
Subtracting (2.34) from (2.32), we deduce that
Z
1
e
?nt n
t dt +
0
Z
?
e?nt tn?1 dt = ?(n) n?n ?
1
1 ?n
e .
n
(2.35)
1 ?n
e .
n
(2.36)
Adding (2.34) to (2.33), we obtain
Z
1
e
0
?nt n?1
t
dt +
Z
?
e?nt tn dt = ?(n) n?n +
1
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Hence, for 0 < x < 1
1
n
?(x + n) =
x+n
<
Z
Z
1
+
0
Z
?
1
1
e
?nt n
t dt +
0
< ?(n) n?n +
Z
?
e?nt tn?1 dt
1
1 ?n
e
n
by (2.36). Similarly,
1
nx+n
?(x + n) > ?(n) n?n ?
1 ?n
e .
n
Hence it follows that
?(n) n?n ?
1 ?n ?(x + n) ?n
1
e <
n < ?(n) n?n + e?n ,
n
nx
n
or
1?
nn?1 e?n
?(x + n)
nn?1 e?n
<
<
1
+
.
?(n)
?(n) nx
?(n)
Letting n ? ?, we conclude (2.31).
Now we turn to the deduction of (2.1) from (2.31). Putting
nx n!
,
x(x + 1) и и и (x + n)
?n (x) =
we have
lim ?n (x) = ?(x)
(2.37)
n??
by (2.31). Expanding
1
x(x+1)иии(x+n)
?n (x) = n
x
into partial fractions, we obtain
n
X
k=0
n
1
,
(?1)
k x+k
k
whose right-hand side can be expressed as
Z 1
x
tx?1 (1 ? t)n dt.
n
0
Hence
?n (x) =
Z
n
t
0
x?1
t
1?
n
n
dt.
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Now choose 0 < ? <
For 0 ? t ? n? we have
1
x+2
and divide the interval [0, n] into [0, n? ]?[n? , n].
t
n log 1 ?
n
and so
Z
= ?t + O t2??1 ,
n
Z n?
t
tx?1 e?t dt + O n?(x+2)?1 ,
tx?1 1 ?
dt =
n
0
n?
0
while for n? ? t ? n,
t
n log 1 ?
? n log 1 ? n??1 ? ?c n? ,
n
c > 0,
so that
Z
n
n?
?
= O n? e?cn .
It follows that
?(x) =
which tends to
Z
n?
tx?1 e?t dt + o(1),
0
R ? x?1 ?t
e dt. Thus from (2.37) we conclude (2.1).
0 t
Equivalent definitions of ? other than (2.1) and (2.31) will be given in
Chapter 5 ((5.21), (5.27)0 , (5.49)).
2.2
The Euler digamma function
As will be discussed in Chapter 5, it is simpler to work with the digamma
function and then shift to the gamma function. Below we shall provide
several important properties of the digamma function. We follow the lines
in Bo?hmer [Bo?h]
The following exercise is similar in spirit to Proposition 7.1.
Exercise 2.9
R x Suppose f (x) is continuous and integrable on R+ = (0, ?).
Let F (x) = 0 f (t) dt. Then for each n ? N prove that
Z ?
Z ?
?nx
f (x) e
dx = n
F (x) e?nx dx
(2.38)
0
0
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and that
lim
n??
Solution
Z
?
f (x) e?nx dx = 0.
(2.39)
0
By assumption, lim F (x) exists and so F (x) = O(1) uniformly
x??
on R+ . Construct the auxiliary function
Z x
G(x) = n
F (t) e?nt dt + F (x) e?nx .
0
Then G(0) = F (0) = 0 and G(?) = n
Z
?
F (t) e?nt dt. Also
0
G0 (x) = n F (x) e?nx + F 0 (x) e?nx ? n F (x) e?nx
= f (x) e?nx
by the fundamental theorem of calculus. Therefore
Z x
Z x
G(x) =
f (t) e?nt dt + G(0) =
f (t) e?nt dt.
0
Hence, in particular
Z ?
Z
f (t) e?nt dt = G(?) = n
0
0
?
F (t) e?nt dt,
0
which is (2.38).
Now, to prove (2.39), we note that F (t) is differentiable on R+ and, a
fortiori, continuous in the right neighborhood of 0. Hence, for any ? > 0,
there is a ? > 0 such that
|F (t)| = |F (t) ? F (0)| < ?, 0 < t < ?.
Now, divide the interval (0, ?) into (0, ?) and (?, ?) to obtain
!
Z ?
Z ?
Z ?
?nt
?nt
?nt
=O n
f
(t)
e
dt
|F
(t)|
e
dt
+
O
n
|F
(t)|
e
dt
0
0
?
?nt ?
?nt ?
= O ?[?e ]0 + O [?e ]?
= O(?) + O(e?n? ),
which is O(?) if n is big enough. This proves (2.39).
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Exercise 2.10 Deduce from (2.39) the integral representation for Euler?s
constant defined by (5.16):
Z ?
f (t, 1) dt,
(2.40)
?=?
0
where
f (t, z) =
e?t
e?zt
?
.
t
1 ? e?t
(2.41)
Prove the Gauss? integral representation for the digamma function ?
defined by (5.18)
Z ?
Z ? ?t
e
e?zt
?(z) =
f (t, z) dt =
?
dt, Re z > 0.
(2.42)
t
1 ? e?t
0
0
R?
Solution We notice that r1 = 0 e?rt dt, r ? N. Hence
n
X
1
r=1
r
=
Z
0
n
?X
r=1
e
?t r
Z
dt =
?
0
1 ? e?nt
dt.
1 ? e?t
By Exercise A.1, (A.2), we have also
Z ?
1 ? e?nt ?t
e dt.
log(n + 1) =
t
0
(2.43)
(2.44)
Subtracting (2.44) from (2.43), we obtain
n
X
1
r=1
r
? log(n + 1) = ?
Z
?
0
?? ?
Z
f (t, 1) 1 ? e?nt dt
?
0
f (t, 1) dt, n ? ?
by Exercise 2.9. Hence, by definition, (2.40) follows.
Let
Z ?
F (z) =
f (t, z) dt,
0
the integral being (absolutely) convergent for Re z > 0. Then
Z ? ?t
e ? e?zt
F (z) + ? =
dt,
1 ? e?t
0
(2.45)
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The theory of the gamma and related functions
whence
F (z + 1) ? F (z) =
Z
?
e?zt dt =
0
1
z
(2.46)
and
F (1) + ? = 0, F (1) = ??.
(2.47)
By the telescoping series technique, we obtain
F (z) + ? = F (z) ? F (1)
=
n?1
X
k=0
(F (z + k) ? F (1 + k) ? (F (z + k + 1) ? F (z + k)))
+ F (z + n) ? F (1 + n)
1
1
+ F (z + n) ? F (1 + n)
?
=
1+k z+k
n?1
X
(2.48)
k=0
by (2.46). Hence
lim (F (z + n) ? F (1 + n)) = 0
(2.49)
n??
is a necessary and sufficient condition
for the convergence of the (telescop
P?
1
1
ing) series k=0 1+k ? z+k .
Since
Z ? ?t
e ? e?zt ?nt
e
dt,
F (z + n) ? F (1 + n) =
1 ? e?t
0
(2.49) follows on account of Exercise 2.9.
It follows from (2.48) and (2.49) that
F (z) = ?? +
? X
k=0
1
1
?
1+k z+k
,
which is the same as the Gaussian representation for ? (cf. (5.17)), whence
(2.42) follows.
Exercise 2.11
Prove Legendre?s (integral) representation
Z 1
1 ? xz?1
?(z) + ? =
dx, Re z > 0.
1?x
0
(2.50)
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Use (2.50) to deduce
?
=
sin ?z
Z
?
0
u?z
du =
1+u
Z
?
0
v z?1
dv,
1+v
(2.51)
and
?
=
cos ?z
Z
?
0
1
uz? 2
du = 2
1+u
Z
?
0
1
1
t2z
dt, ? < Re z <
2
1+t
2
2
(2.52)
(cf. Corollary A.4).
Solution (2.50) follows from (2.42) by the change of variable t = ? log x.
Using (2.50), we have
Z 1 ?1
Z ? ?1
x 2 ? xz?1
t 2 ? t?z
1
=
dx = ?
dt, Re z > 0,
?(z) ? ?
2
1?x
1?t
0
1
(2.53)
where the second expression follows from the first by the change of variable
x = t?1 . Similarly,
Z 1 ?1
Z ? ?1
1
x 2 ? xz?1
t 2 ? t?z
?(1 ? z) ? ?
dt = ?
dx, Re z < 1.
=
2
1?t
1?x
1
0
(2.54)
Subtraction of (2.54) from (2.53) gives
?(z)??(1?z) = ?
Z
? ? 21
t
0
? t?z
dt =
1?t
Z
?
0
1
x? 2 ? xz?1
dx, 0 < Re z < 1.
1?x
If we admit the relation ?(z) = (log ?(z))0 , then from (2.15), we deduce
that
?(z) ? ?(1 ? z) = ?? cot ?z,
(2.55)
an important relation linking the digamma and the trigonometric functions.
Use this to write the above equality as
? cot ?z =
Z
? ? 21 ?t?z
t
0
1?t
dt = ?
Z
?
0
which becomes, on writing t = u2 , x = v 2 , z ?
?
?
cot z =
2
2
Z
?
0
1 ? u1?z
du = ?
1 ? u2
Z
0
?
1
x? 2 ? xz?1
dx,
1?x
z
:
2
1 ? v z?1
dv, 0 < Re z < 2 (2.56)
1?v
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and
?
?
tan z = ?
2
2
Z
?
0
1 ? u1?z
du =
1 ? u2
Z
?
0
1 ? v z?1
dv, ?1 < Re z < 1,
1 ? v2
(2.57)
by the interchange of z by 1 ? z.
Adding (2.56) and (2.57), we deduce (2.51), whence (2.52) follows by
1
changing z by ? z. This completes the proof.
2
We are now in a position to prove the following remarkable result of
Gauss (cf. Theorems 8.1 and 8.2):
Theorem 2.2
?
For integers 1 ? p < q, we have
q?1
X
?
2pk
p
k
p
= ?? ? log q ? cot ? +
cos
log 2 sin ?.
q
2
q
q
q
a=1
(2.58)
Proof. We make the change of variable x = uk in Legendre?s formula
p
(2.50) with z = to obtain
q
Z 1
p
+? =
f (u) du,
?
q
0
(2.59)
where
up?1 ? uq?1
.
uq ? 1
f (u) = q
(2.60)
1
Let ? = e2?i q be the first primitive q-th root of 1. Then the denominator
decomposes into
(u ? 1)
q?1
Y
l=1
u ? ?l ,
and the factor u ? 1 cancels that of the denominator. Hence the partial
fraction expansion for f (u) is of the form
f (u) =
q?1
X
k=1
Ak
,
u ? ?k
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and
Ak = lim (u ? ?k )f (u)
u??k
k p?1
=q
(? )
? (?k )q?1
=
q?1
Q
k
l
(? ? ? )
l=0,l6=k
=
?k
l=0,l6=k
?pk ? 1
= ?pk ? 1,
(?q )k
q
q
?pk ? 1
q??k
q?1
Q
q?1
Q
l=0,l6=k
(1 ? ?l?k )
where we used the identity
q?1
Y
l=1
1 ? ?l = q,
(2.61)
which follows from the decomposition
q?1
Y
l=1
u ? ?l = uq?1 + и и и + 1.
(2.62)
Hence
?
Z
q?1
X
1 1
p
+? =
?pk ? 1
du
q
0 u ? ?k
k=1
or
q?1
X
1 ? ?k
p
?pk ? 1 log
.
+? =
?
q
??k
(2.63)
k=1
Noting that
q?1
Q
(1 ? ?k )
1 ? ?k
q
k=1
log
= log q?1
= log q,
= log
q q?1
k
??
Q
2
??
k
k=1
(?? )
q?1
X
k=1
we may rewrite (2.63) as
q?1
X
2i sin 2?
p
q k
+? =
?pk log
?
? log q,
k
q
?2
k=1
(2.64)
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The theory of the gamma and related functions
which becomes for q ? p in place of p:
q?1
X
2i sin 2?
p
q k
? log q.
?pk log
? 1?
+? =
k
q
?2
(2.65)
k=1
Adding and subtracting (2.64) and (2.65), we deduce that
q?1
X
p
p
2?p
2?
?
+? 1?
+ 2? =
2 cos
k log sin
k ? 2 log q
q
q
q
q
(2.66)
k=1
and
X
q?1
p
k
2?p
p
+? 1?
=
,
k ??i
2i sin
?
q
q
q
q
k=1
or
q?1
p
p
2?p
p
2? X
? cot ? = ?
k.
k sin
?? 1?
=
q
q
q
q
q
(2.67)
k=1
Hence adding (2.66) and (2.67), we obtain
q?1
X
p
2?p
2?
p
2?
cos
+ 2? = 2
k log sin
k ? 2 log q ? ? cot ? ,
q
q
q
q
k=1
whence (2.58) follows, and the proof is complete.
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Chapter 3
The theory of the Hurwitz-Lerch
zeta-functions
Abstract
In this chapter we shall give various contributions to the
theory of the Hurwitz zeta-function. In Д3.2 we shall give
integral representations (for the derivatives as well) which
give a basis for the theory of the gamma and related functions to be developed in Chapter 5. In Д3.3 we shall give a
proof of a formula of Ramanujan whose prototype (? = 1)
was first stated by Ramanujan and elaborated in [KKaY]
In Д3.4 we shall give two more proofs of the closed formula
for the integral of the psi-function, thus recovering the recent result of Espinosa and Moll. Finally, in Д3.5 we shall
give another proof of the functional equation.
3.1
Introduction
We shall consider the partial sum defined by (3.5) of the Hurwitz zetafunction defined by (3.1) and prove the integral representation which turns
out to hold true for ? itself. The proof as presented here is quite simple,
but the result is far-reaching and we may even base the whole theory of the
gamma and related functions on our results (Theorem 3.1 and its corollaries). We shall develop this aspect of our theory in Chapter 5. The special
feature of Theorem 3.1 is that the derivatives may be computed by differentiating with respect to u and the whole results may be inherited (for more
details, cf. the introductory remark at the beginning of Д3.2).
In Д3.3, we are going to give the sixth proof of the far-reaching formula
51
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Fig. 3.1
Riemann
of Ramanujan. This proof, incorporating the structure of the Hurwitz zetafunction as the principal solution of the difference equation, seems one of
the most natural ones.
We give only a simple example. For more summation formulas going
far beyond those in [SC] cf. e.g. [KTTY3].
InR Д3.4, we shall give two more proofs of the closed formula for the intez
gral 0 t? ?(t+a) dt, thus recovering the seemingly most important formula
of Espinosa and Moll [EM1]. We also give two enlightening remarks, the
latter of which speaks out the relation between Espinosa and Moll?s results
and Mikola?s? results.
In Д3.5, we shall sum up the existing proofs of the functional equation
(3.67) for the Hurwitz (Lerch) zeta-function and reveal the hierarchical
relationship among them, referring to Laurinc?ikas and Garunks?tis [LG] for
the Lerch zeta-function aspects. We shall add one more proof of (3.67)
based on the Dirac delta-function. Since from the delta-function, we may
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53
deduce the Poisson summation formula, we might regard our proof more
fundamental.
Thus, we put the existing literature on the Hurwitz zeta-function in
their hierarchical and historical perspective, with our recent contributions
[KKaY], [KKSY], [KTTY3] as touchstones.
We define the Hurwitz zeta-function and its special case, the Riemann zeta-function, by Dirichlet series absolutely convergent for ? > 1
in the first instance. Both are meromorphically continued over the whole
plane with a simple pole at s = 1 as we shall see below.
?(s, a) =
?
X
1
,
(n + a)s
n=0
?(s) = ?(s, 1) =
Re s = ? > 1,
?
X
1
,
ns
n=1
a>0
?>1
(3.1)
(3.2)
We define the counterpart of the Hurwitz zeta-function, the Lerch zetafunction or the polylogarithm function by (cf. Proposition B.1)
ls (a) =
?
X
e2?ina
,
ns
n=1
? > 1,
a ? R (or s = 1, 0 < a < 1)
(3.3)
We note that ?(s, a) satisfies the DE
?(s, a + 1) ? ?(s, a) = a?s .
We shall use the following notation.
Z ?
e?t ts?1 dt,
?(s) =
(3.4)
?>0
0
? the gamma function;
Z
?(s, a) =
a
?t s?1
e t
0
dt,
?(s, a) =
Z
?
e?t ts?1 dt
a
? the incomplete gamma functions of the 1st and the 2nd kind (cf. (3.53),
(3.66)), which satisfy ?(s, a) + ?(s, a) = ?(s);
?(s) =
?0 (s)
d
=
log ?(s)
?(s)
ds
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? the Euler digamma function or the psi function;
Hn = ?(n + 1) ? ?(1) = ?(n + 1) + ? =
n
X
1
k
k=1
? the n-th harmonic number, where ? signifies Euler?s constant defined by
(5.16) (see below as the Laurent constant ?0 (1) = ??(1) = ?).
n X
n
Bn (z) =
Bk z n?k
k
k=0
? the n-th Bernoulli polynomial with Bk the k-th Bernoulli number defined
through
?
X Bk
z
=
zk
z
e ?1
k!
|z| < 2?
k=0
B n (z) = Bn ({z}) = Bn (z ? [z]) for z ? R
? the n-th periodic Bernoulli polynomial, with [x] and {x} signifying the
integral and fractional parts of x, respectively.
We use the following as known:
Bm (z) = ?m ?(1 ? m, z),
m ? N,
? (m) (z) = (?1)m+1 m! ?(m + 1, z),
(cf. (4.1))
m ? N,
(cf. (5.17)).
The Laurent expansion of ?(s, a) at s = 1 is given by (cf. Corollary 3.3)
?(s, a) =
?
X
1
(?1)n ?n (a)
? ?(a) +
(s ? 1)n ,
s?1
n!
n=1
s ? 1.
The addition formula for the Bernoulli polynomial ((A))
n X
n
Bn (x + y) =
Bk (x) y n?k .
k
k=0
3.2
Integral representations
For complex variables u and a and x ? 0 let
X
Lu (x, a) =
(n + a)u ,
0?n?x
(3.5)
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55
denote the partial sum of the Hurwitz zeta-function, where for negative
values of u, the possible value of n for which n + a = 0 is to be excluded.
We shall use the Euler-Maclaurin sum formula (Theorem B.5, i.e. under
Appell?s (D0 )) to prove an integral representation for Lu (x, a), which has
?k
the following far-reaching features shared by the derivatives ?u
k Lu (x, a) as
well, i.e. all statements about the function in u (Lu (x, a) and ?(?u, a)) are
valid for their derivatives as well in the form of (i) below.
(i) It gives an analytic expression for Lu (x, a), which entails an integral
P
k
?k
u
representation for each derivative ?u
k Lu (x, a) =
0?n?x (n+a) log (n+a)
(the differentiation of the integral being carried out under the integral sign).
(ii) It gives an asymptotic formula for Lu (x, a) in x by estimating the
integral trivially, which is feasible for applications in the divisor problems.
(iii) It gives a generic definition of ?(?u, a) for u 6= ?1 (and for ?0 (a) :=
??(a) for u = ?1).
(iv) It gives an integral representation for the associated Hurwitz zeta?k
(k)
(?u, a)) for u 6=
function ?(?u, a) (and its derivatives ?u
k ?(?u, a) = ?
?1, and for u = ?1, it gives an analytic expression for the generalized
Euler constant ?k (a) (the k-th Laurent coefficient of ?(s, a) at s = 1),
which follows by simply putting x = 0 in the integral representation.
(v) The integral representation for ?(s, a) (or ?k (a)) in (iii) yields an
asymptotic formula for the ?(s, a + z) in z with Bernoulli polynomial coefficients (Theorem 2 [Kat1]) as given by Theorem 5.2 below.
?(u+1)
Convention. We sometime use (ur ) r! and ?(u+1?r)
interchangeably,
where the former is suited for easier calculation and the latter for expected
differentiation with respect to u.
Theorem 3.1 (Integral Representations) For any l ? N with l >
Re u + 1 and for any x ? 0, we have the integral representation
Lu (x, a) =
l
X
?(u + 1) (?1)r
B r (x) (x + a)u?r+1
?(u
+
2
?
r)
r!
r=1
Z ?
(?1)l ?(u + 1)
+
B l (t) (t + a)u?l dt
l! ?(u + 1 ? l) x
?
? 1 (x + a)u+1 + ?(?u, a), u 6= ?1,
+ u+1
?log(x + a) ? ?(a),
u = ?1.
(3.6)
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Also the asymptotic formula
Lu (x, a) =
l
X
u
(?1)r
B r (x)(x + a)u?r+1 + O xRe (u)?l
r?1
r
r=1
?
? 1 (x + a)u+1 + ?(?u, a), u 6= ?1
+ u+1
?log(x + a) ? ?(a),
u = ?1
(3.7)
holds true as x ? ?.
Furthermore, the integral representation
l
X
1
(?1)r
u
u+1
?(?u, a) = a ?
a
?
Br au?r+1
u+1
r
r
?
1
r=1
Z ?
l+1 u
B l (t)(t + a)u?l dt,
+ (?1)
l
0
u
(3.8)
which follows from (3.6) by putting x = 0, holds true for all complex u 6=
?1, where l can be any natural number subject only to the condition that
l > Re u + 1.
Proof.
Since the r-th derivative of f (t) = (t + a)u is
u
?(u + 1)
f (r) (t) =
(t + a)u?r ,
r! (t + a)u?r =
?(u + 1 ? r)
r
we see that the terms in the Euler-Maclaurin sum formula (Theorem B.5)
with a = 0 become
?
Z x
Z x
? 1 (x + a)u+1 ? 1 au+1 , u 6= ?1
u+1
f (t) dt = (t + a)u dt = u + 1
?log(x + a) ? log(a),
0
0
u = ?1,
l
o
X
(?1)r n
Br (x)f (r?1) (x) ? Br (0)f (r?1) (0)
r!
r=1
=
l
X
r=1
?(u + 1) (?1)r Br (x)(x + a)u?r+1 ? Br au?r+1 ,
?(u + 2 ? r) r!
and
(?1)l+1
l!
Z
x
Bl (t)f
0
(l)
(?1)l+1 ?(u + 1)
(t) dt =
l!
?(u + 1 ? l)
Z
x
0
Bl (t)(t + a)u?l dt,
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The theory of the Hurwitz-Lerch zeta-functions
respectively. Hence writing Lu (x, a) = au +
P
vista
57
(n + a)u , we obtain
0<n?x
?
?
1
1
(x + a)u+1 ?
au+1 , u 6= ?1
u+1
Lu (x, a) = a + u + 1
?log(x + a) ? log a,
u = ?1
u
+
l
X
r=1
?(u + 1) (?1)r
Br (x) (x + a)u?r+1
?(u + 2 ? r) r!
l
X
(3.9)
r
?(u + 1) (?1)
Br au?r+1
?(u
+
2
?
r)
r!
r=1
Z x
(?1)l+1 ?(u + 1)
Bl (t) (t + a)u?l dt.
+
l!
?(u + 1 ? l) 0
?
Now for any natural number l > Re u + 1, we have by integration by parts,
Z
?
x
Bl (t) (t + a)u?l dt
?
Z
u?l ?
Bl+1 (t)
(t + a)u?l
?
Bl+1 (t)(t + a)u?l?1 dt
l+1
l
+
1
x
Z ?x
Re u?l
u?l?1
=O x
+O
(t + a)
dt = O xRe u?l ,
=
x
whence it follows that
Z
x
Bl (t)(t + a)
u?l
dt =
0
=
Z
Z
?
0
0
?
Bl (t)(t + a)
u?l
dt ?
Z
?
Bl (t)(t + a)u?l dt
x
Bl (t)(t + a)u?l dt + O xRe u?l .
Hence we may replace the integral in (3.9) by
R?
0
Bl (t)(t + a)u?l dt +
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O xRe u?l to obtain
?
? 1 (x + a)u+1 ? 1 au+1 , u 6= ?1
u
u+1
Lu (x, a) = a + u + 1
?log(x + a) ? log a,
u = ?1
+
l
X
r=1
?(u + 1) (?1)r
Br (x) (x + a)u?r+1
?(u + 2 ? r) r!
l
X
?(u + 1) (?1)r
Br au?r+1
?(u
+
2
?
r)
r!
r=1
Z ?
(?1)l+1 ?(u + 1)
Bl (t) (t + a)u?l dt + O xRe u?l .
+
l!
?(u + 1 ? l) 0
(3.10)
?
Now for Re u < ?1 we let x ? ? to obtain for any l ? N,
l
X
1
?(u + 1) (?1)r
au+1 ?
Br au?r+1
u+1
?(u
+
2
?
r)
r!
r=1
Z ?
l
(?1) ?(u + 1)
Bl (t) (t + a)u?l dt,
+
l! ?(u + 1 ? l) x
?(?u, a) = au ?
(3.11)
which is (3.8).
Now for any u 6= ?1, take l ? N such that l > Re u + 1. Then the last
integral in (3.11) is absolutely convergent for Re u < l ? 1 and represents an
analytic function in Re u < l ? 1. Substituting (3.11) in (3.10), we deduce
(3.6) for u 6= ?1.
In the case u = ?1, the Euler-Maclaurin sum formula on the same lines
as above (cf. the proof of (5.39)) gives rise to
Z ?
?1
B l (t) (t + a)?1?l dt
L?1 (x, a) = (?1)l+1
l
0
l X
?1 1
(?1)r Br a?r
?
r
?
1
r
r=1
l X
(3.12)
?1 1
+
(?1)r B r (x) (x + a)?r
r?1 r
r=1
Z ?
l ?1
+ (?1)
B l (t) (t + a)?1?l dt
l
x
+ a?1 + log(x + a) ? log a.
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The theory of the Hurwitz-Lerch zeta-functions
Now for x large and l = 1, we obtain, from (3.12)
X
1
? log(x + a) = f (a) + O x?1 ,
n+a
(3.13)
0?n?x
where f (a) is a function in a only. If we adopt the definition (5.18) of ?(a),
then we must have f (a) = ??(a). Noting that from (3.12) with x ? ?
f (a) = ? log a + a
?1
+
l
X
Br
r=1
r
a
?r
?
Z
?
B l (t) (t + a)?1?l dt,
0
we have the integral representation for the digamma function
Z ?
l
1 ?1 X Br ?r
a +
B l (t) (t + a)?1?l dt.
?(a) = log a ? a +
2
r
0
r=2
(3.14)
Substituting the constant term ??(a) in (3.12), we have the integral representation for the partial sum.
Since the integrals appearing in Theorem 3.1 are analytic in the region
Re u < 1 ? l, we may differentiate (3.6) and (3.8) in u there. We state the
counterpart of (3.7) as the following corollaries (the counterpart of (3.8) to
be read off from them by putting x = 0).
Corollary 3.1
For any complex u and a > 0,
X
d
Lu (x, a) =
(n + a)u log(n + a)
du
(3.15)
0?n?x
=
l
X
(?1)r
r=1
r!
B r (x)(x + a)u?r+1
?(u + 1)
О
{?(u + 1) ? ?(u + 2 ? r) + log(x + a)}
?(u + 2 ? r)
Z
(?1)l ?
B l (t)(t + a)u?l
+
l!
x
?(u + 1)
{?(u + 1) ? ?(u + 1 ? l) + log(t + a)} dt
О
?(u + 1 ? l)
? 1
1
u+1
u+1
?
(x + a)
log (x + a) ?
?
?
2 (x + a)
?u+1
(u + 1)
+
?? 0 (?u, a) ,
u 6= ?1
?
?
?
? 1 {log (x + a)} 2 + ? (a) ,
u = ?1.
1
2
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Vistas of Special Functions
Corollary 3.2
For any complex u and a > 0,
d2
Lu (x, a)
(3.16)
du2
l
X
(?1)r
B r (x)(x + a)u?r+1
=
r!
r=1
2
?(u + 1) h
?(u + 1) ? ?(u + 2 ? r) + log(x + a)
О
?(u + 2 ? r)
i
+ ? 0 (u + 1) ? ? 0 (u + 2 ? r)
Z
(?1)l ?
B l (t)(t + a)u?l
+
l!
x
2
?(u + 1) h
О
?(u + 1) ? ?(u + 1 ? l) + log(t + a)
?(u + 1 ? l)
i
+ ? 0 (u + 1) ? ? 0 (u + 1 ? l) dt
?
u+1
(x + a)
2(x + a)u+1
?
?
{log(x + a)}2 ?
log (x + a)
?
?
?
(u + 1)2
? u+1
2(x + a)u+1
+
+
+ ? 00 (?u, a),
u 6= ?1
?
?
(u + 1)3
?
?
?
1
? {log(x + a)}3 + ? (a),
u = ?1.
2
3
We note that Theorem 3.1 [(3.6), (3.7)] is first obtained by Mellin [Me]
by means of the integral transform under his name (Д7.4) and is the most
?
informative for Lu (x, a), so are Corollaries 3.1 and 3.2 for ?u
Lu (x, a) and
?2
?u2 Lu (x, a), respectively. Formula (3.6) with l = 1, u 6= ?1, Re u < 0
appears as a prototype of the ?approximate functional equation? in Landau
[Lan, 9?10]. Mikolas? [M1] used it with x = 1 to obtain (3.8) with l = 1.
Then he proceeds to deduce (3.8) with l = 2 by integration by parts.
?(u+1)
Care should be taken in interpreting the coefficients like ?(u+1?l)
(?(u +
1) ? ?(u + 1 ? l)) when u is a negative integer; it is to be taken as one
without singularities (e.g. in deducing (5.20)).
Corollary 3.3 The k-th Laurent coefficient of the Hurwitz zeta-function
k
(at s = 1) is given by (?1)
k! ?k (a), where
?
?
X logk (n + a) logk+1 (x + a)
?
?
(3.17)
?k (a) = lim ?
x??
n+a
k+1
0?n?x
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The theory of the Hurwitz-Lerch zeta-functions
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61
and ?k (a) admits the integral representation
?k (a) =
1
1
logk a ?
logk+1 a
2a
k+1
Z ?
B 1 (t) k
k+1
?
log
(t
+
a)
?
k
log
(t
+
a)
dt.
(t + a)2
0
(3.18)
In particular, ?0 (a) = ??(a).
Proof. The following is the simplest possible method known. The starting point is Theorem 3.1 with l = 1 and ?s (s 6= 1, ? > 0) for u:
Z ?
(x + a)1?s
B 1 (x)
B 1 (t)
L?s (x, a) =
+ ?(s, a) ?
+s
dt. (3.19)
1?s
(x + a)s
(t
+
a)s+1
x
Since both sides of (3.19) are analytic in ? > 0, we may compute the k-th
Taylor coefficient around s = 1. The k-th Taylor coefficient of the left-hand
side is
1 ?k
(?1)k X
(n + a)?1 logk (n + a)
L?s (x, a)|s=1 =
k
k! ?s
k!
(3.20)
0?n?x
and that of the right-hand side is
(?1)k
k!
logk+1 (x + a)
B 1 (x)
+ ?k (a) ?
logk (x + a)
k+1
x+a
Z ?
B 1 (t) k
k?1
+
log (t + a) ? k log
(t + a) dt ;
(t + a)2
x
(3.21)
equating (3.20) and (3.21), we conclude that
X
?k (a) =
(n + a)?1 logk (n + a)
0?n?x
logk+1 (x + a) B 1 (x)
+
logk (x + a)
k+1
x+a
Z ?
B 1 (t) k
k?1
?
log
(t
+
a)
?
k
log
(t
+
a)
dt.
(t + a)2
x
?
(3.22)
We now note that (3.22), being valid for any x ? 0, implies both (3.17)
and (3.18) by letting x ? ? and x = 0 respectively, (cf. Berndt [Ber3]).
In the case k = 0, we note that (3.13) and (3.14) correspond to (3.17) and
(3.18), respectively.
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A formula of Ramanujan
In this section we are going to give the sixth proof of the fundamental
summation formula based on the use of finite differences, which has been
applied successfully in recent researches [KKY3], [KKY2].
Theorem 3.2 (Ramanujan)
For 0 ? ? ? Z and |z| < |?| we have
?
? X
X
?(m, ?) m+?
? 0
z
=
? (?k, ? ? z) z ??k
m
+
?
k
m=2
k=0
? ? 0 (??, ?) ?
?
X
1
?(k ? ?, ?) z k
k
(3.23)
k=1
1
+
(?(?) ? H? ) z ?+1 .
?+1
Proof. Let ?? f (?) = f (? + 1) ? f (?) be the difference operator (introduced in (DE) in Chapter 1). We apply this to the sum S on the LHS of
(3.23) to obtain
?? S = ? ?
?
?
?
X
X
X
??m m+?
1 z m
?(m, ?) m+?
.
z
=?
z
= ???
m+?
m+?
m ?
m=2
m=2
m=?+2
The resulting infinite series is nothing but
or
?+1
z X 1 z m
? log 1 ?
?
,
?
m ?
m=1
1 z m
? log(? ? z) ? log ? +
m ?
m=1
?+1
X
!
,
whence
?? S = ?? log(? ? z) ? ?? log ? +
?+1
X
???m m
z .
m
m=1
(3.24)
Rewriting
the first term on the RHS of (3.24) in the form
P?
? ??k
(? ? z)k log(? ? z) and telescoping (3.24), thereby noting
k=0 k z
that
? 0 (s, ? + 1) ? ? 0 (s, ?) = ??s log ?,
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The theory of the Hurwitz-Lerch zeta-functions
we get
S=
? X
?
k=0
k
? 0 (?k, ? ? z) z ??k ? ? 0 (??, ?)
?
X
?(?) ?+1
1
?(k ? ?, ?) z k +
z
+ f (z, ?),
?
k
?+1
(3.25)
k=1
where f (z, ?) is the function satisfying the conditions
?? f (z, ?) = 0
(3.26)
f (0, ?) = 0.
(3.27)
and
H? ?+1
It remains to determine f (z, ?) (to be ? ?+1
z
). First note that
? ?
d 0
? (?k, ? ? z) =
?(s, ? ? z) |s=?k
dz
?s ?z
?
s ?(s + 1, ? ? z) |s=?k
=
?s
(
?(1 ? k, ? ? z) ? k ? 0 (1 ? k, ? ? z), k ? N
=
??(? ? z),
k = 0.
With this in mind, we differentiate (3.25) with respect to z to obtain
??1
X ? ?
S=
(? ? k) ? 0 (?k, ? ? z) z ??k?1
(3.28)
?z
k
k=0
? ? X
X
?
? 0
+
?(1 ? k, ? ? z) z ??k ?
? (1 ? k, ? ? z) z ??k
k
k
k=1
? ?(? ? z) z ? ?
k=1
?
X
k=1
?(k ? ?, ?) z k?1 + ?(?) z ?? +
?
f (z, ?).
?z
We note that the two sums on the RHS of (3.28) containing ? 0 cancel
each other, while the second sum, say S2 , becomes, in view of the addition
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64
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formula,
? X
? 1
Bk (? ? z)z ??k
(3.29)
k k
k=1
k ? ??k X
? ??k
X
X
k
? z
? z
=?
(?z)k
Bl (?) (?z)k?l ?
k
k
l
k
k
S2 = ?
l=1
k=1
k=1
= S2,1 + S2,2 ,
say. Using (1.14) and changing the order of summation in S2,1 , we have
S2,1 = ?
? X
?
l
l=1
Bl (?) z ??l
??l X
? ? l (?1)k
k=0
k+l
k
.
Invoking the formula
K X
K (?1)k
k=0
k+l
k
=
K! ?(l)
,
?(l + K + 1)
we deduce that
S2,1 = ?
?
X
Bl (?)
l=1
l
z ??l =
?
X
?(1?l, ?) z ??l =
l=1
?
X
?(l??, ?) z l?1 . (3.30)
l=1
For S2,2 , we use another formula
? X
? (?1)k
k=1
k
k
= H?
to obtain
S2,2 = H? z ? .
(3.31)
From (3.29), (3.30) and (3.31) it follows that
S2 =
?
X
l=1
?(l ? ?, ?) z l?1 + H? z ? .
(3.32)
Substituting (3.32) in (3.28), we conclude that
?
?
S = ??(? ? z) z ? + ?(?) z ? + H? z ? +
f (z, ?).
?z
?z
(3.33)
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The theory of the Hurwitz-Lerch zeta-functions
On the other hand, from (3.25) we know that
?
S = ? (?(? ? z) ? ?(?)) z ? .
?z
(3.34)
Hence, comparing (3.33) and (3.34), we obtain
?
f (z, ?) = ?H? z ? ,
?z
whence
f (z, ?) = ?
H? ?+1
z
+ C.
?+1
By condition (3.27), C = 0, and
f (z, ?) = ?
H? ?+1
z
.
?+1
(3.35)
Substitution of (3.35) into (3.25) completes the proof.
There are enormous amount of formula (e.g. in [SC], where one third
is devoted to the statement of such formulas) which are consequences of
Theorem 3.2 (cf. [KKY1], [KKY3] and [KTTY3]). We give only a simple
example.
Example 3.1
The formula
log ?(a + 1)
2k
?
X
?
1
?(2k, a + 1) 1
1
1
= a+
log a +
+ a + ? log 2? ?
2
2
2
2k(2k + 1) 2
k=1
is first stated by Wilton [Wil1, Eq.(4)] and is a rather special case of Theorem 3.2. As an asymptotic formula in a, this gives the Stirling formula and
is a special case of Corollary 5.1.
3.4
Some definite integrals
We shall give two proofs of [EM2], Theorem 4.3, which seems the highest
summit of the paper, and coincides with our Corollary 3 (i) [KKY3]; the
first proof depends on a modified form of Theorem 3.2, which we state
as Lemma 3.1 while the second depends on a more antecedent one, i.e.
the intermediate formula toward the proof of Proposition 1 [KKaY, p.10],
which we state as Lemma 3.2.
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Theorem 3.3 ([KKY3, Corollary 3 (i)]=[EM2, Theorem 4.3])
? ? N ? {0}, we have
Z
z
0
=
For
t? ?(t + a) dt
Bk+1 (a + z)
(?1)k z ??k ? 0 (?k, a + z) ? Hk
k+1
k
k=0
B?+1 (a + z)
.
? (?1)? ? 0 (??, a) ? H?
k+1
? X
?
(3.36)
(3.36) should be compared with our previous result ([KKY3, Corollary 3]): (i) For ? ? N ? {0} and |z| < ?,
Z
z
t? ?(? + t) dt
0
= (?1)?
?
X
C? (r, ?) log ?r+1 (? + z)/?r+1 (?)
r=0
+ (?1)
?
?
X
l=1
?
B??l+1 (?)
z ?+1
0
(?1)
? (l ? ?) +
zl +
H? .
l
l(? ? l + 1)
?+1
l
(ii) For ? ? N
?
Z
z
t??1 log ?(? + t) dt
0
= z ? log ?(? + z) ? (?1)?
? (?1)?
?
X
l=1
(?1)l
?
X
C? (r, ?) log ?r+1 (? + z)/?r+1 (?)
r=0
z ?+1
?
B??l+1 (?)
H? ,
zl ?
? 0 (l ? ?) +
l(? ? l + 1)
?+1
l
where ?r (a) signifies the multiple gamma function [SC, p.39].
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Lemma 3.1 ([KKaY], (9))
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We have
?
X
?(m, ?) m+?
z
m+?
m=2
? X
?
=
{? 0 (?k, ? ? z) + Hk ?(?k, ? ? z)} z ??k
k
(3.37)
k=0
? (? 0 (??, ?) + H? ?(??, ?)) +
?(?) ?+1
z
.
?+1
Proof. (First proof of Theorem 3.3) We start from the Taylor expansion
(|z| < ?)
?
?
X
X
? (n) (?) n
z = ?(?) +
(?1)m ?(m, ?) z m?1 .
n!
m=2
n=0
?(z + ?) =
(3.38)
Multiplying both sides of (3.38) by z ? and integrating over [0, z] with respect to z, we deduce that
Z
z
0
=
u? ?(? + u) du
Z
z
u? ?(?) du +
0
(3.39)
?
X
(?1)m ?(m, ?)
m=2
Z
z
u?+m?1 du
0
?
X
z ?+1
?(m, ?)
(?z)m+? +
?(?).
= (?1)?
m+?
?+1
m=2
Substituting (3.37) with ?z in place of z into (3.39), we obtain
Z
z
u? ?(? + u) du
0
= (?1)
?
? X
?
k=0
k
{? 0 (?k, ? + z) + Hk ?(?k, ? + z)} (?z)??k
(3.40)
?(?)
?(?) ?+1
(?1)? (?z)?+1 +
z
?+1
?+1
? (?1)? (? 0 (??, ?) + H? ?(??, ?)) ,
+
which is (3.36).
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Lemma 3.2
(?1)?+1
=
1
s?1
?
Z
For 0 ? ? ? Z we have
z
u? ?(s, a + u) du
0
? X
k=0
? ?(2 ? s) k!
?(s ? k ? 1, a + z) (?z)??k
k ?(k + 2 ? s)
1 ?! ?(2 ? s)
?(s ? ? ? 1, a).
s ? 1 ?(? ? 2 ? s)
Proof. (Second proof of Theorem 3.3) Subtracting
?+1
(?1)
s?1
the left-side, and
(?1)
?+1
Z
z
u
0
?
?+1
z
?+1
(?1)?+1
s?1
Rz
0
(3.41)
u? du from
from the right-side, of (3.41), we deduce that
1
?(s, a + u) ?
s?1
du =
1
F (s),
s?1
(3.42)
where
? X
? ?(2 ? s) k!
?(s ? k ? 1, a + z) (?z)??k
F (s) =
k ?(k + 2 ? s)
k=0
?(2 ? s)
(?z)?+1
? ?!
?(s ? ? ? 1, a) ?
.
?(? + 2 ? s)
?+1
(3.43)
We are to take the limit of (3.42) as s ? 1. For this we first contend
that F (1) = 0. Indeed,
F (1) = ?
Rewriting
that
? X
? Bk+1 (a + z)
k=0
?
1
k k+1
F (1) = ?
k+1
k
as
?+1
1
?+1 k+1
(?z)??k +
B?+1 (a) (?z)?+1
?
.
?+1
?+1
(3.44)
and writing k for k+1, we derive from (3.44)
?+1 1 X ?+1
B?+1 (a)
Bk (a + z) (?z)?+1?k +
,
?+1
k
?+1
(3.45)
k=0
where we incorporated the last term in (3.39) in the first sum of (3.45).
Noting that the fist sum of (3.45) is nothing but the expansion of the
Bernoulli polynomial B?+1 (a + z ? z) = B?+1 (a), we conclude F (1) = 0.
Hence, we
R z may take the limit as s ? 1 of (3.37). On the left side we
have (?1)? 0 u? ?(a + u) du by the Laurent expansion of ?(s, a + u), and
on the right-side we just differentiate F (s) with respect to s, thereby noting
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The theory of the Hurwitz-Lerch zeta-functions
the formula
0 ?(2 ? s)
?(k + 1 ? s) s=1
?(2 ? s)
1
=
=
(??(2 ? s) + ?(k + 2 ? s))
Hk ,
?(k + 2 ? s)
k!
s=1
(3.46)
to obtain
F 0 (1) =
? X
?
k=0
k
{? 0 (?k, a + z) + Hk ?(?k, a + z)} (?z)??k
(3.47)
? {? 0 (??, a) + H? ?(??, a)} ,
Rz
which is equal to (?1)? 0 u? ?(a + u) du. By multiplying by (?1)? completes the proof.
Remark 3.1
In the notation of [EM2, (3.1), (3.28)],
1
((k + 1)? 0 (?k, q) ? Hk Bk+1 (q))
k+1
= (k + 1)! ? (?k?1) (q),
? 0 (?k, q) + Hk ?(?k, q) =
and our Theorem 3.3 coincides with Theorem 4.3 of Espinosa and Moll.
Remark 3.2
form
(i) Espinosa and Moll [EM1] developed the Hurwitz transZ
1
f (u)?(s, u) du
0
and deduced several results for special types of f (u) which can be expanded
into Fourier series as consequences of their Theorem 2.2, which in turn is
a consequence of the ?Fourier series?:
?(s, u) = 2 ?(1 ? s)
?s ,
(2?n)s?1 sin 2?nu +
2
n=1
?
X
(3.48)
or, more commonly known as the Hurwitz formula (cf. (5.56)). We note
that Mikola?s? [M3] gave the simplest proof of (3.48) as the Fourier series,
whereby he computed the Fourier coefficients
Z 1
?(1 ? s)
?(s, u) e?2?i?u du =
,
(3.49)
1?s
(2?i?)
0
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0 < s < 1, 0 6= ? ?
/ Z. From (3.49) we immediately deduce
Z 1
Z 1
?(s, u) cos(2??u) du ? i
?(s, u) sin(2??u) du
0
0
Z 1
1
1
(2?)s ? s?1
?2?i?u
,
?i
?(s, u) e
du =
=
4?(s)
cos ?2 s
sin ?2 s
0
whence follows Formulas (2.2) and (2.3) of Espinosa and Moll.
(ii) Espinosa and Moll [EM1] refer to Mikola?s? paper [M2] and quote
the result
s
Z 1
?(2s) (a, b)
2
?(1 ? s, {aq}) ?(1 ? s, {bq}) dq = 2 ? (s)
(2?)2s [a, b]
0
((a, b) = g.c.d. of a and b, and [a, b] = l.c.m.)
for Re (1 ? s) < 12 . We note that Mikola?s [M1] obtained the result on the
basis of Fourier analysis (the Parseval formula):
Z 1
?(s, u) ?(s0 , u) du
0
?
0
(s ? s0 ) ?(2 ? s ? s0 )
= 2 (2?)s+s ?2 ?(1 ? s) ?(1 ? s0 ) cos
2
for max{0, Re s} + max{0, Re s0 } < 1; the region of validity wider than that
of Espinosa and Moll who have s < 0, s0 < 0.
This result of Mikola?s?, combined with our recent developments of the
product of zeta-functions [KTY1], may shed some new light on the asymptotic formula for mean square of zeta-functions. In fact, it looks like the
region is one of the excluded one in Katsurada [Kat] and Katsurada and
Matsumoto [KM]. For recent developments, cf. [Hashimoto] and [KTZ2].
3.5
The functional equation
In [KKSY] statements were made about the proof of the functional equation, or the Hurwitz formula (3.48), for the Hurwitz zeta-function, using the
absolutely convergent Fourier series for B 2 (t) rather than the boundedly
convergent Fourier series for B 1 (t). Meanwhile the book of Laurinc?ikas and
Garunks?tis [LG] has appeared which has rich contents about rather wide
P? e2?in?
spectrum of the theory of the Lerch zeta-function ?(?, a, s) = n=0 (n+a)
s
(cf. (8.21)), and we can do no better than referring to it regarding various
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71
proofs of the functional equation for ?. We shall therefore review mostly
those papers which were not quoted in [LG].
As mentioned in Remark 3.2, Mikola?s [M3] made use of the Fourier
series to deduce the functional equation for ?(s, a) and in the subsequent
paper [M4], he applied the same method to prove the functional equation
for ?(?, a, s).
Berndt [Ber3] used the boundedly convergent Fourier series to deduce
(3.48), which he further applied to ?(?, a, s) to deduce the functional equation in [Ber4], where he gave another proof for it, which was reproduced by
[LG].
Fine [Fine] applied Riemann?s second method, i.e.
the thetatransformation formula (for ?3 ), or what amounts to the same thing,
the Poisson summation formula, to prove (3.48), while Apostol [Ap1] deduced (3.48) from the functional equation and the distribution property for
?(?, a, s).
Apostol?s paper [Ap2] (cf. also [Ap3]) contains the seemingly most natural proof of the functional equation for ?(?, a, s) based on the transformation
formula and the difference equational structure of ?.
As has been developed rather fully in [KTY7], the theta-transformation
formula or the modular relation a la? Bochner and the functional equation
are equivalent. In this respect, Fine and Apostol would lead to Bochner and
may be considered as the prototype of manifestation of the zeta-function
associated to prehomogeneous vector spaces.
We remark, however, that although in the above mentioned papers, Lipschitz [Li], Lerch [Le], Hurwitz [H] are referred to, but are neither Malmste?n
[Ma] nor Schlo?milch [Sch], who gave the functional equation for some Lfunctions (the L-function modulo 4, to be precise), nor the paper of Euler.
In this regard we must take into account Weil?s paper [We], which gives
a translation and comments on Eisenstein?s copy of Gauss? Disquisitiones,
especially the last page (dated 1849) inserted by the binder. On that page,
Eisenstein made an ?unmotivated? application of the Poisson summation
formula to prove the functional equation for ?(?, a, s) from which he deduces that for L-function mod 4. His argument precedes Oberhettinger
[Ob] by 107 years in that he uses the Fourier transform
Z ?
e2?ixy xq?1 dx,
0
while Oberhettinger produces the proof by using the Laplace transform.
The Fourier transform is also the basis of Mikola?s? proof [M3]. Here we
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may enjoy a happy encounter of some of the greatest unsimultaneous mathematicians of all time, Gauss, Eisenstein and Weil. We are also fascinated
by Weil?s imagination on the source of Riemann?s paper.
We are indebted to Sato?s paper [Sa] for this paper of Weil; without
Sato?s, we may have missed it, and indeed, in no other places, this discovery
of Eisenstein has been presented. E.g. in Grosswald, the Lipschitz transformation formula (i.e. the functional equation) is proved by the Poisson
summation formula, which is in principle the same as Eisenstein?s proof.
Sato?s paper (cf. [KTY7] as well) contains a very nice list of functional
equations that follow from the theta-transformation formula and some other
deep insight.
We can present a high-brow proof using the Fourier series for the Dirac
delta function ?(s) by completing the incomplete gamma functions.
Our starting point is thus the combination of (41) and (43) of [KKSY]
(where we write s for ?u), which we state as (3.65) below. To derive it, we
shall make full use of Formula (3.8) with l = 2 (even though any l ? 2 will
work here):
1
1
1
au+1 + au ?
u au?1
u+1
2
12
Z
u(u ? 1) ?
B 2 (t)(t + a)u?2 dt,
?
2!
0
?(?u, a) = ?
(3.50)
valid for Re s < 2, which coincides with [UN, Formula (24)].
Substituting the absolutely convergent Fourier series (cf. (1.9)):
?
?
1 X e2?int + e?2?int
1 X cos(2?nt)
B 2 (t) =
= 2
2? 2 n=1
n2
? n=1
n2
(3.51)
in the integral in (3.8) in the same context as in Rademacher [R, p.83], Pan
and Pan [PP, p.125], Kanemitsu [Kan], and Ueno and Nishizawa [UN], we
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The theory of the Hurwitz-Lerch zeta-functions
infer after simplification that
Z ?
B 2 (t)(t + a)u?2 dt
0
=
(3.52)
Z ?
?
1 X 1
?2?ina
1?u
e
(?2?in)
e?x xu?2 dx
2? 2 n=1 n2
?2?ina
Z ?
2?ina
1?u
?x u?2
+e
(2?in)
e x
dx
2?ina
?
1 X 1 ?2?ina
e
(?2?in)1?u ?(u ? 1, ?2?ina)
=
2? 2 n=1 n2
+ e2?ina (2?in)1?u ?(u ? 1, 2?ina) ,
where ?(s, z) designates the incomplete gamma function of the second
kind defined by
Z ?
?(s, z) =
e?x xs?1 dx.
(3.53)
z
The function ? (s, z) can be expressed in terms of the confluent hypergeometric function as follows [Erd, p.266, Eq. 6.9 (21)]:
?(s, z) = e?z ?(1 ? s, 1 ? s; z),
where [Erd, p.255, Eq. 6.5(2)]
Z ?
1
e?zt ta?1 (1 + t)c?a?1 dt,
?(a, c; z) =
?(a) 0
(3.54)
min{Re a, Re z} > 0,
(3.55)
is a solution of the differential equation:
z
d2 w
dw
+ (c ? z)
? aw = 0
dz 2
dz
(3.56)
and is denoted by U (a, c; z) in Slater [Sla].
Using [Erd, p.257, Eq. 6.5(6)]
?(a, c; z) = z 1?c ?(a ? c + 1, 2 ? c; z)
(3.57)
ez z ?s ?(s, z) = ?(1, 1 + s; z).
(3.58)
in (3.54), we get
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Substituting (3.58) in (3.52), we may write
Z ?
B 2 (t)(t + a)u?2 dt
0
=
?
1 X 1 u?1
a
{?(1, u; ?2?ina) + ?(1, u; 2?ina)} ,
2? 2 n=1 n2
(3.59)
which, when substituted in (3.50), gives rise to
?(?u, a)
1
1
1
=?
au+1 + au ?
u au?1
u+1
2
12
?
u(u ? 1) au?1 X 1
?
{? (1, u; ?2?ina) + ?(1, u; 2?ina)} .
2
2? 2 n=1 n2
(3.60)
To deduce the Ueno-Nishizawa formula [UN, Formula (25)], we take
l = 1 and argue in the same way. We just state here the corresponding
formulas:
Z ?
1
1
B 1 (t)(t + a)u?1 dt,
(3.61)
au+1 + au + u
?(?u, a) = ?
u+1
2
0
? ?
1 X e2?int ? e?2?int
1 X sin (2?nt)
B 1 (t) = ?
=?
,
2?i n=1
n
? n=1
n
Z
?
B 1 (t)(t + a)u?1 dt
0
=?
and
(3.62)
?
1 X au ?2?ina
e
(?2?ina)?u ?(u, ?2?ina)
2?i n=1 n
? e2?ina (2?ina)?u ?(u, 2?ina) ,
1
1
au+1 + au
u+1
2
u au X 1
?
?(1, u + 1; ?2?ina),
2?i
n
?(?u, a) = ?
n6=0
which is [UN, Formula (25)].
(3.63)
(3.64)
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The theory of the Hurwitz-Lerch zeta-functions
We shall use the combination of (3.61) and (3.63):
? e?2?ina
1 X
?(s, a) = s?1
?(1 ? s, ?2?ina)
a
(?2?ina)1?s
n=1
e2?ina
?(1 ? s, 2?ina)
+
(2?ina)1?s
1
1
1
+ s + s?1
.
2a
a
s?1
(3.65)
We use the incomplete gamma function ?(s, a) of the first kind
Z a
Z 1
?u s?1
s
?(s, a) =
e u
du = a
e?au us?1 du
(3.66)
0
0
and complete ?(1 ? s, a) to write ?(1 ? s, a) = ?(1 ? s) ? ?(1 ? s, a). Thus
? X
1
1
e2?ina
1
e?2?ina
+ s + s?1
+
?(s, a) = ?(1 ? s)
1?s
1?s
(?2?in)
(2?in)
2a
a
s
?
1
n=1
Z
?
1 X ?2?ina 1 2?inau ?s
e
u du
? s?1
e
a
0
n=1
Z 1
2?ina
?2?inau ?s
+e
e
u du .
0
We invert the order of summation and integration in the last term and
P0 ?
?2?ina(u?1)
consider the series
as the Fourier series for ?(a(u ?
n=?? e
1)) ? 1. Then we are left with the integration (? < 0)
Z 1
Z 1
1
1
?s
?
?(a(u ? 1)) u du +
u?s du = ? ?
.
2a
s
?
1
0
0
1
1
Hence the last term is ? 2a1s ? s?1
as?1 , which cancels the second term and
we finally arrive at the Hurwitz formula
?(1 ? s) 1?s ?i
? 1?s
2
2 ?i l
l
(1
?
a)
+
e
(a)
,
(3.67)
?(s, a) =
e
1?s
1?s
(2?)1?s
which is equivalent to (3.48).
Finally, we introduce a class of functions ?n (x), n ? N (due to Milnor [Mi]) defined by
?1?t (x) =
?
?(t, x)
?t
(3.68)
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Note that
?(x)
?1 (x) = ? 0 (0, x) = log ?
.
2?
Exercise 3.1
Prove the Kubert identity
q?1 X
x+k
s?1
?(1 ? s, x) = q
? 1 ? s,
q
(3.69)
k=0
for each n ? N (s 6= 0). Also prove the modified Kubert identity
q?1
X
1
x+h
n?1
?n (x) = (log q) Bn (x) + q
.
?n
n
q
(3.70)
k=0
Solution
By (8.12),
?(s, a, 1) = ?(s, a).
(3.71)
Hence (8.13) reduces to (3.69).
To prove (3.70), we differentiate
q s ?(s, x) =
q?1 X
x+k
? s,
,
q
(3.72)
k=0
with respect to s to obtain
? 0 (1 ? n, x) = ?(log q) ?(1 ? n, x) + q n?1
q?1 X
x+k
? 0 1 ? n,
.
q
k=0
This leads to (3.70) on appealing to (4.1).
For more information on Kubert identities, the reader is referred to
Sun [Su1].
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Chapter 4
The theory of Bernoulli polynomilas
via zeta-functions
Abstract
In this chapter we shall deduce some of the basic properties
of Bernoulli polynomials from those of the Hurwitz zetafunction. The basis is the relation (4.1). We may develop
the theory of Euler polynomials in the same spirit. This
is due to the fact that the Euler number En corresponds
to the special value L(?n, ?4 ) of the Dirichlet L-function
L(s, ?4 ) with the unique odd character mod 4 (cf. Chapter 8), which therefore is not presented (cf. e.g. [SC]).
Exercise 4.1
Under (U ) deduce
?(1 ? n, x) = ?
1
B n (x),
n
n?N
from (3.8).
Solution
Formula (3.8) with u = n ? 1, l = n reads for 0 < a < 1,
?(1 ? n, a) = an?1 ?
n
1 n X (?1)r n ? 1
a ?
Br an?r .
n
r
r
?
l
r=1
This can be transformed, on using (cf. (1.13) and (1.16))
Br = (?1)r Br ,
r ? 2,
77
1
B1 = ? ,
2
(4.1)
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into
n 1 n 1 n?1 1 X n
?
?(1 ? n, a) = ? a + a
Br an?r
n
2
n r=2 r
n 1X n
Br an?r ,
=?
n r=0 r
which is (4.1) in view of (1.7).
Remark 4.1 Since (3.8) depends on (1.1), we have deduces (4.1) under
(D?) and (U). But since (D?), (A) and (U) are equivalent as giving Taylor
coefficients, we may choose any one of them as a definition and assume
other two valid.
We may also define the Bernoulli polynomial Bs (x) as an integral function of s through the relation
Bs (x) = ?s ?(1 ? s, x)
(4.2)
as in [Ca] or [Mi] and develop the whole theory independently of Chapter 1
(which procedure will be sketched below), and thus we shall take (4.1) for
granted and deduce other properties from (3.8) etc.
As a special case of Exercise 4.1, we have (n ? N)
(
? n1 Bn ,
n ? 2,
1
?(1 ? n) = ? Bn (1) =
1
n
B1 = ? , n = 1,
(4.3)
2
by (1.16), which in turn is a consequence of (1.9).
Exercise 4.2
Solution
Deduce (1.9) from (3.67) and (A) (and (4.1)).
For 0 < x < 1, (3.67) reads for s = 1 ? n, n ? N
?(1 ? n, x) =
?(n) ? ?i n
?i
{e 2 ln (x) + e 2 n ln (1 ? x)},
n
(2?)
(4.4)
so that
B n (x) = ?n ?(1 ? n, x)
=?
?(n + 1)
{(?i)n ln (x) + in ln (1 ? x)}
(2?)n
and
B n (1 ? x) = ?
?(n + 1)
{(?i)n ln (1 ? x) + in ln (x)}.
(2?)n
(4.5)
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The theory of Bernoulli polynomilas via zeta-functions
Comparing these completes the proof of (1.9) in the case 0 < x < 1.
If, in general, n ? 1 < x < n, n ? Z, then [x] = n and [1 ? x] = ?n, and
therefore
x?
/ Z.
[1 ? x] = ?[x],
(4.6)
By (A),
Bn (x) = Bn (x + [x]) =
n X
n
k=0
k
B n?k (x)[x]k .
Substituting (1.9) for 0 < x < 1 and (4.6). we conclude (1.9) for x ?
/ Z.
For x ? Z, (1.9) follows from continuity.
Proposition 4.1 The difference relation (DE) for Bernoulli polynomials
is a consequence of that for the Hurwitz zeta-function (3.4).
Proof.
This follows immediately from (3.4) (under (4.1)):
Bn (x + 1) ? Bn (x) = ?n{?(1 ? n, x + 1) ? ?(1 ? n, x)}
= nxn?1 .
Proposition 4.2 The functional equation (3.67) for the Hurwitz zetafunction implies the Fourier expansion (H) for the Bernoulli polynomials.
Proof.
(4.5) reads
n!
B n (x) = ?
(2?i)n
(
?
X
e2?ikx
k=1
kn
?
X
e2?i(?k)x
+ (?1)
kn
n
k=1
)
,
which gives (H) for n ? 2, since, then, the series are absolutely convergent.
In the case n = 1, the sum is to be taken symmetrically:
lim
N ??
2?ikx
N
N
X
X
e?2?ikx
e
sin 2?kx
?
,
= lim 2i
N ??
k
k
k
k=?N
k6=0
k=1
whence
?
1 X sin 2?kx
B 1 (x) = ?
?
k
k=1
as in (7.9).
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Proposition 4.3 The Kubert identity (3.69) for the Hurwitz zetafunction implies the Kubert identity (1.8) for Bernoulli polynomials.
Proof.
This follows from (3.69) on substituting (4.1).
Now that we have established (DE), (H) and (K) for Bernoulli polynomials, we may trace the logical path given at the end of Chapter 1 to
complete the theory of Bernoulli polynomials.
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Chapter 5
The theory of the gamma and related
functions via zeta-functions
Abstract
In this chapter we shall give a new foundation of the theory
of the gamma and related functions. The core of the idea
lies in appealing to Lerch?s formula (5.4) through which
we may transfer the results on the Hurwitz zeta-function
to the gamma function (cf. [Mi]), as was the case with the
Bernoulli polynomials and the function ?(0, x) in Chapter 4.
We shall give two (three if we count the uniqueness theorem
as one) proofs of Lerch?s formula with minimum possible
assumptions: the integral representation (5.1) for ?(s,z)
(which is a corollary to Theorem 3.1 and the value ? 12 ).
Then we continue to keep the assumption minimum by
defining the digamma function by either of the conditions
in Lemma 5.1 and the gamma function as its integral.
5.1
Derivatives of the Hurwitz zeta-function
Notation: s = ? + it is the complex variable; z is another complex variable,
used interchangeably with s;
?(s) = ? lim
N ??
N
X
1
? log(N + s)
n+s
n=0
81
!
vista
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?the digamma function (cf. (5.18));
log ?(s) =
Z
s
?(u) du
1
?the log-gamma function;
? = ??(1)
?the Euler?s constant (cf. (5.16));
?(s, a) =
?
X
1
(n
+
a)s
n=0
?the Hurwitz zeta-function, where ? > 1, a ? C, a 6= non-negative integer;
?(s) = ?(s, 1) =
?
X
1
,
s
n
n=1
?>1
?the Riemann zeta-function. For x ? 0, a ? C, a 6= non-negative integer,
u ? C,
X
Lu (x, a) =
(n + a)u
0?n?x
?the partial sum ((3.5)) of the Hurwitz zeta-function ?(?u, a);
B k (t) = Bk (t ? [t])
?the k-th periodic Bernoulli polynomial;
Bk (t) =
k X
k
r
r=0
Br tk?r
?the k-th Bernoulli polynomial (cf. (1.7)); Bk -the k-th Bernoulli number;
[t] ?the integral part of t.
Theorem 5.1
and ?(z):
If we suppose the integral representations for ?(s, z)
1
1
z 1?s + z ?s ? s
?(s, z) =
s?1
2
Z
?
B 1 (t) (t + z)?s?1 dt,
0
1
?(z) = log z ? z ?1 +
2
Z
?
0
B 1 (t) (t + z)?2 dt,
? > ?1, (5.1)
(5.2)
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and also the value
?
1
1
= ?2
2
vista
83
(5.3)
as known, then we have Lerch?s formula
?(z)
? 0 (0, z) = log ? .
2?
(5.4)
Integrating (5.2) from 1 to z, we obtain
Z z
Z ?
Z z
1
log ?(z) =
dz +
B 1 (t) (t + z)?2 dz dt
log z ?
2z
1
0
1
1
= z log z ? z ? log z + 1
2
Z
Z
Proof.
?
?
?
B 1 (t) (t + z)?1 dt +
0
(5.5)
B 1 (t) t?1 dt.
1
On the other hand, differentiation of (5.1) gives
Z ?
1
0
B 1 (t) (t + z)?1 dt.
? (0, z) = z log z ? z ? log z ?
2
0
Comparing (5.5) and (5.6), we see that
Z
? 0 (0, z) = log ?(z) ? 1 ?
?
B 1 (t) t?1 dt
(5.6)
(5.7)
1
and it remains to evaluate the last integral. For this we differentiate the formula ? s, 12 = (2s ? 1) ?(s) to obtain
1
= 2s (log 2) ?(s) + (2s ? 1) ? 0 (s).
(5.8)
? 0 s,
2
Hence in view of ?(0) = ? 21 , a consequence of (5.1),
1
1
? 0 0,
= (log 2) ?(0) = ? log 2.
2
2
in (5.7) and use the value of ?( 12 ) to obtain
Z ?
?
1
? log 2 = log ? ? 1 ?
B 1 (t) t?1 dt.
2
1
Now put z =
(5.9)
1
2
Hence
1+
Z
?
1
B 1 (t) t?1 dt = log
?
2?,
(5.10)
(5.11)
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which is of some interest in its own right.
(5.6) and (5.11) combine to give (5.4). This completes the proof.
Exercise 5.1
If we assume the integral representation (5.1) and the value
?
? 0 (0) = ? log 2?
(5.12)
as known, then we have Lerch?s formula (5.4).
Differentiation of (5.11) with respect to s gives rise to
Proof.
? 0 (s, x)
=
x1?s
1
?1
x1?s ?
log x ? x?s log x ?
2
(s ? 1)
s?1
2
Z ?
?s
B 1 (t) (t + x)?s?1 (? log(t + x)) dt,
Z
?
B 1 (t) (t + x)?s?1 dt
0
0
whence
1
log x ?
2
Z
1 1 1
?2 0
? (0, x) = +
?2
?x2
x 2 x2
Z
? 0 (0, x) = ?x + x log x ?
?
B 1 (t) (t + x)?1 dt.
(5.13)
B 1 (t) (t + x)?3 dt.
(5.14)
0
Now
?
0
Now the last integral on the right-hand side of (5.14) is the sum of the
terms
Z n+1
t ? n ? 21 (t + x)?3 dt
n
=
=
Z
n+1
n
Z n+1
n
(t + x) ? n + x +
1
2
(t + x)?2 ? n + x +
1
2
(t + x)?3 dt
(t + x)?3 dt
n+1
1
1
+
n + x + 12 (t + x)?2
t+x 2
n
1
1
=
?
n+x n+x+1
1
1
n + 1 + x ? 21 (n + 1 + x)?2 ?
n + x + 12 (n + x)?2
+
2
2
1
1
?
=
n+x n+x+1
= ?
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The theory of the gamma and related functions via zeta-functions
1
+
2
Hence
Z
?
1
1
?
n+1+x n+x
1
?
4
1
1
+
(n + 1 + x)2
(n + x)2
.
B 1 (t)(t + x)?3 dt
0
? ?
1
1
1X
1X
1
1 1
=
?
?
+
2 n=0 n + x n + x + 1
2 n=0 (n + x)2
4 x2
=
?
1X
1
11 1 1
+
?
.
2
2x 4x
2 n=0 (n + x)2
Substituting this in (5.14), we obtain
?
X
?2 0
1
1 1 1
1 1 1
?
(0,
x)
=
+
?
?
+
?x2
x 2 x2
x 2 x2 n=0 (n + x)2
= ?(2, x).
It is essential to notice that
?
X
d ?0
d
1
d2
log
?(x)
=
(x)
=
?(x)
=
,
2
dx
dx ?
dx
(n + x)2
n=0
the last being due to (5.17) below.
Hence
?2 0
d2
? (0, x) =
log ?(x).
2
?x
dx2
(5.15)
(5.15) gives rise to
? 0 (0, x) = log ?(x) + ax + b.
First
? 0 (0, 1) = a + b
and
? 0 (0, 2) = 2a + b.
Recalling ?(s, x + 1) = ?(s, x) ? x?s , we see that ? 0 (s, x + 1) = ? 0 (s, x) +
x log x, whence ? 0 (0, 2) = ? 0 (0, 1). Hence a = 0.
The value of b = ? 0 (0, 1) = ? 0 (0) is determined by (5.12) and we have
?
? 0 (0, x) = log ?(x) ? log 2?,
?s
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i.e. Lerch?s formula.
Lemma 5.1
Under the definition of Euler?s constant
!
N
X
1
? = lim
? log(N + z)
N ??
n
n=1
(5.16)
for any z other than negative integers, the two definitions for ? are equivalent:
? X
1
1
,
(5.17)
?
?(z) + ? =
n z+n?1
n=1
the Gaussian representation (cf. (5.41)), and
?(z) = ? lim
N ??
!
1
? log(N + z) ,
n+z
n=0
N
X
(5.18)
for any z other than negative integers, the generic definition.
Proof.
Substituting (5.16) in (5.17) in the form
N X
1
1
?(z) + ? = lim
?
N ??
n z+n+1
n=1
N X
1
? log(N + z)
= lim
N ??
n
n=1
!
N
X
1
?
? log(N + z)
,
z+n?1
n=1
we deduce that
?(z) + ? = ? ? lim
N ??
N X
n=0
1
? log(N + z) ,
z+n
whence (5.18).
On the other hand, (5.18) may be written as
N X
1
1
?
?(z) = lim
N ??
n
z
+
n
?1
n=1
?
X
1
1
=
? ?,
?
n z+n?1
n=1
?
N X
1
n=1
n
? log(N + z)
!
,
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87
i.e. (5.17).
Remark 5.1 In (5.16), z is usually taken to be 0, but can be any number
as in (5.16) because
log(N + z) ? log(N + w) ? 0,
N ? ?.
The absolute
convergence of the series in (5.17) is clear because each term
is O n12 , and the existence of limits in (5.16) and (5.18) follows from the
comparison with the corresponding integral, or the Euler-Maclaurin formula
(cf. Chapter 2).
We shall illustrate the far-reaching power of Theorem 3.1 by the first
?
Lu (x, a) or ?? 0 (?u, a)) in the special case of u = m, m ?
derivative ( ?u
N ? {0}. For N 3 l > m + 1, Corollary 3.1 eventually yields (cf. [KTTY3])
?? 0 (?m, a) = lim
N ??
N
X
(n + a)m log(n + a)
(5.19)
n=0
1
1
(N + a)m+1 log(N + a) +
(N + a)m+1
m+1
(m + 1)2
m+1
X m Br
1
? (N + a)m log(N + a) ?
r ? 1 r!
2
r=2
!
1
1
m?r+1
и
+иии +
+ log(N + a) (N + a)
.
m
m?r+2
?
and
? 0 (?m, a)
(5.20)
1
1
1
1
am+1 log a ?
am+1 ? am log a +
am?1 log a
=
m+1
(m + 1)2
2
12
?
?
m+1
r?2
X Br X
m
1
? (?1)j m
+
+
log a?
r
?1
r
r
?
1
?
j
j
r=4
j=0
?
?
r?1
l
X
X
1
r?m?2
1 ? m?r+1
+
Br ? (?1)j
a
j
m + 1 r=m+2
r
?
j
j=0
?
?
Z ? X
l?1
1
? (?1)j l ? m ? 1
B l (t)(t + a)m?l ? dt,
+ (?1)l+1
l
?
j
j
0
j=0
where (5.19) and (5.20) correspond to (3.7) and (3.8), respectively.
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Exercise 5.2 Show that, in view of Lerch?s formula (5.4), (5.19) with
m = 0 gives Euler?s product formula (or Weierstrass? canonical product of
genus 1) (5.21) for ?(a).
? Y
1 ?a a
1
1+
1+
,
=a
?(a)
n
n
n=1
Solution
(5.21)
Indeed, (5.19) with m = 0 reads
N
X
?? 0 (0, a) = lim
N ??
n=0
log(n + a) ?
1
log(N + a)
2
(5.22)
!
? (N + a) log(N + a) + N + a .
For a = 1, (5.22) with N + 1 replaced by N gives
0
?? (0) = lim
N ??
N
X
1
log n ? N +
2
n=1
log N + N
!
.
(5.23)
Substituting
1
1
log(N + a) = log N + o(1),
2
2
(N + a) log(N + a) = N log N + a log N + a + o(1),
we transform (5.22), under (5.4), into
?(a)
? log ?
2?
!
N
X
1
= lim
log N + N ? a log N .
log(n + a) ? N +
N ??
2
n=0
(5.24)
Subtracting (5.23) from (5.24) yields, on using (2.20),
? log ?(a) = lim
N ??
N
X
n+a
? a log N
log a +
log
n
n=1
!
.
(5.25)
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The theory of the gamma and related functions via zeta-functions
PN ?1
By expressing log N as
n=1
log n+1
n , we may write
n + a n + 1 ?a
log
? log ?(a) = lim log a +
и
N ??
n
n
n=1
?
X
a
1 ?a 1+
,
= log a +
log
1+
n
n
n=1
N
X
!
(5.26)
whence (5.21).
Remark 5.2 Our procedure is a reverse to that of Berndt [Ber2] in which
he starts from one of the equivalent definitions of the gamma function given
by
N ! (N + 1)a
,
N ?? a(a + 1) и и и (a + N )
(5.27)
?(a) = lim
Euler?s interpolation formula or
log ?(a) = lim
N ??
?
N
X
log(n + a) +
n=0
N
X
log n + a log(N + 1)
n=1
!
(5.27)0
and deduces Lerch?s formula by comparing (5.27)0 with (5.22) (and (5.23)).
Of course, we can cover (5.27)0 in the same way as above. Indeed, from
(5.23) and (5.24), we deduce that
0
0
N
X
?? (0, a) + ? (0) = lim
N ??
n=0
log(n + a) ?
N
X
n=1
log n ? a log N
!
,
which?reduces to (5.27)0 by Lerch?s formula, save for the value ? 0 (0) =
? log 2?; this value is found in Exercise 2.6 with the Stirling?s formula
being taken for granted.
We now turn to recover Deninger?s Theorem 2.3 [D], especially, the
Gaussian representation
?? 00 (0, a) = ?? 00 (0) ? log2 a
+ lim
N ??
2
a log N ?
N
X
n=1
2
2
log (n + a) ? log n
!
.
(5.28)
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Indeed, choosing u = 0 and x = N ? N in Corollary 3.2, we get
N
X
log2 (n + a) =
n=0
Z ?
1
B 1 (t)
log2 (N + a) ?
log2 (t + a) dt
(5.29)
2
t
+
a
N
+ (N + a) log2 (N + a) ? 2 log(N + a) + 2 + ? 00 (0, a).
Put a = 1 and write N for N + 1 in (5.29) to get
N
X
Z
?
B 1 (t)
log2 t dt
t
(5.30)
N
n=1
2
00
+ N log N ? 2 log N + 2 + ? (0).
Noting that log(N + a) = log N + Na + O N12 , we have log2 (N + a) =
1
log2 N + 2a
N log N + O N 2 . Hence
log2 n =
1
log2 N ?
2
(N + a){log2 (N + a) ? 2 log(N + a) + 2}
= N log2 N ? 2N log N + 2N + a log2 N + O
1
N
.
Hence (5.29) may be written as
N
X
1
log2 N ?
2
log2 (n + a) =
n=1
Z
?
N
+N log2 N ? 2N log N + 2N
+ a log2 N ? log2 a + ? 00 (0, a) + O
1
N
(5.29)0
.
From (5.29)0 and (5.30) it follows that
?? 00 (0, a) = ?? 00 (0) + a log2 N ? log2 a ?
?
Z
?
B1 (t)
N
2
N
X
n=1
2
log (t + a) log t
?
t+a
t
log2 (n + a) ? log2 n
dt + O
1
N
,
or
?? 00 (0, a) = ? ? 00 (0) ? log2 a + a log2 N
?
N
X
n=1
log2 N
log (n + a) ? log n + O
N
2
2
upon estimating the integral. Now (5.31) implies (5.28).
(5.31)
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We may also recover the Weierstrass representation ([D, (2.3.2)]) by
Corollary 3.1 with u = ?1, a = 1 and x = N ? N (we write N for N + 1):
or
Z ?
N
X
1 ?1
1
log n
B 1 (t)
= N log N ?
log t dt + log2 N + ?1
2
n
2
t
2
N
n=1
?1 =
N
X
log n 1
log N
.
? log2 N + O
n
2
N
n=1
(5.32)
(cf. [KKSY, (8)]).
Solving (5.32) for log2 N and substituting it in (5.31), we deduce that
?? 00 (0, a) = ? ? 00 (0) ? 2 ?1 log a ? log2 a
(5.33)
N
X
log n
log2 N
+O
.
log2 (n + a) ? log2 n ? 2
?
n
N
n=1
5.2
Asymptotic formulas for the Hurwitz and related zetafunctions in the second variable
0
In this section we shall show that our formula (3.8) below coincides with
Katsurada?s formula (2.2) (Theorem 1 of [Kat1]) in the special case when
? = 1. Since our formula (3.44) with confluent hypergeometric function
coefficients readily extends to a general l, it suffices to show that the main
terms coincide with each other.
We suppose that u 6= ?1
Formula
(3.8) with ? + z in place
1and apply
u
u+1
1
of a, and the relations r?1 r = u+1 r and
(
Bk ,
k 6= 1
k
(?1) Bk = Bk (1) =
B1 + 1, k = 1,
(cf. (1.16)) in order to obtain
l 1 X u+1
?(?u, ?+z) = ?
(?+z)u?r+1 Br +O |z|Re (u)?l . (3.8)0
u + 1 r=0
r
Supposing further that |?| < |z|, we infer, by the binomial expansion,
0
that the right-hand side of (3.8) can be written as
S + O |z|Re (u)?l ,
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where
S := ?
l
l 1 XX u+1 u+1?r
Br ?k?r z u?k+1 ,
u + 1 r=0
r
k?r
k=r
since, for k ? l ? r, we have Re (u) ? r ? k ? Re (u) ? l.
Using (1.14), we obtain
l k 1 X u + 1 u?k+1 u?k+1 X k
z
z
S=?
Br ?k?r ,
u+1
k
r
r=0
k=0
whose innermost sum is precisely Bk (?).
Hence we conclude that
S=?
l?1
X
(?1)r+1
1
z u+1 +
(?u)r Br+1 (?) z u?r .
u+1
(r
+
1)!
r=0
(5.34)
Substituting (5.34) into (3.8)0 , we obtain the special case of Katsurada?s
result [Kat1, p.168, Theorem 1] when ? = 1.
Theorem 5.2
For any integer l ? 0 and any z in | arg (z) | < ?,
l?1
?(s, ? + z) =
X (?1)r+1
1
z 1?s +
Br+1 (?) (s)r z ?s?r
s?1
(r
+
1)!
r=0
?Re (s)?l
+ O |z|
.
(5.35)
Remark 5.3 Formula (5.35), in conjunction with a generalization of Formula (3.64) will yield the aforementioned special case of Katsurada?s main
result [Kat1, p.168, Formula (2.2)].
The method of proof of Theorem 5.2 readily extends to ?? 0 (s, ?) and
? (s, ?) contained in Corollary 3.1 and Corollary 3.2, respectively. Thus
we have the following consequences of Theorem 3.1 corresponding to Katsurada?s Corollary 1 and Corollary 2 in [Kat1].
We restate these results (see Corollary 3.3 and Corollary 3.2) in terms of
log ?(z) and R(z) in view of Lerch?s formula (5.4) and Deninger?s definition:
00
R(z) = ? ? 00 (0, z).
Corollary 5.1
(5.36)
For any integer l and any fixed ? > 0, the generalized
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Stirling?s formula:
log
?(z + ?)
?
2?
1
= z+??
log z ? z
2
+
l?1
X
(?1)r+1
r=1
r(r + 1)
Br+1 (?) z ?r + O |z|?l
holds true for z ? ? and | arg (z) | ? ? ? ? (? > 0).
For any integer l ? 1 and fixed ? > 0,
Corollary 5.2
1
R(z + ?) = z + ? ?
2
+
l?1
X
(?1)r+1
r=1
r(r + 1)
+ O |z|
?l
2
(log z) ? 2z log z + 2z
Br+1 (?) z
log(|z| + 1)
r?1
X
1
log z ?
h
?r
h=1
!
(5.37)
valid for z ?? ? and |arg (z)| ? ? ? ? (? > 0)
5.3
An application of the Euler digamma function
In this section what we are going to mainly use is the case u = ?1 of
Theorem 3.1, which we restate as the following:
For x ? 0, a ? C, a 6=non-negative integer, we have
Theorem 5.3
L?1 (x, a) = log(x + a) ? ?(a)
?
l
X
1
r=1
r
B r (x) (x + a)?r +
Z
?
B l (t) (t + a)?1?l dt.
(5.38)
x
Corollary 5.3 (i) (3.14) is a special case of (5.38) with x = 0.
(ii) (5.18) is a special case of (5.38) as x ? ?.
(iii) ?(a) admits the Gaussian representation
N
?(a) = lim
N ??
1 X
a
? ?
?z +
N
a
k=1
1
1
?
k+a k
!
.
(5.39)
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Proof. Only (5.39) needs a proof, which goes on the similar lines as those
for the case u = ?1 of Theorem 3.1. Formula (5.38) gives
X
1
1
?(a) = log(x + a) ?
,
+O
n+a
x
0?n?x
which we slightly rewrite as
N
N +a X
?
?(a) = log
N
n=1
1
1
?
n+a n
N
X
1
1
+ log N ?
.
+O
n
N
n=1
Recalling (5.16) (which fact is also contained in Theorem 5.3), we deduce
(5.39), on taking the limit as N ? ?.
Now we shall see what formula (5.39) means in the light of the
Dufresnoy-Pisot type uniqueness theorem ( cf. [D]).
Lemma 5.2
satisfies
If the function g : R+ ? R (R+ meaning positive reals)
lim (g(x + n) ? g(n)) = 0,
n??
0 < x ? 1,
(5.40)
then for any ? ? R there is at most one function f : R+ ? R with the
following properties:
(a) f (1) = ?
(b) f is convex on some interval (A, ?), A > 0
(c) f is a solution of the difference equation (DE)
f (x + 1) ? f (x) = g(x),
x ? R+ .
If such a function exists, it is given by the Gaussian representation
(cf. (5.17))
!
n?1
X
f (x) = lim ? + x g(n) ? g(x) ?
(g(x + k) ? g(k)) .
(5.41)
n??
k=1
Theorem 5.4 (i) The digamma function ?(a) defined by (5.18) is a
unique solution (convex for large argument) of the DE
f (x + 1) ? f (x) =
1
,
x
x ? R+ .
(5.42)
(ii) (5.39) is exactly (5.41), furnished by the Dufresnoy-Pisot type theorem, which already entails Assertion (i).
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Proof. For curiosity, we prove (i) without assuming (5.39). With g(x) =
1
x , (5.40) is satisfied. (a) follows from the definition and (c) follows from
(5.18); only (b) remains.
We differentiate (5.2) to obtain
Z ?
1
1
B 1 (t) (t + a)?4 dt.
(5.43)
? 00 (a) = ? 3 ? 2 + 6
a
a
0
To express the last integral inR closed form is an easy exercise. Indeed,
n+1
it is the sum of integrals of type n . Since
1
B 1 (t)(t + a) dt =
t+a? a+n+
(t + a)?4 dt
2
n
n
Z n+1
Z n+1
1
=
(t + a)?3 dt ? a + n +
(t + a)?4 dt
2
n
n
1
= ? (n + 1 + a)?2 ? (n + a)?2
2
1
1
n+a+
(n + 1 + a)?3 ? (n + a)?3
+
3
2
1
1
1
= ? (n + 1 + a)?2 ? (n + a)?2 +
(n + a + 1)?2 ? (n + 1 + a)?3
2
3
2
1
1
(n + a)?2 + (n + a)?3
?
3
2
1
1
= ? (n + 1 + a)?2 ? (n + a)?2 ?
(n + a + 1)?3 + (n + a)?3 ,
6
6
Z
n+1
Z
?4
n+1
summing these for n = 0, 1, 2, 3, 4, и и и , we obtain
Z
?
B 1 (t) (t + a)?4 dt =
0
?
1 ?2 1 ?3 1 X
1
a ? a ?
.
6
6
6 n=0 (n + 1 + a)3
Hence
? 00 (a) = ?
2
? ?(3, a + 1) < 0,
a3
and (b) follows, whence uniqueness follows from Lemma 5.2.
Corollary 5.4
(5.44)
(i) For |z| < 1 we have
?
X
k=2
?(k, a) z k?1 = ??(a ? z) + ?(a).
(5.45)
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?(z)
is the unique solution (convex for large argument) of the DE
(ii) log ?
2?
f (x + 1) ? f (x) = log x.
(5.46)
Proof. (i) is well-known and best viewed as the Taylor expansion of
?(s, a) in the second variable a (cf. [Klu], [KKaY], [SC]) and the proof is
immediate as follows.
The left side of (5.45) is
k
? ? 1 XX
z
1
=
(n + a)k
z n=0
n+a
n=0
k=2
k=2
2
z
?
?
X
1 X n+a
1
=
=
z
z n=0 1 ? n+a
(n
+
a
?
z)(n + a)
n=0
?
?
X
1
1
?,
= lim ?
? log(x + a ? z) + log(x + a) ?
x??
n+a?z
n+a
?
X
z k?1
?
X
0?n?x
which is the right side of (5.45), in view of (5.18).
Assertion (ii) is known as the Bohr-Mollerup theorem, and is a consequence of Lemma 5.2. We could cover (ii) also by our Theorem 5.3, (ii)
if only we assume we know the value (5.12) ? 0 (0) = log ?12? . We may also
regard (ii) as Lerch?s formula (5.4) ([Ber2]).
Corollary 5.5
We have the duplication formula
1
1
1
+ log 2,
?(2z) = ?(z) + ? z +
2
2
2
(5.47)
and, a fortiori,
?(2z) = 2
2z?1 ? 21
?
1
?(z) ? z +
2
.
(5.48)
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Indeed, using (5.18) in the form
1
1
1
?(z) + ? z +
2
2
2
1
1
= lim
log(x + z) x + z +
x?? 2
2
Proof.
X
?
X
1
?
2n + 1 + 2z
0?2n+1?2x+1
0?2n?2x
?
?
2
X
1
1
1 ?
= lim ? log 4 x + z +
? log 2 ?
,
x??
2
2
n + 2z
?
1
?
2n + 2z
0?n?2x+1
which is ?(z) ? log 2. (5.48) follow from (5.47) if we use ?
1
2
=
?
?.
Remark 5.4 The property in Corollary 5.3 is a special case of the Kubert identity (or distribution property) shared by a wide class of functions
(cf. (8.13), [Mi], [Su1]).
5.4
The first circle
Proposition 5.1
The product representation for the gamma function
?(z + 1) = e??z
z ?1
z
en 1 +
n
n=1
?
Y
(5.49)
is a consequence of (5.45).
Exercise 5.3
Solution
Deduce (5.49) from (5.45).
Integrating (5.45) from 0 to z, we obtain
?
X
?(k, a) k
z = log ?(a ? z) ? log ?(a) + ?(a) z,
k
(5.50)
k=2
which is also a well-known formula (cf. e.g. [SC]). We need only the special
case of (5.50) with a = 1, z replaced by ?z:
?
X
(?z)k
k=2
k
?(k) = log ?(z + 1) + ? z.
(5.51)
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We now apply a procedure similar to that of the proof of Corollary 5.4. We
see that the left side of (5.51) becomes
? X
?
X
1 z k
?
,
k
n
n=1
k=2
the inner sum of which can be summed by the elementary formula (|r| < 1)
?
X
1 k
r = ?r ? log(1 ? r).
k
k=2
Hence
?
X
(?z)k
k=2
k
?(k) =
z .
? log 1 +
n
n
? X
z
n=1
(5.52)
Combining (5.50) and (5.51) completes the proof of (5.49).
Proposition 5.2
The reciprocal relation
?(s) ?(1 ? s) =
?
,
sin ?s
(5.53)
for the gamma function is a consequence of the asymmetric form of the
functional equation
?s ?(1 ? s) = 21?s ? ?s ?(s) cos
?(s),
(5.54)
2
for the Riemann zeta-function.
Proof.
Changing s by 1 ? s in (5.54), we deduce its counterpart
?s ?(s) = 2s ? s?1 ?(1 ? s) sin
?(1 ? s).
(5.55)
2
Multiplying (5.54) and (5.55) and canceling the common factor
?(s) ?(1 ? s), we arrive at (5.53).
Remark 5.5 The proof of Proposition 5.2 is modeled on Eisenstein?s
1849 proof (cf. [We]) of the functional equation for the Hurwitz-Lerch zetafunction. A standard proof (cf. e.g. [Leb] and Exercise
R ? 1?z2.3) is via the beta
function. Use is made of the integral formula 0 x1+x dx = sin??z , 0 <
Re z < 1.
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Lemma 5.3 The asymmetric form (5.54) of the functional equation for
the Riemann zeta-function is a consequence of the functional equation for
the Hurwitz zeta-function (or the Hurwitz formula) (0 < a < 1)
?
?
?(s, a) = ?i (2?)s?1 ?(1 ? s) e 2 is l1?s (a) ? e? 2 is l1?s (1 ? a) , (5.56)
?
X
e2?ina
stands for the polylogarithm function (3.3), Forns
n=1
mula (5.56) (which already appeared as (3.48) and (3.67)) in the long run,
is a consequence of (5.1).
where ls (a) =
A recent proof of (5.56) based on the Fourier expansion of the Dirac
delta function can be found in [BKT] or [KTTY3] and is sketched in Д3.5.
A more laborious but easier proof can be found in [R] (for the Riemann
zeta) and [PP] (for the general case). It amounts to completing the integral
in
1
1
1
a1?s + a?s +
s a?s?1
?(s, a) =
s?1
2
12
Z
(5.57)
s(s + 1) ?
?
B 2 (t) (t + a)?s?2 dt, ? > ?2,
2
0
R?
in the from ?a B 2 (t)(t + a)?s?2 dt, then using the absolutely converging
Fourier series for B 2 (t) and appealing to a formula for the Mellin transform.
We refer to the above references.
Proposition 5.3
The product representation for the sine function
? Y
sin ?z
z2
(5.58)
=
1? 2 ,
?z
n
n=1
is a consequence of (5.49) and (5.53).
Proof.
Writing ?z for z in (5.49), we get
z
z ?1
.
e? n 1 ?
n
n=1
(5.59)
Multiplying (5.49) and (5.59), we deduce that
?1
? Y
z2
z ?(z) ?(1 ? z) =
1? 2
,
n
n=1
(5.60)
?(1 ? z) = e?z
?
Y
where we used the formula ?(z + 1) = z?(z), which is also a consequence
of (5.49) and (5.16). Plugging (5.53) in (5.60) gives (5.58).
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Proposition 5.4
tion
The partial fraction expansion for the cotangent func?
2z X
1
1
?
?z
? n=1 n2 ? z 2
? 1
1X
1
1
=
,
+
?
?z
? n=1 n + z
n?z
cot ?z =
(5.61)
is a consequence of (5.58).
Proof.
This follows immediately by logarithmic differentiation.
Remark 5.6
again.
Comparing (5.17) and (5.61), we cover formula (2.55)
Proposition 5.5 The partial fraction expansion for the hyperbolic cotangent function and (5.61) are equivalent:
?
1
1
1
xX
1
1
coth ?x = 2?x
+ =
+
,
2
e
?1 2
2?x ? n=1 n2 + x2
Re x ? 0.
(5.62)
Proof. This follows by putting ix = z in (5.61) (i.e., we move from the
right half-plane into the upper half-plane).
Lemma 5.4 The partial fraction expansion for coth x and the (symmetric
form) functional equation
s
1?s
? s2
? 1?s
2
? ?
?
?(s) = ?
?(1 ? s),
(5.63)
2
2
are equivalent.
Proof.
This can be found in [KTTY4], [Ko].
Remark 5.7 Historically, (5.62) was first used to deduce (5.63) (the fifth
proof of [Tit], where an appeal to a formula (cf. Corollary A.4) for the
Mellin transform is needed). Then Koshlyakov [Ko] deduced (5.62) from
(5.63).
Supplementarily, we state a result which allows us to skip above propositions and deduce (5.61) directly from (5.56) or rather its equivalent under
(5.53):
o
?(1 + s) n ?is
?is
2 ?(1 + s, x) ? e? 2 ?(1 + s, 1 ? x)
. (5.56)0
l?s (x) = i
e
(2?)1+s
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Proposition 5.6 The functional equation (5.56) for the Hurwitz zetafunction implies the partial fraction expansion (5.61) for the cot-function.
Proof. We remark that the functional equation (5.56) for ?(s, x) may be
expressed on the basis of (5.53) as (5.56)0 .
First we assume that Im x > 0. Then the sum for l0 (x) converges for
every s ? C, and the left-hand side is
l0 (x) =
?
X
e2?inx =
n=1
e2?ix
1
= (?1 + i cot ?x) .
2?ix
1?e
2
(5.64)
By analytic continuation, this holds true for every x ? R ? Z.
We consider the limit as s ? 0, s > 0 on the right hand side of (5.56)0 .
First we note that
o
n ?is
?is
e 2 ?(1 + s, x) ? e? 2 ?(1 + s, 1 ? x) ? {?(1 + s, x) ? ?(1 + s, 1 ? x)}
?is
?is
= e 2 ? 1 ?(1 + s, x) ? e? 2 ? 1 ?(1 + s, 1 ? x)
o
?is
?s n ?is
e 4 ?(1 + s, x) + e? 4 ?(1 + s, 1 ? x)
= 2i sin
4
o
n ?is
sin ?s
?is
4 ?
= 2i ?
e 4 s ?(1 + s, x) + e? 4 s ?(1 + s, 1 ? x) ,
4
4s
which tends to ?i as s ? 0 on account of lims?0 s ?(1 + s, x) = 1.
Secondly, since
? X
1
1
?(1 + s, x) ? ?(1 + s, 1 ? x) =
?
,
(n + x)1+s
(n + 1 ? x)1+s
n=0
?>0
we get
lim
(?(1 + s, x) ? ?(1 + s, 1 ? x))
?
1
1 X 2x
1
?
= +
.
n+x n+1?x
x n=1 x2 ? n2
s?0,s>0
? X
=
n=0
Hence the limit of the right-hand side of (5.56)0 as s ? 0 through positive
values is
i
?
1 X 2x
+
.
x n=1 x2 ? n2
Combining this with (5.64), we conclude (5.61).
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Proposition 5.7 The functional equations in symmetric form (5.63) and
in asymmetric form (5.54) are equivalent under (5.48) and (5.53).
Proof is immediate.
Lemma 5.5 The functional equation (5.63) for the Riemann zetafunction and (5.56) for the Hurwitz zeta-function are equivalent.
Proof. This can be found in [KTTY4], [BKT] and is a manifestation of
the most far-reaching modular relation principle.
We are now in a position to state the main result of this chapter.
Theorem 5.5 Under some known formulas, all formulas (5.53), (5.54),
(5.56), (5.58), (5.61), (5.62) and (5.63) are equivalent in the sense of the following logical scheme (the portion including (5.65) is due to Theorem 5.6):
(5.17)
=?
(5.49)
(5.56) ? (5.54) ? (5.53)
?
?
?
(5.58) ? (5.61) ? (5.62) ? (5.63) ? (5.56)
m
m
(5.65) ? (5.56)
(5.54)
Lemma 5.6 (Berndt) The functional equation (5.56) for the Hurwitz
zeta function implies Kummer?s Fourier series for log ?(x), which reads
1
1
?(x)
= ? log(2 sin ?x) + (? + log 2?)(1 ? 2x)
log ?
2
2
2?
?
X
1
log n
+
sin 2?nx,
? n=1 n
(5.65)
which implies the reciprocal relation (5.49).
Proof is given by Berndt [Ber2], which depends on Lerch?s formula (5.4)
and the integral representation (5.1) for ?(s, z) with z = 0.
Theorem 5.6 Kummer?s Fourier series for log ?(x) is equivalent to the
functional equation (5.63) for the Riemann zeta-function.
Exercise 5.4
Deduce Euler?s identity
2 ?(2m)
B2m
= (?1)m?1
,
(2m)!
(2?)2m
m?1
(5.66)
from the partial fraction expansion (5.62) for the hyperbolic cotangent function.
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Solution
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Rewriting (5.66) in the form
?
X
1
2?x
2
+
?x
=
1
+
2x
2 + x2
e2?x ? 1
n
n=1
and putting 2?x = z, we obtain
?
X
z
1
1
2
=
1
?
z
+
2z
z
2
e ?1
2
z + (2?n)2
n=1
1
= 1 ? z + 2z 2 ? z 2
2
say, where
?(w) =
?
X
w + 4? 2 n2
n=1
?1
(5.67)
.
(5.68)
Since
?(r) (w) = (?1)r r!
?
X
r=0
w + 4? 2 n2
?r?1
,
we see that
?
X
?(r) (0)
(?1)r
1
= (?1)r
=
?(2r + 2).
2r+2
r!
(2?n)
(2?)2r+2
n=1
Hence
?
X
1
(?1)r
z
=
1
?
z
+
2
?(2r + 2) z 2r+2 ,
2r+2
ez ? 1
2
(2?)
r=0
and so
?
X
z
1
2 (?1)m?1
=
1
?
z
+
?(2m) z 2m .
2m
ez ? 1
2
(2?)
m=1
Recalling the expansion
?
X
1
B2m 2m
z
=
1
?
z
+
z ,
z
e ?1
2
(2m)!
m=1
we conclude (5.66).
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Remark 5.8 As is proved above, (5.62) and the functional equation for
the Riemann zeta-function are equivalent, whence we see that Euler?s iden2
tity (5.66), and in particular the solution to the Basel problem ?(2) = ?6 ,
is a consequence of the functional equation.
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Chapter 6
The theory of Bessel functions and
the Epstein zeta-functions
Abstract
In this chapter we study an energy invariant ? the
Madelung constant associated to a crystal lattice through
the lattice zeta-function, which is manifested as the Epstein zeta-function. We take into account the lattice structure (crystal symmetry) in our study through the functional equation of the Epstein zeta-function (zeta symmetry).
6.1
Introduction and the theory of Bessel functions
In this chapter we are going to study an energy invariant associated to a
crystal lattice, called the Madelung constant about which numerous papers have appeared so far. Main references in book form are [Bor], [GZ]
and [Ter1]. The main feature of our treatment in this regard is that we incorporate the lattice structure in its full extent, especially, the relationships
between mutually dual lattice structures are revealed as those between the
associated lattice zeta-functions, which are in turn manifested as the Epstein zeta-functions. That is, unlike previous work (save for Terras), we
are going to express the distance and ion charges of the crystal structure
in the form of a quadratic form and construct the Epstein zeta-functions
associated to it, and then apply decomposition of the coefficient matrices
to the Epstein zeta-function as is seen in Terras [Ter1]. For more details
we refer to [KTTY2].
As the second main feature, we shall present a rather complete version
of the theory of Epstein zeta-functions, which include generalizations of the
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theory of Berndt [Ber6], Chowla-Selberg and Terras as well as a unification
of the theory of lattice zeta-values developed so far. They are manifested
as the special values like ?(1/2), ?(1/2), where ?(s) = L(s, ?4 ) referred to
in Abstract of Chapter 4, and we may efficiently incorporate our recent
results on special values (see [KTY7]), using a perturbed Dirichlet series,
or the Mellin-Barnes integrals (6.70) (cf. Paris-Kaminski [PK]).
Definition 6.1 The n-th Bessel function Jn (z)
is defined as the n-th
1
z
Laurent coefficient of the function exp 2 (w ? w ) in w, viz.
exp
Proposition 6.1
tegral)
?
X
1
z
w?
=
Jn (z) wn .
2
w
n=??
We have the integral representation (called Bessel?s in-
Jn (z) =
Proof.
(6.1)
1
?
Z
?
0
cos(z sin ? ? n?) d?.
(6.2)
By Theorem A.11, we have the integral representation
Z
exp z2 (w ? w1 )
1
dw.
Jn (z) =
2?i |w|=1
wn+1
By the parametric expression for the curve |w| = 1: w = ei? , 0 ? ? ? 2?,
we may rewrite the above as
Z 2?
1
ei(z sin ??n?) d?
(6.3)
Jn (z) =
2? 0
on noting that w ? w1 = 2i sin ?. Dividing the interval [0, 2?] into [0, ?] and
[?, 2?] and making the change of variable in the integral over [?, 2?], we
obtain
Z ?
1
Jn (z) =
e?i(z sin ??n?) d?.
2? 0
Adding this to the integral over [0, ?], we conclude (6.2).
Proposition 6.2
The ?-th Bessel function J? (z) may be defined by
J? (z) =
z ?+2n
(?1)n
.
n! ?(? + n + 1) 2
n=0
?
X
(6.4)
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Proof. For ? ? Z, J? (z) is also defined as the ?-th coefficient of the
z
z 1
product of two power series for e 2 w and e? 2 w . Hence the ?-th term is
given by
?
z 2
X z ? 1 (?1)n z ? X
(?1)n
,
=
2
m! n!
2 n=0 ?(? + n + 1) n! 2
m?n=?
as claimed. For other values of ?, we understand (6.4) as the definition. We note that from (6.2) it results
J?n (z) = (?1)n Jn (z)
Exercise 6.1
Solution
(6.5)
Viewing (6.1) as a Fourier series, deduce (6.2).
With w = ei? (? ? R), (6.1) reads
eiz sin ? =
?
X
Jn (z) ein? ,
n=??
which is a Fourier series converging to the left-hand side member (in view of
Theorem 7.2) and the Fourier coefficient Jn (z) may be computed by (7.2):
Z ?
1
ei(z sin ??n?) d?,
Jn (z) =
2? ??
which is (6.3), and we may argue as in the proof of Proposition 6.1.
Exercise 6.2
Use (6.5) to deduce for n even
Z
1 ?
cos(z sin ?) cos(n?) d? = Jn (z)
? 0
1
?
Z
1
?
Z
(6.6)
?
sin(z sin ?) sin(n?) d? = 0
0
while for n odd,
1
?
Z
?
cos(z sin ?) cos(n?) d? = 0
0
?
sin(z sin ?) sin(n?) d? = Jn (z)
0
(6.7)
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Solution Jn (z) + J?n (z) is 2Jn (z) for n even and 0 for n odd, and by
(6.2), this is
Z
1 ?
(cos(z sin ? ? n?) + cos(z sin ? + n?)) d?,
? 0
whence the first and the second identities follow.
Considering Jn (z) ? J?n (z), we deduce the third and the fourth identities.
In case ? ?
/ Z, the Bessel functions J? (x) and J?? (x) are two independent solutions to the Bessel differential equation
d2 y
1 dy
n2
+
+ 1 ? 2 y = 0.
(6.8)
dx2
x dx
x
For n ? Z, the fundamental solutions to (6.8) are given by Jn (x) and
Yn (x), the Weber function, relevant to analytic number theory. J? (z) and
Y? (z) are often referred to as the Bessel function of the first kind and
of the second kind, respectively.
Equally relevant to number theory are modified Bessel functions. The
modified Bessel function I? (z) of the first kind is defined by
I? (z) =
z ?+2n
1
,
n! ?(? + n + 1) 2
n=0
?
X
(6.9)
whence
In (z) = i?n Jn (iz)
for n ? Z.
The modified Bessel function K? (z) of the second kind is defined
by
K? (z) =
? I?? (z) ? I? (z)
2
sin ??
(6.10)
(the limit is to be taken for ? ? Z), which has the integral representation
Z
?
1 ? ? 12 z(t+ 1t ) ??1
1
K? (z) =
(6.11)
e
t
dt, Re ? > ? , | arg z| < .
2 0
2
4
This appears in the proof of Theorem 6.1 in the context of Mellin inversion
(sometimes referred to as the inverse Heaviside integral)
Z ? х
1
х+?
х??
? s+
? s+
x?s ds = 2 x 2 K? 2 x , (6.12)
2?i (c)
2
2
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for c + Re х+?
2 ? Re ? > 0.
Bessel functions are, in a sense, generalizations of exponential functions,
and they reduce to these functions for half-integral order, e.g.
r
r
2
2
J 21 (z) =
sin(z), J? 12 (z) =
cos(z)
(6.13)
?z
?z
and
K 21 (z) = K? 21 (z) =
6.2
r
? ?z
e .
2z
(6.14)
The theory of Epstein zeta-functions
We now introduce the notation (from Terras [Ter1]) concerning the Epstein
zeta-functions, which will be used throughout in what follows.
Notation. Let g, h ? Rn be n-dimensional real vectors which (in the
first place) give rise to the perturbation and the (additive) characters, respectively.
Let Y = (yij ) be a positive definite n О n real symmetric matrix. Define
the Epstein zeta-function associated to the quadratic form
Y [a] = a и Y a = t aY a =
n
X
yij ai aj ,
(6.15)
i,j=1
where a = (a1 , . . . , an ) ? Rn and ?и? means the scalar product, by
Z(Y, 0, 0, s) =
X
a?Zn
a6=0
1
,
Y [a]s
?>
n
,
2
(6.16)
where ? = Re s.
For g, h ? Rn define the general Epstein zeta-function (of HurwitzLerch type) by
Z(Y, g, h, s) =
X
a?Zn
a+g6=0
e2?ihиa
,
Y [a + g]s
?>
n
,
2
(6.17)
and incorporate the completion
?(Y, g, h, s) = ? ?s ?(s) Z(Y, g, h, s),
(6.18)
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which satisfies the functional equation of the form (5.63) with an additional
factor and replacement of parameters (proof given in Д6.4):
n
1
e?2?igиh ? Y ?1 , h, ?g, ? s .
?(Y, g, h, s) = p
(6.19)
2
|Y |
In what follows we always denote the special vector t 12 , 21 , 21 by c0 :
?1?
2
? ?
c0 = ? 12 ? .
1
2
We shall now give some illustrative examples.
Example 6.1 The relationship between the Madelung constants of the
NaCl and CsCl structure.
In [Hautot, p.1724], it is stated that the cations of CsCl are at a ?
3
3
1
2
?
?2 Z
. The Madelung constant
Z
+
and
anions
are
at
a
?
2
3
3
MCsCl is defined, in the first place, by
?
?
3 X
3 X
?1
MCsCl =
|a + c0 | ?
|a|?1 ,
(6.20)
2
2
3
3
a?Z
a?Z
a6=0
which, in our notation above, is
?
?
1
1
3
3
Z I, c0 , 0,
Z I, 0, 0,
?
2
2
2
2
(6.21)
and is in turn equal to
?
1
? 3 Z B, 0, c0 ,
2
,
(6.22)
where
?
and
10
I = ?0 1
00
?
0
0?
1
?
(identity matrix),
?
3 ?1 ?1
B = ??1 3 ?1? .
?1 ?1 3
(6.23)
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111
Hautot [Hautot], without giving any reasons, transforms (6.20) into the
form
2
? MCsCl = 2MN aCl
3
?1/2
X X X
2
2
2
(2l) + (2m + 1) + (2n + 1)
+6
(6.24)
?1/2 ? (2l)2 + (2m + 1)2 + (2n)2
,
and then proceeds to transfer the triple sum using the Schlo?milch series
technique (cf. [KTZ] also). Thus (6.24) suggests that there may be a relationship between MN aCl and MCsCl structure. This suggestion is strengthened by the comparison of numerical values
MN aCl = 1.74756459463 . . .,
MCsCl = 1.76267477307 . . ..
(6.25)
The real situation is the following duality relations (6.26) and (6.27),
which can be found only through the study of lattice structures.
Between
the Madelung constants MN aCl = ?Z(I, 0, c0 , 21 ) and MCsCl
?
= ? 3 Z(B, 0, c0 , 12 ), the duality relations hold (under the notation (6.23)):
2
1
MN aCl = ?Z I, 0, c0 ,
= ? {Z(B, 0, 0, 1) ? Z(B, 0, c0 , 1)} (6.26)
2
?
and
?
?
1
3
=?
MCsCl = ? 3 Z B, 0, c0 ,
{Z(I, 0, 0, 1) ? Z(I, 0, c0 , 1)}
2
2?
(6.27)
(cf. Formula (1.8) on p.721 of [KTTY1]; proof given in Example 6.2 below).
In the case of MZnS , Hautot states another relation corresponding to
(6.24), again without giving any reason why the Madelung constants MZnS
and MCsCl should be related:
XXX
1
2
4
? MZnS = ? MCsCl ? 6
p
,
2
3
3
(2l) + (2m + 1)2 + (2n + 1)2
(6.28)
where we note that the Madelung constant MZnS is to be defined by
?
?
1
1
1
3
3
Z A, c0 , 0,
Z A, 0, 0,
MZnS =
?
,
(6.29)
2
2
2
2
2
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where
?
21
A = ?1 2
11
?
1
1? .
2
(6.30)
Comparing (6.27) and (6.29) and the numerical values (6.25) and (6.31)
below does not give much to expect a relation between them;
MZnS = 1.63805505338 . . ..
Surprisingly enough, there holds a remarkable relation
?
1
3
MN aCl + MCsCl .
MZnS =
4
2
(6.31)
(6.32)
For a proof see [KTTY2].
In Example 6.2 we are going to reveal those identities given in Example 6.1 as special cases of zeta-function relations.
We introduce the general principle.
Principle. Suppose L is a lattice with basis e1 , e2 , e3 :
L = Ze1 ? Ze2 ? Ze3 .
(6.33)
With M = (e1 , e2 , e3 ), the associated Gram matrix Y is defined by t M M :
?
?
e1 и e 1 e1 и e 2 e1 и e 3
Y = t M M = ?e2 и e1 e2 и e2 e2 и e3 ? .
(6.34)
e3 и e 1 e3 и e 2 e3 и e 3
Let f 1 = e2 + e3 , f 2 = e3 + e1 , f 3 = e1 + e2 , and
?
?
011
J = ?1 0 1? .
110
(6.35)
Then the matrix tJ Y J = J Y J is the Gram matrix associated to the
sublattice L1 = Zf 1 ? Zf 2 ? Zf 3 of L, and we have
1
1
1
(6.36)
L = L 1 ? L1 + f 1 + f 2 + f 3 .
2
2
2
We appeal to the fact to be proved in Д6.3 (Proposition 6.3) that the
zeta-function of a lattice coincides with the Epstein zeta-function of the
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corresponding Gram matrix under suitable identification. It follows that
?(Y, 0, c0 , s) = ? tJ Y J, 0, 0, s ? ? tJ Y J, c0 , 0, s ,
(6.37)
and
?(Y, 0, 0, s) = ? tJ Y J, 0, 0, s + ? tJ Y J, c0 , 0, s .
(6.37)0
Now note that the inverse matrix (tJ Y J)?1 is the Gram matrix associated to the dual lattice L01 (?
= Hom(L1 , Z)) or recall the functional equation
(6.19) to transform the right-hand side of (6.37) further into
1
3
3
p
? (tJ Y J)?1 , 0, 0, ?s ? ? (tJ Y J)?1 , 0, ?c0 , ?s
,
2
2
|tJ Y J|
so that
?(Y, 0, c0 , s)
1
=p
t
| J Y J|
3
3
t
?1
t
?1
? ( J Y J) , 0, 0, ?s ? ? ( J Y J) , 0, ?c0 , ?s
.
2
2
(6.38)
Now we apply the above principle to some lattice sums.
?
?
211
Example 6.2 First choose tJ Y J = A = ?1 2 1? ((6.30)). Then Y = I
112
and (6.37) reads
Z(I, 0, c0 , s) = Z(A, 0, 0, s) ? Z(A, c0 , 0, s).
(6.39)
This explains the reason why the proper definition (given in [KTTY2])
of the Madelung constant MN aCl as the value at s = 21 of
Z(A, c0 , 0, s) ? Z(A, 0, 0, s)
coincides with the value at s = 12 of ?Z(I, 0, c0 , s) i.e.
1
1
1
MN aCl = Z A, c0 , 0,
? Z A, 0, 0,
= ?Z I, 0, c0 ,
.
2
2
2
Next, we choose Y = 14 B ((6.23)). Then tJ Y J = I, and
1
B, 0, c0 , s = Z(I, 0, 0, s) ? Z(I, c0 , 0, s).
Z
4
(6.40)
(6.41)
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We make an important remark, which will be in effect in treating the
Abel mean in [KTTY2], that for c > 0
Z(c Y, g, h, s) = c?s Z(Y, g, h, s),
(6.42)
i.e. we may incorporate the parameter c in Y by just multiplying by the
factor c?s .
Using (6.42) and (6.41), we have for s = 21
? ?
1
1
3
1
Z I, c0 , 0,
? Z I, 0, 0,
= ? 3 Z B, 0, c0 ,
, (6.43)
2
2
2
2
which asserts that (6.21) and (6.22) are equal.
We turn to the proof of duality relations (6.26) and (6.27).
As we deduced (6.39), we choose tJ Y J = A, and so Y = I. Then (6.38)
gives
3
1
3
? A?1 , 0, ?c0 , ? s ? ? A?1 , 0, 0, ? s
??(I, 0, c0 , s) = p
.
2
2
|A|
Since A?1 = 41 B, we apply (6.42) to obtain
3
3
1 3 ?s
2
? B, 0, ?c0 , ? s ? ? B, 0, 0, ? s
,
??(I, 0, c0 , s) = 4
2
2
2
or
? ? ?s ?(s) Z(I, 0, c0 , s)
(6.44)
3
3
3
2?2s ?( 23 ?s)
=2
?
?
?s
Z B, 0, ?c0 , ? s ? Z B, 0, 0, ? s
,
2
2
2
which in turn gives (6.26) for s = 12 .
Also, for the choice Y = 14 B, tJ Y J = I, (6.38) reads, as in (6.43),
3
3
?s
? I, 0, 0, ? s ? ? I, 0, ?c0 , ? s
?(B, 0, c0 , s) = 4
,
2
2
or
? ?s ?(s) Z(B, 0, c0 , s)
(6.45)
3
3
3
3
= 4?s ? ?( 2 ?s) ?
?s
Z I, 0, 0, ? s ? Z I, 0, ?c0 , ? s
,
2
2
2
which gives (6.27) for s = 12 .
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Lattice zeta-functions
In this section we shall clarify the relationship between the zeta-functions
mentioned in the title and apply to the study of Madelung constants.
Let L be a lattice, i.e. a free Abelian group of finite rank (n, say) with
biadditive form ( , )L . We form the zeta-function ZL (s) = Z(L, 0, 0, s)
corresponding to (6.16) by
X
1
,
(6.46)
ZL (s) = Z(L, 0, 0, s) =
(x, x)sL
x?L
x6=0
absolutely convergent for ? > n2 .
If, in particular, L ? Rm and (a, b)L means the scalar product a и b =
Pm
t
ab = i=1 ai bi , then
X
1
.
(6.47)
Z(L, 0, 0, s) =
(x21 + и и и + x2m )s
x?L
x6=0
As usual, let L0 denote the dual lattice of L: L0 = Hom(L, Z). Then
for lattice elements p, q with real coefficients, p ? L ? R, q ? L0 ? R,
we introduce the general lattice zeta-function Z(L, p, q, s) corresponding to
(6.17) by
Z(L, p, q, s) =
X
x?L
x+p6=0
absolutely convergent for ? >
through
n
2.
e2?iq(x)
,
(x + p, x + p)sL?R
(6.48)
Here we understand the meaning of q(x)
L0 ? R ?
= Hom(L, R) ?
= HomR (L ? R, R)
and the completion corresponding to (6.18):
?(L, p, q, s) = ? ?s ?(s) Z(L, p, q, s).
(6.49)
We recall the Principle in Д1 in the following form.
Associated to a lattice L with basis e1 , . . . , en , L = Ze1 ? и и и ? Zen , is
its Gram matrix
?
?
(e1 и e1 )L и и и (e1 и en )L
?
?
..
..
..
(6.50)
Y = tM M = ?
?,
.
.
.
(en и e1 )L и и и (en и en )L
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where M = (e1 , . . . , en ).
Let ? be the canonical isomorphism
? : Zn ?? L, x = ?(a) = a1 e1 + и и и + an en ,
(6.51)
for a = (a1 , . . . , an ), or ?(a) = M a. Through ?, we may interpret the
bilinear form (x, x)L as (?(a), ?(a))L , which we may think of as Y [a].
Thus,
ZL (s) = Z(L, 0, 0, s) = Z(Y, 0, 0, s).
(6.52)
We may extend ? to the isomorphism
? : Rn ?? L ? R, x = ?(a) = a1 e1 + и и и + an en ,
for a = (a1 , . . . , an ) ? Rn . Then we have
(?(a), ?(a))L?R = Y [a].
If further we put p = ?(g) and q(x) = q ? ?(a) = h и a (a ? Rn ), then
Z(L, p, q, s) =
X
e2?iq??(a)
(?(a + g), ?(a + g))sL?R
X
e2?ihиa
= Z(Y, g, h, s).
Y [a + g]s
a?Zn
a+g6=0
=
a?Zn
a+g6=0
(6.53)
Thus, we have
Proposition 6.3
Under above notation, we have
?(L, p, q, s) = ?(Y, g, h, s).
Hence, whenever we speak about a lattice zeta-function, we may do well
with the corresponding Epstein zeta-function with the Gram matrix.
Example 6.3
(i) The simple cubic (s.c.) structure, Z3
? ?
1
= Z ?0? ?
0
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The theory of Bessel functions and the Epstein zeta-functions
?
?
? ?
? ?
100
0
0
Z ?1? ? Z ?0? with Gram matrix I = ?0 1 0?. The zeta-function is
001
1
0
X
Z Z3 , 0, 0, s = Z(I, 0, 0, s) =
3
a?Z
a6=0
Fig. 6.1
1
.
|a|2s
(6.54)
the simple cubic (s.c.) structure
(ii) The face-centered cubic structure (f.c.c.),
? ?
? ?
? ?
0
1
1
n
o
Lf = a ? Z3 (?1)a1 +a2 +a3 = 1 = Z ?1? ? Z ?0? ? Z ?1?
1
1
0
?
?
211
with Gram matrix A = ?1 2 1? ((6.30)).
112
The zeta-function is
ZLf (s) = Z(A, 0, 0, s)
X
1
.
=
2
2
2
(2a1 + 2a2 + 2a3 + 2a1 a2 + 2a2 a3 + 2a3 a1 )s
3
a?Z
a6=0
With c0 = t ( 21 , 12 , 12 ), Z(I, 0, c0 , s) =
as
P
a?Z3
a6=0
(?1)a1 +a2 +a3
I[a]s
Z(I, 0, c0 , s) = 2Z(A, 0, 0, s) ? Z(I, 0, 0, s).
can be written
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Solving in Z(A, 0, 0, s), we have
Z(A, 0, 0, s) =
Fig. 6.2
1
(Z(I, 0, 0, s) + Z(I, 0, c0 , s)) .
2
the face-centered cubic structure (f.c.c.)
(iii) The body-centered cubic structure (b.c.c.),
Lb = a ? Z3 |a2 + a3 , a3 + a1 , a1 + a2 ? 2Z
= a ? Z3 |(?1)a2 +a3 + (?1)a3 +a1 + (?1)a1 +a2 = 3
?
?
3 ?1 ?1
with Gram matrix B = ??1 3 ?1? ((6.23)). The zeta-function is
?1 ?1 3
ZLb (s) = Z(B, 0, 0, s)
X
1
=
2 + 3a2 + 3a2 ? 2a a ? 2a a ? 2a a )s .
(3a
1 2
2 3
3 1
1
2
3
3
a?Z
a6=0
(6.55)
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119
Since
? ?
? ?
? ?
?1?
1
1
0
2
2
2
? ?
? ?
? ?
3 Z I, 0,? 12 ?, s = Z I, 0,? 21 ?, s + Z I, 0,? 0 ?, s + Z I, 0,? 12 ?, s
1
1
0
0
2
2
X (?1)a2 +a3 + (?1)a3 +a1 + (?1)a1 +a2
,
=
I[a]s
3
a?Z
a6=0
we get, on resorting to the definition of Lb ,
? ?
1
X
X 3
?1
? 21 ?
+
3 Z I, 0,? 2 ?, s =
s
s
I[a]
I[a]
a?Lb
a?Z3 ?Lb
0
a6=0
= 4 Z(B, 0, 0, s) ? Z(I, 0, 0, s),
whence
Z(B, 0, 0, s) =
Fig. 6.3
?
?
1?
4?
?
Z(I, 0, 0, s) +
? ?
?
1
?
?
? 21 ?
.
3 Z I, 0,? 2 ?, s
?
0
?
the body-centered cubic structure (b.c.c.)
(6.56)
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Using ?square root? of A and B, i.e. J from (6.35) and
?
?
?1 1 1
K = ? 1 ?1 1 ?
1 1 ?1
(6.57)
(J 2 = A, K 2 = B, J ?1 = 21 K), we obtain generalizations of formulas (6.55)
and (6.56):
Z(J Y J, g, h, s)
?
?1? ?
?
2
?
?
?
1
?1?
?1
?1
=
Z(Y, Jg, J h, s) + Z Y, Jg, J h + ? 2 ?, s
?
2?
?
?
1
(6.58)
2
Z(KY K, g, h, s)
?
? ?
?
0
?
1
?1?
?1
?1
Z(Y, Kg, K h, s) + Z Y, Kg, K h + ? 2 ?, s
=
4?
?
1
(6.59)
2
? ? ?
1
?
?
2
2
?
?
?1
?1
.
+ Z Y, Kg, K h + ? 0 ?, s + Z Y, Kg, K h + ? 12 ?, s
?
1
?
0
2
?1?
Example 6.4 For the notation and more details, cf. [KTTY2].
(i) The N aCl (Sodium Chloride) structure. Here the data is
n+ = n? = 1,
S++ = S?? = a ? Z3 |a1 + a2 + a3 ? 2Z (f.c.c.),
S+? = S+? = a ? Z3 |a1 + a2 + a3 ? 2Z + 1 (f.c.c.),
so that by (6.40)
ZN aCl (s) = Z(A, c0 , 0, s) ? Z(A, 0, 0, s)
= ?Z(I, 0, c0 , s).
Formula (6.60) justifies the first equality in (6.26).
(6.60)
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Fig. 6.4
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121
the N aCl (Sodium Chloride) structure (s.c.)
Thus, by (6.40), the Madelung constant MN aCl is given by
1
MN aCl = ?Z I, 0, c0 ,
2
X (?1)a1 +a2 +a3
= 1.7475645849 . . .
=?
|a|
3
a?Z
a6=0
as stated in (6.25) and (6.26).
(6.61)
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(ii) The CsCl (Caesium Chloride) structure. Here the data is
n+ = n? = 1,
S+? = S+?
3
2
? Z (s.c.),
3
3
1
2
?
Z+
=
(s.c.),
2
3
S++ = S?? =
and the zeta-function is, as discussed in Example 6.2, (6.41)?(6.43),
4
4
I, c0 , 0, s ? Z
I, 0, 0, s
ZCsCl (s) = Z
3
3
s
s
3
3
=
Z(I, c0 , 0, s) ?
Z(I, 0, 0, s)
4
4
= ?3s Z(B, 0, c0 , s).
Fig. 6.5
the CsCl (Caesium Chloride) structure (b.c.c.)
(6.62)
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123
Hence
MCsCl
?
?
1
1
3
3
=
?
Z I, c0 , 0,
Z I, 0, 0,
2
2
2
2
?
1
= ? 3 Z B, 0, c0 ,
2
(6.63)
as in (6.43), whence (6.27) ensues.
(iii) The ZnS (Zincblende) structure. The data is
n+ = n? = 1,
S++ = S??
S+? = S+?
)
2 3
= ? a a ? Z , a1 + a2 + a3 ? 2Z (f.c.c.),
3 )
(
3
2 1
1
(f.c.c.),
= ? aa? Z+
, a1 + a 2 + a 3 ? 2 Z +
2
2
3 (
=
Fig. 6.6
the ZnS (Zincblende) structure (diamond)
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and the zeta-function is
4
4 1
A, c0 , 0, s ? Z
A, 0, 0, s
ZZnS (s) = Z
3 2
3
s s
3
1
3
=
Z A, c0 , 0, s ?
Z(A, 0, 0, s) ,
4
2
4
(6.64)
i.e. (6.29). Further in a similarly way as we prove (6.67) in [KTTY2], we
may prove
s
1 3
1
(6.65)
ZZnS (s) =
ZN aCl (s) + ZCsCl (s),
2 4
2
whence as in (6.32) and (6.31)
MZnS
?
3
1
MN aCl + MCsCl
=
4
2
= 1.63805805338 . . ..
(6.66)
(iv) The CaF2 (Fluorite) structure. The data:
n+ = 1, n? = 2,
)
(
2 3
S++ = ? a a ? Z , a1 + a2 + a3 ? 2Z (f.c.c.),
3 3
1
2
(s.c.)
Z+
S+? = ?
2
3
3
2
S?? = ? Z (s.c.)
3
)
(
3
1
1
2 , a1 + a2 + a3 ? 2Z +
(f.c.c.).
S?+ = ? a a ? Z +
2
2
3 The zeta-function is
1
4
4
ZCaF2 (s) =
Z
I, c0 , 0, s ? 2 Z
A, 0, 0, s
2
3
3
4
4 1
1
A, c0 , 0, s ? Z
I, 0, 0, s
2Z
+
2
3 2
3
s
s
1 3
1 3
=
Z(I, c0 , 0, s) ?
Z(I, 0, 0, s)
2 4
2 4
s s
3
1
3
+
Z A, c0 , 0, s ?
Z(A, 0, 0, s),
4
2
4
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=
Fig. 6.7
the CaF2 (Fluorite) structure
for which the following is proved in Example 6.1.
s
1 3
ZCaF2 (s) =
ZN aCl (s) + ZCsCl (s),
2 4
(6.67)
whence
MCaF2
6.4
?
3
=
MN aCl + MCsCl
4
= 2.51939243992 . . ..
(6.68)
Bessel series expansions for Epstein zeta-functions
In this section we shall prove a Bessel series expansion of Chowla-Selberg
type (Theorem 6.2) for the Epstein zeta-function ?(Y, g, h, s) corresponding
to a block decomposition of the matrix Y . The proof depends on another
Bessel series expansion (Theorem 6.1) for the generalized Epstein zetaP
e2?ihиa
function a?Zn (Y [a+g]+b)
s for b > 0, which is interesting in its own right
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and which we call the Mellin-Barnes type, being dependent on the MellinBarnes integrals.
As a corollary to Theorem 6.1, we shall prove the Benson-Mackenzie
formula (Corollary 6.1), and for applications of Theorem 6.2, we refer to
[KTTY2].
Theorem 6.1 (Mellin-Barnes type formula)
above, we have for b > 0,
Notation being as
X
e2?ihиa
(Y [a + g] + b)s
a?Zn
r
s? n
2
p
?1 [a + h]
X
2
Y
=p
Ks? n2 2 Y ?1 [a + h]b ?
e?2?igи(a+h)
b
|Y |
n
? ?s ?(s)
a?Z
a+h6=0
1 ?(s ? n2 ) 1
+ ?(h) p
n
n ,
|Y | ? s? 2 bs? 2
(6.69)
where Ks (z) signifies the modified Bessel function of the second kind defined
by (6.10).
Proof. This is Formula (1.25) [KTY7] with the term ?(g)(?b)?s ?(s) incorporated in the left-side member. There the proof depended on the modular relation, i.e. the Poisson summation modified so as to suit the case. We
refer to Terras for a similar but subtler proof using the Poisson summation
formula.
We may deduce (6.69) from the functional equation (6.19) via the
Mellin-Barnes integral
Z
1
?(s ? z) ?(z) ?z
(1 + x)?s =
x dz
(6.70)
2?i (c)
?(s)
for x > 0, 0 < c < ?, which has been used extensively in various context
(cf. e.g. [KTZ], [KTTY1], [Matsumoto] and [PK]). The proof starts from
expressing the sum in the form of the integral (6.70), applying the functional
equation, and then finally appealing to (6.12).
Corollary 6.1 (Benson-Mackenzie
(cf. Borweins? [Bor])) Let as
?1?
?
?
!
100
2
1
1
0
?
?
2
1
, c1 = 1 . Then we
before I = ?0 1 0?, c0 = ? 2 ? and I2 =
01
2
1
001
2
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The theory of Bessel functions and the Epstein zeta-functions
have
6 ? s+1
Z(I, 0, c0 , s) =
?(s + 1)
X
X
2
b (?1)
2
a?(Z+ 12 ) b?Z
b6=0
b
r
s
p
I2 [a]
2?
2
I
K
[a]b
s
2
b2
(6.71)
and
1
Z I, 0, c0 ,
2
= ?12?
Since for ? >
Proof.
?
? X
X
sech
a1 = 21 a2 = 21
q
a21
+
a22
?
2
,
(6.72)
3
2
Z(I, 0, c0 , s) =
X (?1)a1 +a2 +a3 (a2 + a2 + a2 )
1
2
3
,
(a21 + a22 + a23 )s+1
3
a?Z
a6=0
we may write
Z(I, 0, c0 , s) = 3
XX
b?Z a?Z2
b6=0
(?1)a1 +a2 +b b2
,
(a21 + a22 + b2 )s+1
whence
Z(I, 0, c0 , s) = 3
X
2
b (?1)
b
X
a?Z2
b?Z
b6=0
e2?ic1 иa
(I2 [a] + b2 )s+1
!
.
(6.73)
We apply Theorem 6.1 to the inner sum on the right of (6.73) to obtain
Z(I, 0, c0 , s)
=3
X
b?Z
b6=0
2 ? s+1 X
b (?1)
?(s + 1)
2
2
b
a?Z
r
!
s
p
I2 [a + c1 ]
2
,
Ks 2 I2 [a + c1 ]b ?
b2
(6.74)
which is (6.71).
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Now put s = 21 and recall Formula (6.14) to deduce that
p
X X
1
|b| (?1)b exp ?2 I2 [a] |b| ?
Z I, 0, c0 ,
= 6?
2
1 2
a?(Z+ 2 ) b?Z
b6=0
?
X X
= 12 ?
a?(Z+ 21 )2 b=1
(6.75)
b
p
b ? exp ?2 I2 [a] ?
The inner sum can be evaluated to be
p
q
2
? exp ?2 I2 [a] ?
1
,
sech
a21 + a22 ?
p
2 = ?
4
1 + exp ?2 I2 [a] ?
2
a1
? Z + 21 . Hence, splitting the sum over a1 , a2 into 4 parts, we
a2
conclude (6.72), completing the proof.
a=
To state Theorem
6.2, we introduce new notation.
A B
be a block decomposition with A an n О n matrix
Let Y = t
BC
and B an m О m matrix. Set
D = C ? t BA?1 B.
In
accordance
with this decomposition, we decompose the vectors g =
g1
h1
,h =
, g 1 , h1 ? Z n , g 2 , h2 ? Z m .
g2
h2
Theorem 6.2 (generalized Chowla-Selberg type formula cf.
[Ter1, Example 4, p.208]) Under the above notation, we have
?(Y, g, h, s)
(6.76)
1
n
= ?(g 2 ) e?2?ig2 иh2 ?(A, g 1 , h1 , s) + ?(h1 ) p
? D, g 2 , h2 , s ?
2
|A|
?2?ig 1 иh1
X
X
?1
2e
p
+
e2?i(?g1 иa+h2 иb) e?2?iA B(b+g2 )и(a+h1 )
|A|
a?Zn
b?Zm
О
s
a+h1 6=0 b+g 2 6=0
A?1 [a
+ h1 ]
D[b + g 2 ]
s? n
2
p
Ks? n2 2 A?1 [a + h1 ] D[b + g 2 ] ? .
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The theory of Bessel functions and the Epstein zeta-functions
Proof. The case g = h = 0, n = m = 1 is due to Chowla and Selberg
[CS], [SC] (cf. also Bateman and Grosswald [BG]), the case m = 1 is due to
Berndt [Ber6] and the general case with g = h = 0 is due to Terras [Ter1].
The proof in our most general case runs as follows.
Noting that
X
?(Y, g, h, s) = ? ?s ?(s)
a?Zn ,b?Zm
(a+g 1 ,b+g 2 )6=0
e2?i(h1 иa+h2 иb)
,
Y [(a + g 1 , b + g 2 )]s
we distinguish three cases: b + g 2 = 0 (g 2 = 0 or not) and b + g 2 6= 0:
X
?(Y, g, h, s) = ?(g 2 ) ? ?s ?(s)
a?Zn
a+g 1 6=0
X
+ ? ?s ?(s)
X
e2?i(h1 иa?h2 иg2 )
Y [(a + g 1 , 0)]s
b?Zm a?Zn
b+g 2 6=0
(6.77)
e2?i(h1 иa+h2 иb)
.
Y [(a + g 1 , b + g 2 )]s
We now apply the formula
Y [(a, b)] = A[a + A?1 Bb] + D[b],
a ? Z n , b ? Zm
(6.78)
to transform (6.77) into
?(Y, g, h, s)
= ?(g 2 ) ? ?s ?(s) e?2?ih2 иg2
+ ? ?s ?(s)
X
m
X
a?Zn
a+g 1 6=0
e2?ih2 иb
b?Z
b+g 2 6=0
X
a?Zn
e2?ih1 иa
A[(a + g 1 ]s
(A[a+g1
(6.79)
e2?ih1 иa
+A?1 B(b+g
2 )]+D[b+g 2 ])
s
.
The first sum on the right of (6.79) is ?(A, g 1 , h1 , s) and to the inner
sum in the second term, we apply Theorem 6.1. Then the second term on
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the right of (6.79) becomes
X
2
e2?ih2 иb p
|A|
b?Zm ,b6=0
b+g 2 6=0
О
+
s
A?1 [a + h1 ]
D[b + g 2 ]
X
b?Zm
b+g 2 6=0
e
2?ih2 иb
s? n
2
X
e?2?i(g1 +A
?1
B(b+g 2 ))и(a+h1 )
n
a?Z
a+h1 6=0
p
?1
Ks? n2 2? A [a + h1 ] D[b + g 2 ]
?(h1 ) p
? s?
|A| D[b + g 2
n
2
n
n
] s? 2 ? s? 2
,
which are the third and second terms on the right of (6.76), whence the
result follows.
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Chapter 7
Fourier series and Fourier transforms
Abstract
This chapter contains elementary facts about Fourier series
and transforms; Theorem 7.2 gives sufficient conditions for
a Fourier series to converge to the given function f (t), i.e.
that f (t) is piecewise of C 1 , which is superfluous but sufficient for most of the purpose. In the proof the Fourier series (7.9) for the first periodic Bernoulli polynomial B 1 (x)
is essentially used. Because of its importance, we give two
different proofs for (7.9), one by Theorem 7.2, the other
by Abel?s and Dirichlet?s theorem in Appendix B. Regarding integral transforms, we emphasize the case of complex
Fourier transforms (and its equivalent form, the Laplace
and Mellin transforms). The reader can familiarize oneself
with many worked-out concrete examples. This chapter
can be read parallel to Chapter 8.
7.1
Fourier series
Suppose f is a periodic function with period 2T (T > 0) and integrable
over [?T, T ], and that the procedures made below are all valid.
In analogy with the Laurent expansion (=the Taylor expansion with
P?
denominator) n=?? cn z n , we wish to express f (t) in terms of a series
?
in ei T t :
?
X
n=??
?
cn ei T t
n
=
131
?
X
n=??
cn ei
n?
T t
,
(7.1)
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called the Fourier series S[f ] of f .
We want to determine the coefficients cn . Integrate the series term by
term, thereby noting that
(
Z T
1, n = 0
1
t
i n?
e T dt =
2T ?T
0, n 6= 0,
n?
the orthogonality of the sequence {ei T t }n?Z of complex exponentials, we
conclude that
Z T
Z T
?
X
(m?n)?
1
1
t
?i n?
T
cm
dt =
f (t) e
ei T t dt = cn .
2T ?T
2T ?T
m=??
Thus the n-th Fourier coefficient of f should be defined as
Z T
1
n?
cn =
f (t) e?i T t dt,
2T ?T
(7.2)
and we may express the Fourier series of f which we denote by S[f ], as
a complex exponential series, in the form
?
X
S[f ] =
cn ei
n?
T t
cn ei
n?
T t
.
(7.3)
n=??
We then write
f (t) ?
Sn (t) =
?
X
.
n=??
n
X
k=?n
k?
ck ei T t ,
(n ? N)
(7.4)
is called the n-th partial sum.
Many of the results in the theory of Fourier series may be treated, from
a more general standpoint, as those for the orthogonal systems in a vector
space. In view of this fact and in anticipation of future progress, let us
try to develops a method based on linear algebra. Needless to say, one can
come to the same conclusions by direct computation of integrals.
Definition 7.1 The set V of all complex-valued piecewise continuous
periodic functions f with period 2T forms a complex vector space. In V
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we define the positive semidefinite Hermitian form
Z T
1
(f, g) =
f (t) g(t) dt
2T ?T
(cf. Exercise 7.1). (Note that we take the complex conjugationpof g).
Then V becomes a semi-normed complex vector space with kf k = (f, f )
as its semi-norm (length). If (f, g) = 0, then we say that f and g are
orthogonal and write f ?g. A subset S ? V consisting of vectors of
positive length is called an orthogonal system if for ?f, g ? S, f 6= g,
we have f ?g. An orthogonal system with all its vectors having length 1 is
called an orthonormal system (ONS). Namely, S ? V is an orthonormal
system if and only if
(
1, f = g
.
(f, g) =
0, f 6= g
An orthonormal system S ? V is called a complete orthonormal system
if an addition of one vector having positive length invalidates the orthogonality, or in other words, S is a complete system if and only if an element
f of V is orthogonal to all elements of S, then kf k = 0.
Exercise 7.1 Prove that (f, g) defined above satisfies the conditions (i)?
(iv) of the positive semidefinite Hermitian form.
(i)
(ii)
(iii)
(iv)
(f, f ) ? 0; (f, f ) = 0 ? f (t) ? 0 (save for discontinuities)
(f, g) = (g, f )
(f + g, h) = (f, h) + (g, h), (f, g + h) = (f, g) + (f, h)
(?f, g) = ? (f, g), (f, ?g) = ?? (f, g) (? ? C)
Example 7.1
An example of an orthonormal system.
n
n? o
exp i t
T
n?Z
is a complete orthonormal system in V . Indeed, we may directly check that
Z T
Z T
m? n? (m ? n)
1
1
exp i
exp i
t exp i t dt =
t dt = ?mn .
2T ?T
T
T
2T ?T
T
Lemma 7.1 If {?1 , . . . , ?n } ? V (?i 6= ?j for i 6= j) is an orthonormal
system, then for any c1 , . . . , cn ? C, we have
2
n
n
X
X
|ck |2 .
c k ?k =
k=1
k=1
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Exercise 7.2
Prove Lemma 7.1.
Lemma 7.2 Let {?1 , . . . , ?n } ? V (?i 6= ?j for i 6= j) be an orthonormal
system and for any f ? V , put ck = (f, ?k ). Then we have
2
n
n
X
X
|ck |2 .
ck ?k = kf k2 ?
f ?
k=1
k=1
From Exercise 7.1 and Lemma 7.1 it follows that
!
n
n
X
X
LHS = f ?
c k ?k , f ?
c k ?k
Proof.
k=1
2
= kf k ?
n
X
k=1
k=1
ck (?k , f ) ?
n
X
k=1
2
n
X
ck ?k = RHS.
c?k (f, ?k ) + k=1
Corollary 7.1 (Bessel?s inequality)
Let {?j }j?J ? V (?i 6= ?j for
i 6= j) be an orthonormal system and for any f ? V , put cj = (f, ?j ). Then
cj = 0 except for countably many j and we have
X
2
2
|cj | ? kf k .
j?J
In particular, we have
P
j?J
2
|cj | < ?.
Remark 7.1 cj = (f, ?j ) is called the j-th Fourier coefficient of f with
respect to the ONS {?j }j?J .
Corollary 7.2 If f is a continuous periodic function of period 2T and
is piecewise of class C 1 , then for the Fourier coefficients (7.2) of f , the
estimate
?
?2 X 2
2
n |cn |2 ? kf 0 k < ?
T 2 n=1
holds.
Proof. f 0 being piecewise continuous, belongs to V and therefore for its
Fourier coefficients
Z T
1
n?
?n =
f 0 (t) e?i T t dt,
2T ?T
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we have Bessel?s inequality
?
X
n=1
2
2
|?n | ? kf 0 k .
(7.5)
Integrating by parts, thereby using the periodicity of f , we get
Z
n?
n? T
in? T
n?
1 ?n =
c n,
f (t) e?i T t dt = i
f (t) e?i T t ?T +
2
2T
2T ?T
T
whence, substituting these in (7.5), we conclude the assertion.
Exercise 7.3
Compute the right-hand side of
Z T
2
|f (t) ? Sn (t)| dt
?T
with the aid of Exercise 7.2 and give a direct proof of Bessel?s inequality in
Corollary 7.1.
Now we shall express (7.3) in a trigonometric form: We write
cn =
1
(an ? i bn )
2
(n ? Z),
with
a?n = an , b0 = 0, b?n = ?bn
(n ? N),
Definition 7.2 Let f be a periodic function of period 2T (T > 0)
f (t + 2T ) = f (t), t ? R. Then we call
Z
Z
1 T
1 T
n?t
n?t
dt, bn =
dt, 0 ? n ? Z
an =
f (t) cos
f (t) sin
T ?T
T
T ?T
T
(7.6)
the n-th Fourier cosine coefficient and Fourier sine coefficient, respectively. We have
?
? X
a0 X
n?t
n?t
S[f ] =
An (t) =
(7.7)
+
an cos
+ bn sin
2
T
T
n=0
n=1
and
Sn (t) =
n
X
k=0
n a0 X
k?t
k?t
Ak (t) =
.
+
ak cos
+ bk sin
2
T
T
k=1
(7.8)
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Exercise 7.4
Prove that
an = cn + c?n , bn = i(cn ? c?n )
and that (7.3) and (7.7) are equivalent.
Exercise 7.5 Prove that if f is a periodic even [odd] function, then
bn = 0 [an = 0] and that
"
#
?
?
X
a0 X
n?t
n?t
S[f ] =
.
+
S[f ] =
bn sin
an cos
2
T
T
n=1
n=1
This is called the Fourier cosine (respectively, the Fourier sine) series.
Remark 7.2 Since f is a periodic function of period 2T , we may choose
any interval of length 2T as the interval for integration to define the Fourier
coefficients.
We often use [?T, T ] or [0, 2T ]. Also, by the change of variable
?
t
u = T t u = 2T
, we may assume the period of f to be 2? [1]. Subsequently,
we shall solely consider periodic functions of period 2T . The Fourier series
S[f ] for f is just a trigonometrical (exponential) series formed from f and
it is not known a priori whether it is convergent or, if convergent, whether
it coincides with the original f (t). However, if f (t) is of good-natured, like
piecewise smooth (cf. Theorem 7.1), then S[f ](t) converges to f (t) at the
continuity points of f .
Exercise 7.6
Define
f (t) = t ?
1
2
(0 < t < 1),
f (0) = f (1) = 0
and extend the domain of definition to all reals by continuing with period 1.
/ Z (the first
Then find the Fourier series of f . Indeed, f (t) = B 1 (t), t ?
periodic Bernoulli polynomial in Chapter 1).
Solution Since this is the most fundamental, we shall compute the
Fourier coefficients an , bn , and cn . By integration by parts, we have
Z 1
1
an = 2
t?
cos(2?nt) dt = 0,
2
0
bn = 2
Z 1
0
1
t?
2
sin(2?nt) dt = ?
1
,
?n
n ? N.
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Similarly, the n-th Fourier coefficient cn (n 6= 0) is given by
Z 1
cn =
B 1 (x) e2?inx dx,
0
which by integration by parts becomes
1
Z 1
1
1 2?inx
1
1
?
cn =
x?
e
e2?inx dx = ?
i,
2?in
2
2?inx
2?n
0
0
or
an = 0, bn = ?
1
,
?n
as above. Hence
f (t) ? ?
?
1 X sin(2?nt)
.
? n=1
n
Since f is piecewise smooth, we should have the equality (by Theorem 7.1
below)
f (t) = ?
?
1 X sin(2?nt)
,
? n=1
n
?t ? R.
Hence we have (1.9) in the case of n = 1, which we restate for convenience as
?
1 X sin 2?nx
, x?
/ Z.
B 1 (x) = ?
? n=1
n
(7.9)
Once (7.9) is established, it is quite easy to deduce Fourier expansion
of other linear functions. E.g. consider the function f (x) defined as x ? ?
for 0 ? x < 2? and continued to a periodic function of period 2?. Since
x
2?B 1
= f (x), we immediately obtain
2?
f (x) = ?2
?
X
sin nx
, x?
/ 2?Z.
n
n=1
(7.10)
Proposition 7.1 (The Riemann-Lebegues Lemma) Suppose f is
piecewise continuous on the interval [a, b]. Then all of the following holds
true:
Z b
Z b
lim
f (t) sin(Rt) dt = 0, lim
f (t) cos(Rt) dt = 0,
R??
a
R??
a
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Fig. 7.1
Fig. 7.2
Fig. 7.3
lim
R??
Z
b
f (t) eiRt dt = 0.
a
Proof. We may assume that f is continued on the whole interval [a, b]
(Exercise 7.7). Then, being continuous on the compact set [a, b], f is
bounded (Weierstrass? Theorem), i.e. f (t) = O(1), t ? [a, b] and uniformly
continuous, whence
?
f u+
? f (u) = o(1),
R
R ? ?.
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?
, then
We shall prove the first assertion. Putting t = u + R
Z b
Z b??/R ?
I :=
f (t) sin(Rt) dt = ?
f u+
sin(Ru) du.
R
a
a??/R
Hence
Z
?
f u+
sin(Ru) du
R
a??/R
Z b
Z b??/R
? sin(Ru) du
f (u) ? f u +
=
f (u) sin(Ru) du +
R
b??/R
a
Z a
?
f u+
?
sin(Ru) du
R
a??/R
1
1
+ o(1) + O
= o(1), R ? ?.
=O
R
R
2I = I ?
b??/R
Exercise 7.7 (i) Divide [a, b] into subintervals to prove that the above
proof can be reduced to the case where f (t) is continuous on the whole
interval. (ii) Prove the remaining assertions of Proposition 7.1.
Theorem 7.1 If f is a periodic function of period 2T and is piecewise
smooth as well as continuous on any finite interval, then its Fourier series
(7.3) converges to f (t) uniformly on any finite interval.
Proof. First we show that {Sn } is convergent, where Sn = Sn (t) is the
n-th partial sum defined by (7.4):
For integers N > M > 0, we have
N
X
1
k? i T t
k ck e
|SN ? SM | = k
|k|=M +1
v
v
u X
u X
u N
u N
1 i k? t 2
2
2
t
?
k |ck | t
e T k2
|k|=M +1
|k|=M +1
v
u N
u X 1
T
0 t
? kf k
?
k2
k=M +1
by the Canchy-Schwarz inequality and the Bessel inequality (Corollary 7.1).
Hence |SN ? SM | ? 0 as N, M ? ?, and the Canchy criterion applies,
ensuring the convergence of {Sn }. We now show that its limit is f (t).
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Recalling (7.3), we may express (7.4) as
Z T
n
X
k?
k?
1
Sn =
e?i T u ei T t f (u) du
2T ?T
k=?n
Z T
n
1 X i k? (t?u)
1
f (u) du.
e T
=
T ?T 2
k=?n
Now the Dirichlet kernel is
n
k?
1 X
+
cos
(t ? u),
2
T
k=1
sin n + 12 T? (t ? u)
which by Exercise 7.8, (2),
.
2 sin T? t?u
2
Hence
Z
1 T sin n + 12 T? (t ? u)
Sn (t) =
f (u) du
T ?T
2 sin T? t?u
2
Z
1 T ?t sin n + 21 T? v
=
f (t + v) dv
T ?T ?t
2 sin T? v2
Z
1 T sin n + 12 T? u
f (t + u) du
=
T ?T
2 sin T? u2
(7.11)
by change of variable and the periodicity of f .
R T sin(n+ 1 ) ? u
Subtracting f (t) = T1 ?T 2 sin 2? uT f (t) du from (7.11) and using
T 2
1
sin n + 2
sin T? u2
?
Tu
=
sin n T? u cos T?
sin T? u2
u
2
?
+ cos n u,
T
we find that
Sn (t) ? f (t) =
1
2T
1
+
2T
Z
Z
T
?T
T
?
g(u, t) sin n u du
T
?
(f (u + t) ? f (u)) cos n u du,
T
?T
where
g(u, t) =
(f (u + t) ? f (u)) cos T?
sin T? u2
u
2
whose possible discontinuity, save for those of f , is at u = 0.
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However, by the piecewise continuity of f 0 (t),
f (u + t) ? f (u) T? u2 cos T? u2 2T
u
sin T? u2
?
?
2T 0
?
?
f (t+), as u ? 0+
?
?
?
? 2T f 0 (t?), as u ? 0 ? .
?
g(u, t) =
Hence g(u, t) is also piecewise continuous and the Riemann-Lebesgue
Lemma (Proposition 7.1) shows that
lim Sn (t) = f (t).
n??
Theorem 7.2 If f is periodic of period 2T , piecewise continuous and
piecewise of C 1 , then the Fourier series for f (t) converges to 12 (f (t+) +
f (t?)).
Proof. We may consider the case where 0 is the only discontinuity of f ,
other cases being reduced to this. Then the function
F (t) = f (t) + (f (0+) ? f (0?)) g(t)
is piecewise of C 1 and continuous except possibly at 0, where
(
t
B 1 2T
, t?
/ 2T Z
g(t) =
0,
t ? 2T Z.
But
lim F (t) = f (0?) +
t?0?
and
lim F (t) = f (0+) ?
t?0+
1
1
f (0+) ? f (0?) = f (0+) + f (0?)
2
2
1
1
f (0+) ? f (0?) = f (0+) ? f (0?) ,
2
2
and therefore F (t) is also continuous at 0.
Hence we may apply Theorem 7.1 to conclude that F has the Fourier
series which converges to F (t) everywhere. But g(t) has the Fourier series
?
?
1 X sin T? nt
? n=1
n
converging to g(t) everywhere (cf. Exercise 7.6).
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Hence it follows that f (t) also has the Fourier series converging to it
at continuities and to F (0) = 21 (f (0+) ? f (0?)), thereby completing the
proof.
Exercise 7.8
For x ?
/ Z prove that
n
X
e2?ikx = e?i(n+1)x
k=1
sin ?nx
sin ?x
(7.12)
and deduce from (7.12) that
n
X
cos 2?kx +
k=1
1
sin(2n + 1)?x
=
2
2 sin ?x
(7.13)
and
n
X
sin 2?kx = sin((n + 1)?x)
k=1
Solution
cos ?x ? cos ?(2n + 1)x
sin ?nx
=
sin ?x
2 sin ?x
(7.14)
We have
Sn =
n
X
k=1
e
2?ikx
e2?ix e2?inx ? 1
e2?inx ? 1
=
=
.
e?ix ? 1
1 ? e?2?ix
We factor out e?inx (resp. e??ix ) from the numerator (resp. denominator)
to get
Sn = e?i(n+1)x
e?inx ?e??inx
2i
e?ix ?e??ix
2i
,
which is (7.12). The real part of (7.12) is
n
X
k=1
cos 2?kx =
cos(n + 1)?x sin ?nx
sin ?(2n + 1)x ? sin ?x
=
,
sin ?x
2 sin ?x
which proves (7.13). The imaginary part of (7.12) is the same as (7.14).
Another proof uses the formulae
cos ? sin ? =
and
sin ? sin ? = ?
1
sin(? + ?) ? sin(? ? ?)
2
1
cos(? + ?) ? cos(? ? ?) .
2
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The first gives
n
X
cos 2?kx sin ?x
k=1
n
1X
(sin(2k + 1)?x ? sin(2k ? 1)?x)
2
=
k=1
or
1
=
sin(2n + 1)?x ? sin ?x ,
2
n
X
1 sin(2n + 1)?x 1
? ,
2
sin ?x
2
cos 2?kx =
k=1
which is (7.13).
The second gives
n
X
sin 2?kx sin ?x
k=1
n
1X
=?
(cos(2k + 1)?x ? cos(2k ? 1)?x)
2
k=1
which is (7.14).
Example 7.2
1
= ? cos(2n + 1)?x ? cos ?x ,
2
Let f (x) be defined for ?1 ? x < 1 as
f (x) =
(
0, ?1 ? x < 0
1, 0 ? x < 1,
and then defined periodically with period 2:
f (x) = f (x + 2n), n ? Z.
(cf. Fig. 7.4 for its graph).
This function can be represented as
x+1
f(x) = [x + 1] ? 2
.
2
(7.15)
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y
1
-1
0
1
2
3
x
Fig. 7.4
1
1
Recalling B 1 (x) = x ? [x] ? , we have [x] = x ? B 1 (x) ? , so that
2
2
x+1 1
1
x+1
? B1
?
f (x) = x + 1 ? B 1 (x + 1) ? ? 2
2
2
2
2
1
x+1
= + 2B 1
? B 1 (x + 1).
2
2
Applying the Fourier expansion (7.9) for B 1 (x) below, we see that
?
?
1
2 X sin 2?n x+1
1 X sin 2?nx
2
f(x) = ?
+
2 ? n=1
n
? n=1
n
=
=
?
?
1
1X2
1 X sin 2?nx
+
(?1)n?1 sin ?nx +
2 ? n=1 n
? n=1
n
?
1
2 X 2
+
(?1)2m?1 sin 2?mx
2 ? m=1 2m
+
=
?
?
2
1 X
1X1
(?1)2m?2 sin(2m ? 1)?x +
sin 2?nx
? m=1 2m ? 1
? n=1 n
?
1
1
2 X
+
sin(2m ? 1)?x, x ?
/ Z.
2 ? m=1 2m ? 1
For x ? Z, the Fourier series converges to 0. To deduce (7.15) we argue
in the following manner. First y = [x + 1] has the graph (Figure 7.8):
this graph by 2n, i.e. by
? 1 ? x < 2n + 1, we have to pulldown For 2n
x+1
x+1
x+1
< n + 1 implies
since n ?
= n.
2
2
2
2
Example 7.3
There is a method much subtler and more ingenious than
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Fig. 7.5
Fig. 7.6
Fig. 7.7
the customary one for deducing (7.9), which appeals to Abel?s continuity
theorem.
We wish to apply this theorem to the Maclaurin expansion for
? log(1 ? z).
The expansion can most easily be obtained by termwise integration of the
absolutely convergent power series
?
X
n=0
zn =
1
,
1?z
|z| < 1.
(7.16)
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y
3
2
1
1
0
-1
3
2
x
Fig. 7.8
Indeed, integrating from 0 to z, we obtain
? log(1 ? z) =
Z
z
0
? Z z
?
X
X
1
1 n
dz =
z n dz =
z ,
1?z
n
n=0 0
n=1
where we have taken the principal branch of the logarithm.
?
X
1 in?
Now, for z = e , ? ?
/ 2?Z, we contend that the series
e
is
n
n=1
convergent, as shown in Example B.2. Hence, Abel?s theorem B.6 allows
us to write
i?
?
X
1 i? n
? log 1 ? ei? =
e
,
n
n=1
??
/ 2?Z.
By Euler?s identity, the right-hand side is
?
?
X
X
1
1
cos n? + i
sin n?,
n
n
n=1
n=1
(7.17)
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while the left-hand side may be transformed into
?
?
ei 2 ? e?i 2 ?2i
? log 1 ? ei? = ? log
и
?
2i
e?i 2
i?
?
= ? log sin ? log(?2i) + log e? 2
2
?
i
= ? log 2 sin ? (? ? ?),
2 2
?
since log(?2i) = log 2 ? i. Hence, comparing the real and imaginary
2
parts, we conclude that
? log 2 sin
?
X
1
?
=
cos n?
2 n=1 n
? ? ? = ?2
?
X
1
sin n?.
n
n=1
(7.18)
(7.19)
Thus we have not only recovered (7.10) again but obtained the Fourier
expansion of the log sin function.
(
0, ?1 ? x < 0
can be expressed
Example 7.4 The function f (x) =
x, 0 ? x < 1
as 12 (x + |x|), the positive part f + (x) of f (x) = x, where the positive part
of f (x) is defined to be
f + (x) =
f (x) + |f (x)|
.
2
The function f (x) (for its graph, see Fig. 7.9) obtained from f (x) by
continuing it periodically with period 2 is
1
x+1
x + 1 f (x) =
x?2
+ x ? 2
(7.20)
.
2
2
2
x+1
< n + 1,
Indeed, for 2n ? 1 ? x < 2n + 1 (n ? Z), we have n ?
2 x+1
x+1
= n. Hence for 2n ? 1 ? x < 2n, we have x ? 2
=
and so
2
2
x ? 2n < 0,so that f (x) = 21 (x ? 2n + 2n ? x) = 0. For 2n ? x < 2n + 1, we
x+1
have x ? 2
= x ? 2n ? 0, and therefore f (x) = 2(x ? 2n) = x ? |x|.
2
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y
1
0
-1
1
2
3
x
Fig. 7.9
x+1 = x ? 2B 1 x+1
, we deduce that
2
x+1
x + 1 1
f (x) =
2B 1
+ 2B 1
(7.21)
2
2
2
x+1
x + 1 + B 1
= B1
2
2
x + 1 We remark that the graph of y = B 1
is the directly connected
2
infinite tents (see Fig. 7.10).
Noting that 2
2
y
1
-1
0
1
2
3
x
Fig. 7.10
Of course, if we recall this fact first, then the expression (7.21) would
follow by inspection.
We also note that if we replace the period 2 by 2?, we get
1
x + 21
x + 21 1 1
f 1 (x) =
+ x?2
x?2
.
(7.22)
2 ?
2?
?
2? Motivated by the above function, we shall digress here into an equivalent
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statement of the celebrated Riemann Hypothesis (abbreviated as RH) to
the effect that the Riemann zeta-function does not vanish on the central
line ? = 21 .
We need some preparations.
Let t(x) denote the tent function
t(x) =
(
0?x?
2x,
1
2
2 ? 2x,
1
2
? x ? 1.
We consider the (k + 1)-th iterate of t divided by 2k+1 :
f2k (x) =
1
2k+1
tk+1 (x),
where by an iterate we mean the successive composition of t:
tk (x) = t tk?1 (x) ,
t1 (x) = t(x),
t0 (x) = 1.
Denoting the directly connected n tents by fn (x) with length n1 and height
1
k
k
2n , we note that f2k (x) = t (x) in the case n = 2 .
The Farey sequence Fx of order [x] is defined to be the increasing sequence of irreducible fractions ?? between 0 and 1 (0 exclusive) with denominators ? x. This may be constructed from the lower order one by
inserting the mediants of adjacent fractions until the denominator reaches
P
[x]. The total number of elements of Fx is n?x ?(n) = ?(x), say, where
?(n) indicates the Euler function (cf. (8.27)).
For any even integrable core function f on [0, 1] we define the error term
?(x)
Ef (x) =
X
?=1
f (?? ) ? ?(x)
Z
1
f (x) dx,
0
where by an even function we mean that it satisfies f (x) = f (1 ? x),
which we may assume on symmetry grounds. Ef (x) is defined to be 0 for
0 < x < 1.
We consider the Mellin transform (cf. Д7.4) of Ef (x):
F (s) = s ?(s)
Then we have
Z
?
Ef (x) x?s
0
dx
.
x
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Lemma 7.3 ([BKY, Lemmas 2.1 and 2.2]) (i) For the directly connected n tents fn (x), the associated Mellin transform F (s) is given by
F (s) =
1
Fn (s),
12n
Fn (s) =
?
X
cn (m)
ms+1
m=1
where
and
cn (m) = (m, 2n)2 ? (2m, 2n)2 .
This Fn (s) can be written down as follows:
1
Fn (s) = ?3 1 ? s+1 ?(s + 1) Cn (s),
2
where
Cn (s) =
X
d1?s
d|n
X х(?)
.
?2
?|d
d:odd
(ii) For n = 2k we have
C2k (s) =
which does not vanish for
1
2
1 ? 2(1?s)(1+k)
1 ? 21?s
< ? < 1.
Theorem 7.3 ([BKY, Theorem 2.1 (i)])
?(x)
X
?=1
f2k (?? ) =
The asymptotic formula
1 1
?(x)
+
O
x 2 + ,
2k
for every > 0, is equivalent to the Riemann Hypothesis.
Example 7.5 The function defined for ?? ? x < ? by f (x) = x and
continued to be a periodic function with period 2? is
1
1
f (x) = ? B 1
.
x+
2?
2
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Fig. 7.11
7.2
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laplace
Integral transforms
Irrespectively of whether it is in pure or applied areas, a great majority
of important special functions that appear in applications of analysis are
given in the form of an improper integral of a real function
Z ?
(Kf ) (s) =
K(x, s) f (x) dx
??
Z !
Z
?2
t
=
+ lim
lim
?1 ???
?1
?2 ??
K(x, s) f (x) dx,
t
where s = ? + it signifies the complex variable. We call (Kf) (s) an
integral transform of f (with respect to the kernel function K) in view
of the fact that it is obtained by integrating after multiplying f by the
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(complex-valued) kernel function K(x, s). In contrast to differential operators D, ?, ? that we have already learned, K, being an operator exerting
the integration of f , is called an integral operator. The integral transform
of f is denoted by the corresponding capital letter F and is often denoted
by FK (s) in order to suggest the (commonly accepted name of ) the kernel:
Z ?
K(x, s) f (x) dx.
(Kf) (s) = FK (s) =
??
We refer to FK (s) as the resulting integral transform of f . When we know
which integral transform, we may suppress K and simply write F (s). Integral transforms and operators are useful in solving Boundary Value Problems and Differentiable [integral] Equations. However, we have to impose
stringent conditions for the improper integrals to converge. Since the integrals being intrinsically linear, so are the improper ones in the region of
their convergence, and a fortiori so are the integral operators. Namely, for
c1 , c2 ? C,
K c1 f1 + c2 f2 = c1 Kf1 + c2 Kf2 .
The following integral transforms are most frequently used.
Z ?
(F f) (s) = f?(s) =
e?ixs f (x) dx (exponential) Fourier transform
??
(Cf ) (s) = FC (s) =
Z
Z
(LII f ) (s) = FLII (s) =
cos(xs) f (x) dx
Fourier (cosine) transform
??
Z
(Sf) (s) = FS (s) =
(LI f ) (s) = FLI (s) =
?
?
sin(xs) f (x) dx
Fourier (sine) transform
??
?
e?xs f (x) dx
(one-sided) Laplace transform
0
Z
?
e?xs f (x) dx
(two-sided) Laplace transform
??
(M f) (s) = FM (s) =
Z
?
xs?1 f (x) dx
Mellin transform
0
We may express a function that describes the state of a phenomenon at
some time interval as a real function f (t) in the time variable t. We call a
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function a causal function if it is not affected by the conditions before the
initial time t = 0, i. e. if f (t) = 0, t < 0. It follows from Euler?s identity
that
(Cf )(s) =
1
1
1
1
(F f )(s) + (F f )(?s), (Sf )(s) = (F f )(s) ? (F f )(?s),
2
2
2i
2i
whence we see that both the cosine and sine Fourier transforms are special
cases of the exponential Fourier transforms.
The region of convergence of the exponential Fourier transform (if it
exists) must be a strip ?1 < Im s < ?2 containing the real axis (because t
takes values both positive and negative). However, whenever we speak of
a Fourier transform, we mean the case s = y being a real variable, we shall
refer to this as a Fourier transform
Z ?
(F f )(y) = f?(y) =
e?ixy f (x) dx, y ? R,
??
and the exponential Fourier transform as the complex Fourier transform. Then the two-sided Laplace transform is no other than the general
Fourier transform
(LII f )(is) = (F f )(s)
in the complex variable s, rotated clockwise by 90? , and so their theory
is almost parallel and can be translated word-for-word into each other. E.
g. the region of convergence (for two-sided Laplace transform) is a strip
?1 < Re s < ?2 containing the imaginary axis (this is analogous to the fact
stated in Д A.1 that if f (z) is analytic on the unit circle |z| = 1, then it can
be expanded into the Laurent series and
g(?) = f ei?
can be expressed as a Fourier series. If the annulus does not contain the
unit circle, then we cannot write g(?) = f ei? even if g(?) can be expanded
into Fourier series). There is a theory of operators developed on the basis of
two-sided Laplace transforms (cf. [Pa]). It looks as if the Laplace transforms
have driven out Fourier transforms in applied analysis (on the ground that
the former seem to have a wider applicability than the latter), but they
are essentially the same. In comparison with the real Fourier transforms,
the condition that the improper integral for (LII f )(s) be convergent in the
strip ?1 < Re s < ?2 restricts the class of functions than the condition that
the improper integral for (F f )(y) be convergent for any y ? R.
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Indeed, putting s = ? + i t and ?? (x) = e?x? f (x), then we have
Z ?
(LII f )(s) = (LII f )(? + i t) =
e?ixt ?? (tx) dx = (F ?? )(t),
??
? being regarded as a parameter, so that ?the Laplace transform, the range
of whose variable is restricted to a vertical line in its region of convergence,
is a special case of the Fourier transform and vice versa?. More restrictive
it is, the merit of Laplace transforms is that the variable ranging in a strip
without restriction, we may employ the powerful theory of analytic functions. For a causal function, its two-sided and one-sided Laplace transforms
are the same and the region of convergence (if it exists) is the half-plane
Re s = ? > ? with a wide range of applications. In what follows, for a
causal function f (x) (f (x) = 0, x < 0) , we shall call its one-sided Laplace
transform simply the Laplace transform;
Z ?
e?xs f (x) dx, s ? C.
(Lf )(s) = F (s) =
0
In electrical engineering, with ? being the angular cycle, the variable is
often denoted by s = ? + j? (j being the imaginary unit) or p = ? + j?.
??
Use being made of the Fourier transform f(y),
the original function f (x)
which behaves differently on different parts of the x-axis may be expressed
by a unique formula
Z ?
1
f?(y) eiyx dy.
(7.23)
f (x) =
2? ??
This is called the Fourier Integral Theorem. To be more precise, if f, f 0
are piecewise continuous and
Z ?
|f (x)| dx < ?,
??
then the theorem holds in the following form:
Z ?
1
1
? eiyx dy.
{f (x + 0) + f (x ? 0)} =
f(y)
2
2? ??
If we define the inverse Fourier transform F ?1 f of f by
Z ?
1
?1
f (x) eixy dx,
F f (y) =
2? ??
(7.24)
(7.25)
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then we may express the Fourier Integral Theorem as F ?1 F f = f or
b
fb(?x) = f (x). (7.23) may be deduced formally from Theorem 7.2 as follows. If f is piecewise smooth and continuous on [?T, T ], then it can be
expanded into a Fourier series:
?
X
f (t) =
cn e
i?n t
1
cn =
T
,
n=??
?n =
Letting T ? ?,
2?
n,
T
Z
T /2
f (x) e?i?n x dx,
?T /2
|t| <
T
.
2
n ? ?, we may contend that
T cn ?
Z
?
f (x) e?i?n x dx = f? (?n )
??
and therefore
f (t) =
?
1 X
T cn ei?n t (?n+1 ? ?n )
2? n=??
Z ?
?
1 X ?
1
i?n t
? eity dy.
?
f (?n ) e
??n ?
f(y)
2? n=??
2? ??
Viewing this as
f (t) =
Z
?
??
1
2?
Z
?
f (x) e?ixy dx eiyt dy,
??
this may be thought of as giving the motivation for the definition of f?(x).
Also using the defining equation
Z
?
f (x) ?(t ? x) dx = f (t)
??
for the delta function and one of its well known properties
1
2?
Z
?
eixt dx = ?(t),
??
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we can give the following simple proof.
Z ?
Z ?
Z ?
1
1
iyt
?ixy
(F f ) (y) e dy =
e
f (x) dx eiyt dy
2? ??
?? 2? ??
Z ?
Z ?
1
=
f (x) dx
eiy(t?x) dy
2? ??
??
Z ?
=
f (x) ?(t ? x) dx = f (t).
??
This is a legitimate proof if the inversion of the order of integration is
justified. A rigorous proof can be given in the spirit of Proof of Theorem 7.2,
1
in the Fourier
which is omitted. In view of the appearance of the factor 2?
integral theorem, we often introduce normalization by distributing it to
both transforms. The symmetric pair of the Fourier transform and the
inverse Fourier transform is
Z ?
1
e?ixy f (x) dx,
(F f ) (y) = f?(y) = ?
2? ??
Z ?
1
f (x) eixy dx.
F ?1 f (y) = ?
2? ??
Then the Fourier cosine transform takes the form for f even,
r Z ?
r
Z ?
1
2
cos(xs) f (x) dx =
cos(xs) f (x) dx
(Cf ) (s) =
2? ??
? 0
and if f is odd, then the Fourier sine transform becomes
r
r Z ?
Z ?
1
2
(Sf) (s) =
sin(xs) f (x) dx =
sin(xs) f (x) dx.
2? ??
? 0
Hereafter we take for granted the Fourier integral theorem for complex
Fourier transforms: F ?1 F f = f and as its equivalent statements, L?1 Lf =
f and M ?1 M f = f for granted.
We shall give some illustrative examples.
The formula
?
(? > 0, ? = Re s > 0)
(7.26)
L[sin(?t)](s) = 2
s + ?2
and its inverse
L
?1
?
(t) = sin(?t)
s2 + ? 2
(7.27)
are very important in the application of the theory of Laplace transforms.
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The customary proof of (7.26) is the following. By integration by parts,
we have
Z
Z
? ?st
?2
1 ?st
?st
sin(?t) ? 2 e
cos(?t) ? 2 e?st sin(?t) dt,
e
sin(?t) dt = ? e
s
s
s
whence it follows that
Z
1 ?st
? ?st
s2
?st
? e
sin(?t) ? 2 e
cos(?t) . (7.28)
e
sin(?t) dt = 2
s + ?2
s
s
R?
Under the condition ? > 0, the infinite integral 0 e?st sin(?t) dt is absolutely convergent, and by (7.28) is equal to
?
1 ?st
s2
? ?st
?
?
e
sin(?t)
?
e
cos(?t)
.
= 2
2
2
2
s +?
s
s
s + ?2
0
Similarly we may prove the corresponding formula for cosines:
L[cos(?t)](s) =
L
?1
s2
s
(? > 0, ? > 0)
+ ?2
s
(t) = cos(?t).
s2 + ? 2
(7.29)
(7.30)
The following proof is, however, much more concise and instructive.
Suppose first that s = ? > 0 and invoke Euler?s identity to deduce that
Z ?
L[sin(?t)](s) = Im
e?st ei?t dt
0
?
1
?(s?i?t)
= Im ?
e
s ? i?
0
1
s + i?
?
= Im 2
= 2
.
= Im
s ? i?t
s + ?2
s + ?2
Now, L[sin(?t)](s) is an analytic function in s for ? > 0 since the integral
?
. Hence, by
is absolutely convergent there, and so is the function 2
s + ?2
the principle of analytic continuation (Theorem A.9), they must coincide
in the region ? > 0, and this proves Formula (7.26).
Note that the above argument also gives a proof of (7.29), since
Z ?
s + i?
s
L[cos(?t)](s) = Re
e?st ei?t dt = Re 2
= 2
(? > 0)
2
s +?
s + ?2
0
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Other examples are
h
i
?
1
L erfc at?1/2 (p) = e?2a p
p
(7.31)
where the error function is defined by
?
1
,z
2
=
?
? erfc
? z ,
(7.32)
or more generally,
h a i
?
?
?
(p) = 2a 2 p 2 ?1 K? (2 ap) ,
L ? ?,
t
(7.33)
K? (z) indicating the modified Bessel function of the second kind defined
by (6.10), which for ? = 21 reduces to (7.31) in view of (6.14) and
L[?(?, at)](p) =
7.3
?(?)
p
1?
1
(1 + ap )?
.
(7.34)
Fourier transform
In this section, we shall show that the one-sided (complex) Fourier transform and the Laplace transform have the same function by illustrating
Examples 7.6 and 7.7. We use the following data and scheme.
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Fourier series and Fourier transforms
One-sided Fourier Transform
Variable z ? C
the region of absolute convergence
Im z < ?? (? ? R) if |y(t)| =
O(e?t ).
Laplace Transform
Variable s = iz ? C
Region of absolute convergence
Re s > ? if |y| = O(e?t )
y?(z) = F+ [y](z)
Z ?
1
y(t) e?izt dt.
= ?
2? 0
Y (s) = L[y](s)
Z ?
y(t) e?st dt
=
When y (t) is continuous for t > 0
and lim y 0 (t) e?izt = 0
When y (t) is continuous for t > 0
and lim y 0 (t) e?st = 0,
y(+0)
F+ [y 0 ](z) = izF+ [y](z) ? ?
.
2?
L[y 0 ](s) = sL[y](s) ? sy(+0).
0
t??
0
0
t??
Under similar conditions
Under similar conditions
00
L[y 00 ](s) = s2 L[y](s)
2
F+ [y ](z) = (iz) F+ [y](z)
1
? ? (izy(+0) + y 0 (+0)).
2?
F+?1
?
1
(t) = 2? i ei?t
z??
? sy(+0) ? y 0 (+0).
L?1
1
(t) = e?t
s??
L[e?t ](s) =
1
1
F+ ei?t (t) = ?
2? i z ? ?
Example 7.6
1
s??
We solve the differential equation
y 00 + ? 2 y = a,
in y = y(t) with initial conditions
y(0) = b, y 0 (0) = c,
by the Fourier transform method. The above scheme reads:
F+ [y 00 ] + ? 2 F+ [y] = aF+ [1]
1
1 1
(iz)2 y?(z) ? ?
izy(+0) + y 0 (+0) + ? 2 y?(z) = a ?
2?
2?i z
(7.35)
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1
(b i z 2 + cz ? ia)
(iz)2 + ? 2 y?(z) = ?
2? z
1 ?ibz 2 ? cz + ia
1
y?(z) = ?
=?
2? z(z ? ?)(z + ?)
2?
A
B
C
+
+
z
z?? z+?
ia
a
?ibz 2 ? cz + ia
=
= ?i 2
2
z?0
z?0 (z ? ?)(z + ?)
??
?
?ibz 2 ? cz + ia
B = lim (z ? ?)y?(z) = lim
z??
z??
z(z + ?)
2
?ib? ? c? + ia
b? 2 ? ic? ? a
=
= ?i
? и 2?
2? 2
?ibz 2 ? cz + ia
C = lim (z + ?)y?(z) = lim
z??
z??
z(z ? ?)
2
2
?ib? + c? ? a
?ib? + c? + ia
=
=
?? и (?2?)
2? 2
A = lim z y?(z) = lim
1
1
1
1
+ BF+?1
+ CF+?1
AF+?1
y=?
z
z??
z+?
2?
?
?
1 ?
=?
A 2? i + B 2? i ei?t + C 2? i e?i?t
2?
a
b ? 2 ? ic? ? a i?t b ? 2 + ic? ? a ?i?t
= i и (?i)
+
e +
e
?2
2? 2
2? 2
b ? 2 ? a ? ic? i?t
a
e
= 2 + 2 Re
?
2? 2
a
1
= 2 + 2 (b ? 2 ? a) cos(?t) + c? sin(?t) ,
?
?
which is the solution.
Example 7.7 We solve the same differential equation (7.35) under the
same initial conditions as in Example 7.6 by the Laplace transform method.
The above scheme reads.
L[y 00 ] + ? 2 L[y] = aL[1]
s2 Y (s) ? s y(0) ? y 0 (0) + ? 2 Y (s) = a
a
s2 + ? 2 Y (s) = b s + c +
s
1
s
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Y (s) =
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161
bs2 + cs + a
A
B
C
= +
+
s(s2 + ? 2 )
s
s ? i? s + i?
a
bs2 + cs + a
= 2
s?0
s?0
s2 + ? 2
?
bs2 + cs + a
B = lim (s ? i?) Y (s) = lim
s?i?
s?i? s(s + i?)
b(i?)2 + c(i?) + a
b ? 2 ? ic? ? a
=
=
i? и 2i?
2? 2
bs2 + cs + a
C = lim (s + i?) Y (s) = lim
s??i?
s??i? s(s ? i?)
2
?b ? ? ic? + a
=
?i? и (?2i?)
A = lim s Y (s) = lim
y = AL?1
1
1
1
+ BL?1
+ CL?1
s
s ? i?
s + i?
= A + B ei?t + C e?i?t
b ? 2 ? a ? ic? i?t b ? 2 ? a + ic? ?i?t
a
+
e +
e
2
?
2? 2
2? 2
a
1
= 2 + 2 (b ? 2 ? a) cos(?t) + c? sin(?t) ,
?
?
=
which is the solution.
7.4
Mellin transform
In connection with the theory of Fourier and Laplace transforms, we shall
state basics of the theory of Mellin transforms. If we put
e?t = x, f (t) = g(x), FII (s) = GM (s),
in the two-sided Laplace transform
(LII f ) (s) =
Z
then we get the Mellin transform
Z
M (f )(s) =
?
e?ts f (t) dt,
??
?
xs?1 f (x) dx
0
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and the inverse Mellin transform
Z
1
f (s) x?s ds
M ?1 (f ) (x) =
2?i (c)
(where the integral is taken along the vertical Bromwich path Re s = c)
are called the pair of its Mellin transforms. We shall give some typical
examples. First, the Mellin transform of cos x is cos ?2 s ?(s), i.e.
Z ?
? s ?(s).
M (cos)(s) =
xs?1 cos x dx = cos
2
0
To see this it is sufficient to compute M ?1 cos ?2 s ?(s) (x) and we are to
show that
Z
? 1
cos(x) =
x?s cos s ?(s) ds
2?i (c)
2
and this can be easily proved by the Residue Theorem (Theorem A.11), the
convergence of the integral taken for granted (which can be checked by the
Stirling formula (Corollary 5.1))
RHS =
=
?
X
m=0
?
X
Res x?s cos
s=?2m
?
? X
(?1)2m 2m
s ?(s) =
cos m?
x
2
(2m)!
m=0
(?1)m 2m
x = cos x,
(2m)!
m=0
where we used (2.4).
Exercise 7.9
forms.
Similarly as above prove the following pair of Mellin trans-
M (sin)(s) =
Z
?
xs?1 sin x dx = sin
0
1
sin(x) =
2?i
M
1
1+?
Z
(s) =
x?s sin
(c)
Z
?
xs?1
0
1
1
=
1+x
2?i
Z
? s ?(s) ds.
2
?
1
dx =
,
1+x
sin ?s
x?s
(c)
? s ?(s),
2
?
ds
sin ?s
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M e
??
(s) =
e?x =
Z
1
2?i
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163
?
xs?1 e?x dx =?(s),
0
Z
x?s ?(s) ds.
(c)
The second pair appears in Corollary A.4, while the last pair is the most
well-known one appearing as (2.1) above. The improper integrals being all
absolutely convergent for large values of Re s for which the above formulas
hold. Other examples include (7.23) and (B.5).
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Chapter 8
Around Dirichlet?s L-functions
Abstract
In this chapter we shall state rudiments of harmonic analysis on
Z/q Z and its applications to number-theoretic problems, i.e. to
the problems in relation to associated Dirichlet series with periodic coefficients.
In Д8.1 we shall develop the theory of finite Fourier series (or
what amounts to the same, finite Fourier transforms) to such an
extent that is sufficient for our intended number-theoretic applications (based on [IK], [Ka], [Ya]). It is instructive to state the
results in parallel to those in the general theory of Fourier series
(transforms) expounded in Chapter 7.
In Д8.2, we shall establish a remarkable equivalence between
Gauss? formula and a finite expression for the value of the Dirichlet L-function at 1 as expounded in [HKT].
8.1
The theory of periodic Dirichlet series
In what follows q > 1 always indicates the fixed modulus.
Definition 8.1
riod q,
The space C(q) of all arithmetic functions f (n) of pe-
f : Z ? C; f (n + q) = f (n), n ? Z,
165
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Fig. 8.1
Gauss
forms a vector space over C. For f , g ? C(q), we define their inner product
X
(f, g) =
f (a) g(a).
a mod q
Then C(q) becomes a metric vector space with respect to the norm
1
kf k = (f, f ) 2 .
The notions of orthogonality, ONS (orthonormal system), ONB (orthonormal basis) remain the same except that we now speak of the inner
product.
Exercise 8.1 Prove that (1) satisfies the defining properties of a scalar
product (cf. Problem 4.4).
Example 8.1
For each residue class j mod q let ?j be defined by
j
?j (n) = e2?i q n .
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Fig. 8.2
Then
Proof.
1
1
1
?0 , ?1 , и и и , ?q?1
q
q
q
167
Dirichlet
is an ONB of C(q).
This follows from the relation (cf. (8.5))
(
X 2?i j?k
q, j = k
q
e
(?j , ?k ) =
=
0, j =
6 k.
a mod q
Example 8.2
class j mod q:
vista
Let ?j denote the characteristic function of the residue
?j (n) =
(
1, n ? j (modq)
0, n 6? j (modq).
Then {?0 , ?1 , и и и , ?q?1 } is an ONB of C(q).
Definition 8.2
Let (Z/qZ)
О
denote the multiplicative group of reduced
О
[
be an Abelian character, i.e. a
residue classes mod q. Let ? ? Z/qZ
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homomorphism into CО :
?(a?b?) = ?(a?) ?(b?)
and a? = a mod q. We extend the domain of definition of ? by 0-extension,
i.e. we define ?(a) = ?(a?) for (a, q) = 1, and ?(a) = 0 for (a, q) > 1. Then
this ? is a completely multiplicative periodic function called a Dirichlet
character mod q:
? : Z ? C, ?(ab) = ?(a) ?(b), ?(a + q) = ?(a).
О
[
is called
The particular character induced by the identity ?0 of Z/qZ
the principal character mod q and is denoted by ?0 .
? of f with respect to the ONB {?j }:
Define the Fourier transform f(j)
X
? = 1 (f, ?j ) = 1
f (a) ?j (a).
f(j)
q
q
(8.1)
a mod q
Since C(q) =
q?1
M
C?j , we automatically have the Fourier expansion of f :
i=0
f (n) =
q?1
X
? ?j (n)
f(j)
(8.2)
j=0
(cf. Theorem 7.1) for which (8.1) gives, in particular,
??j (a) =
a
1
1
?a (j) = e?2?i q j
q
q
and the Fourier expansion of ?a is
X
?a =
??j (a) ?j ,
(8.3)
(8.4)
j mod q
which is nothing other than the orthogonality of the additive characters
?j ?s:
q?1
X
a=0
?j (a) =
q?1
X
a=0
e
2?i qj a
=
(
q,
j ? 0 (mod q)
0, j 6? 0 (mod q).
(8.5)
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169
For a Dirichlet character ?, we have (8.32) below. Similarly, the Fourier
transform f? of f with respect to {?j } is
X
? = (f, ?j ) =
f(j)
f (a) ?j (a) = f (j).
(8.6)
a mod q
E.g.
??j (n) = ?j (n).
(8.7)
Hence the Fourier expansion of f is
f (n) =
q?1
X
f?(j) ?j (n) =
q?1
X
f (j) ?j (n),
(8.8)
j=0
j=0
which is the same as the decomposition of (positive) integers into residue
classes mod q and holds true for any arithmetic function f : Z ? C, not
necessarily periodic. Especially, in the following form of a series, it is of
great importance:
?
X
n=0
f (n) =
q?1 X
?
X
f (m) ?j (m).
(8.9)
j=0 m=0
The function L(s, ?) defined for a Dirichlet character ? mod q by the
Dirichlet series
?
X
?(n)
,
L(s, ?) =
ns
n=1
?>1
(8.10)
absolutely convergent, is called Dirichlet?s L-function.
For ? 6= ?0 , (8.10) is (conditionally and uniformly) convergent for ? > 1
by Corollary B.1, and we may speak of the value L(1, ?). Formula (8.32)
gives for ? primitive (by (8.40))
q?1
1 X
k
L(s, ?) =
?(k) ls
.
G(?)
q
(8.11)
k=1
Example 8.3
Consider the general Hurwitz-Lerch zeta-function
?(s, a, z) =
?
X
zn
,
(n + a)s
n=0
(8.12)
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a > 0, ? > 1. Then, by (8.9),
?(s, a, z) =
q?1
X
j=0
=
?
X
m=0
m?j ( mod q)
q?1 X
?
X
j6=0
= q ?s
z j+mq
(j + mq + a)s
m=0
q?1
X
j=0
whence
?(s, a, z) = q ?s
zm
(n + a)s
zj
?
X
m=0
(z q )m
m+
j+a
q
s ,
j +a q
z j ? s,
,z ,
q
j=0
q?1
X
(8.13)
which is the most general Kubert identity (or distribution property).
We introduce the vector space of Dirichlet series isomorphic to C(q):
Definition 8.3 Let D(q) denote the space of (formal) Dirichlet series
?
X
f (n)
, with f (n) ? C(q):
ns
a=1
)
(?
X f (n) f ? C(q), ? 1 .
D(q) =
ns n=1
If we impose some growth condition on f , the Dirichlet series is convergent
in some half-plane, ? 1. The spaces C(q) and D(q) are isomorphic
(cf. Yamamoto [Ya]).
We may express (8.8) in terms of the Dirichlet series
?
?
X
X
X
f (n)
?a (n)
f
(a)
=
s
n
ns
n=1
n=1
a mod q
X
=
f (a) ?(s, a mod q),
(8.14)
a mod q
where ?(s, a mod q) signifies the partial zeta-function
X
1
,
?(s, a mod q) =
ns
n?a mod q
(8.15)
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which can be expressed by the Hurwitz zeta-function
a
?s
?(s, a mod q) = q ? s,
,
q
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171
(8.16)
so that it is meromorphic over the whole plane. Hence (8.12) is stated as
?
X
X
f (n)
a
?s
,
(8.17)
=q
f (a) ? s,
ns
q
n=1
a mod q
which continues the left-hand side meromorphically over the whole plane.
Example 8.4
we get
If we trace the above argument in the reverse direction,
X
?
a
?a (n)
=
q ?s ? s,
.
q
ns
n=1
Substituting (8.4) and (8.3) for ?a , we obtain
X
?
X
1
a
=
??j (a) ?j (n) s
q ?s ? s,
q
n
n=1
j mod q
?
X
1 X
j (n)
=
,
?a (j)
q
ns
n=1
j mod q
whence we have
q
?s
a
? s,
q
j
1 X
?a (j) ls
=
,
q
q
(8.18)
j mod q
being valid for all s 6= 1, whose ls (x) indicates the polylogarithm function
defined by (3.3) (cf. Ishibashi [Is, p.447]).
But, if we state (8.18) in the form (the genuine generalization of the
Eisenstein formula)
q?1
X
a
k
a
e?2?i q k ls
= q 1?s ? s,
? ?(s),
(8.19)
q
q
k=1
and interpret the case s = 1 as the limit as s ? 1, then (8.19) is valid for all
s ? C. The limit interpretation of the right-hand side of (8.19) corresponds
to the normalization ls (0) = ?(s) in Milnor [Mi], but even under this, (8.16)
is not valid and only (8.19) stands. We state the limiting case of (8.19) as
Theorem 8.1
The limiting case of (8.19) implies Gauss? formula (8.29).
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For details, cf. [LDH].
Exercise 8.2
Prove that the formula
q?1
r
1 X ?2?i qj r
a+j
? s, a, ? +
, q? =
e
e2?ij? q ?s ? s,
q
q r=0
q
(8.20)
is the two different expressions for the j-component
?
X
e2?im?
?j (s, a, ?) =
?j (m)
(m + a)s
m=0
of
?(s, a, ?) =
q?1
X
?j (s, a, ?),
(8.21)
j=0
where ?(s, a, ?) = ?(s, a, e2?i? ) (cf. (8.12)). Note that one expression is a
consequence of (8.4),
?a (m) =
1 X
1 X 2?i m?a
q j.
a (j) j (m) =
e
q
q
j mod q
Solution
Since ?j (s, a, ?) =
?
X
m=0
m?j ( mod q)
e2?im?
, we may express it, on
(m + a)s
writing m = j + lq, l = 0, 1, и и и , as
?
X
e2?i(lq?+j?)
s
a+j
l=0 l + q
a+j
2?ij? ?s
=e
q ? s,
, q? .
q
?j (s, a, ?) = q ?s
On the other hand, substituting (8.3), we derive that
?j (s, a, ?) =
=
(8.22)
j mod q
q?1
?
X
e2?im? 1 X 2?i m?j
q r
e
s q
(m
+
a)
m=0
r=0
r
q?1
?
1 X ?2?i qj r X e2?i(?+ q )m
e
,
q r=0
(m + a)s
m=0
which is the right-hand side of (8.21).
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Around Dirichlet?s L-functions
We confirm that adding the right-hand side of (8.21) over j mod q, we
obtain ?(s, a, ?), i.e. (8.22) in terms of the right-hand side of (8.21).
As suggested by (5.56)0 , there is a counterpart of (8.11), which is a
decomposition into residue classes mod q:
q?1
1 X
a
L(s, ?) = s
?(a) ? s,
q a=1
q
(8.23)
being valid for any Dirichlet character mod q, not necessarily primitive.
Recall the Laurent expansion for ?(s, x),
?(s, x) =
1
? ?(x) + O(s ? 1),
s?1
s ? 1,
(8.24)
where ?(x) signifies the Euler digamma function (cf. Д5.1)
?(x) =
?0
0
(x) = (log ?(x)) .
?
(8.25)
Also recall the orthogonality of characters
q?1
X
a=1
?(a) =
(
0,
?(q),
if ? 6= ?0 ,
(8.26)
if ? = ?0 ,
where ?0 and ?(q) stand for the principal character mod q and the Euler
function defined by
?(q) =
X
1,
(8.27)
1?a?q
(a,q)=1
respectively.
From (8.23), (8.24) and (8.26) we obtain
q?1
1X
a
L(s, ?) = ?
?(a) ?
+ O(s ? 1),
q a=1
q
s?1
and a fortiori
q?1
a
1X
?(a) ?
L(1, ?) = ?
.
q a=1
q
(8.28)
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For the values of ? pq , we have a formula of Gauss (2.58) which may be
stated by Lemma 8.1 below as
X
p
?
2pk
p
k
= ?? ? log q ? cot ? + 2
cos
?
? log 2 sin ? , (8.29)
q
2
q
q
q
q
k? 2
where ? is the Euler constant (?(1) = ??) ([Bo?h], [Ca], [GR]).
It was D. H. Lehmer [Leh] who first used (8.29) in his study of generalized Euler constants ?(p, q) for an arithmetic progression p mod q. He
deduced (8.29)
[Leh, (11)], and the relation [Leh, Theorem 7], between
from
p
?(p, q) and ? q , and stated ([Leh, p.135]) ?Our proof via finite Fourier
series indicates that Gauss? remarkable result has a completely elementary
basis.?
Our main purpose is to elaborate on this statement of Lehmer and,
on streamlining the argument, to show that (8.29) has a purely numbertheoretic basis and ? is a number-theoretic function. As a converse to this,
we shall also put into practice the statement of Deninger [D, p.180], to the
effect that (8.29) can be used to evaluate L(1, ?). Indeed, Funakura was on
these lines (cf. [Fu, (1)]) but he appealed to the integral representation of
Legendre and applied Lehmer?s argument of using ? log(1 ? e2?ix ), 0 <
x < 1.
8.2
The Dirichlet class number formula
We may now state our main theorem in this chapter.
Theorem 8.2
for L(1, ?).
Gauss? formula (8.29) is equivalent to finite expressions
L(1, ?) =
q?1
a
? X
?(a) cot ?
2q a=1
q
(8.30)
for ? odd and
q?1
a
1 X
?
b(a) log 2 sin ?
L(1, ?) = ? ?
q a=1
q
(8.31)
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175
for ? even, where
?
b(a) =
k
1 X
?(k) e?2?i q a
q
(8.32)
k mod q
is the finite Fourier transform of ? (intimately related to the generalized
Gauss sum G(a, ?), see (8.39) below).
Corollary 8.1
For primitive ?, (8.30) and (8.31) reduce, respectively, to
q?1
?i X
a
L(1, ?odd ) = ?
?(a) B1
G(?) a=1
q
(8.30)0
q?1
1 X
a
L(1, ?even) = ?
?(a) log 2 sin ? ,
G(?) a=1
q
(8.31)0
and
where G(?) = G(1, ?) is the normalized Gauss sum.
Remark 8.1
(i) On symmetry grounds, (8.30) may be stated as
q?1
?i X
a
?
b(a) B1
,
L(1, ?odd ) = ? ?
q a=1
q
(8.30)00
which can be explicitly computed to be (8.30) (cf, e.g. [Fu]).
Although both Funakura [Fu] and Ishibashi-Kanemitsu [IK] treated the
case of periodic functions f (n) of period q, the formulas (8.30)00 and (8.31)
are implicit in Yamamoto?s work [Ya], depending on (8.11) and (7.18) &
(7.19).
(ii) The last statement
of Corollary 8.1 follows, on recalling that the
Kronecker characters |d|
(d < 0) and dи (d > 0) are primitive odd and
и
even characters mod |d|, respectively.
In the course of proof of Theorem 8.2, we shall encounter an interesting
P
number-theoretic function log Nq = ? d|q (х(d) log d) dq which eventually
cancels out in view of the following Theorem 8.3. We believe this function
deserves wider attention and we state
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Theorem 8.3 For q > 1, the number-theoretic function log N (q) =
log Nq admits the following expressions.
log Nq = ?q
X х(d)
d
d|q
log d
?
q?1
X
(8.33)
?
х
q
(a,q)
??
a
??
log?2 sin? ? q
q ?
a=1
(a,q)
q X
?(d) ?
=
d
= ??(q)
(8.34)
(8.35)
d|q
= ?(q)
X log p
,
p?1
(8.36)
p|q
the product extending over all prime divisors of q, where х and ? signify
the Mo?bius function and the von Mangoldt function, respectively.
For a proof cf. [HKT].
8.3
Proof of the theorems
Let f (n) be an arithmetic periodic function of period q:
f : Z ? C;
f (n + q) = f (n),
n ? Z.
We define the parity of f as follows: f is called even if f (?n) = f (n) and
odd if f (?n) = ?f (n).
We prepare some lemmas, of which Lemma 8.1 is repeatedly used in
what follows, without notice.
Lemma 8.1
If f is odd, then
q?1
X
f (a) = 0
a=1
and if f is even, then
q?1
X
a=1
f (a) = 2
X
a? 2q
f (a) = 2
X
a< 2q
f (a) +
1 + (?1)q q .
f
2
2
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Around Dirichlet?s L-functions
In particular, if f and ? mod q are of opposite parity, then
q?1
X
?(a) f (a) = 0
a=1
while if f and ? are of the same parity and q > 2, then
q?1
X
?(a) f (a) = 2
a=1
X
?(a) f (a) = 2
X
?(a) f (a).
a< 2q
a? q2
Lemma 8.2 The ? function satisfies Gauss? multiplicative formula or
the modified Kubert identity
?(x) = log q +
Lemma 8.3
and
q?1 x+a
1 X
?
.
q a=0
q
(8.37)
Let ? denote a Dirichlet character mod q, q ? 3. Then
?
?0
if n 6? ▒1 (mod q),
X
?(n) = ?(q)
?
if n ? ▒1 (mod q).
? even
2
?
?
0
?
?
? ?(q)
X
?(n) =
2
?
?
? odd
?
?? ?(q)
2
if
n 6? ▒1
if
n?1
if
n ? ?1
(mod q),
(mod q),
(mod q),
where the sum is extended over all even and odd characters, respectively.
Proof. For q ? 3, the set {▒1} forms a subgroup of the reduced residue
О
class group G = (Z/qZ) of index 2. Hence the factor group G/{▒1} has
?(q)
order
. Since the set of all even characters coincides with the character
2
group of G/{▒1}, it follows, from the orthogonality of characters, that
X
X
?(a) =
?(n)
? even
\
??G/{▒1}
?
?0
= ?(q)
?
2
if n 6= 1 in G/{▒1},
if n = 1 in G/{▒1},
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which proves the first assertion. The second assertion follows from the first
and the orthogonality relation
(
X
0
if n 6? 1 (mod q)
?(n) =
(8.38)
?(q) if n ? 1 (mod q).
b
? ?G
This completes the proof.
It is instructive to give a proof of Corollary 8.1 first. We introduce the
generalized Gauss sum
G(k, ?) =
q?1
X
a
?(a) e2?i q k
(8.39)
a=1
= q ?(?1) ?
b(k)
(cf. (8.32)) and note that it decomposes into
G(k, ?) = ?(k) G(?)
(8.40)
if and only if ? is primitive ([Ap4], [Da]). We derive (8.30)0 by appealing
to Eisenstein?s formula
q?1 X
p
k
k
?2?i p
q
= B1 ? q B 1
l0
e
q
q
k=1
or rather its converse (cf. [LDH], [Is], [Wa])
q?1
X
k=1
B1
p
p
k
e2?i q k = ?l0
? 1 ? B1
q
q
p
i
= ? cot ?.
2
q
Substituting (8.41) into (8.30), we find that
L(1, ?) =
q?1
?i X
k
B1
G(k, ?).
q
q
k=1
Using (8.38) and other known facts
G(?) = ?(?1) G(?),
we conclude (8.30)0 .
|G(?)|2 = q,
(8.41)
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Around Dirichlet?s L-functions
We may deduce (8.31)0 from (8.31) in a similar way. Substituting (8.32)
into (8.31), we obtain
L(1, ?) = ?
X
q?1
q?1
1X
2?k
k
a
?(a) cos
log 2 sin ?
q
q
q
a=1
(8.42)
k=1
whose inner sum is again G(k, ?). Therefore for ? primitive, we have
q?1
k
G(?) X
?(k) log 2 sin ? ,
L(1, ?) = ?
q
q
k=1
whence (8.31)0 follows in the same way.
We now turn to the proof of Theorem 8.2.
Proof. That (8.29) implies (8.30) and (8.31) is immediate. Indeed, substituting (8.29) in (8.28) and using Lemma 8.1, we obtain (8.30) for ? odd
and (8.42) for ? even, which is the same as (8.31).
Now we are to prove the converse, i.e. we are to deduce (8.29) from
(8.30) and (8.31).
With p, (p, q) = 1, we multiply (8.28) by ?(p?1 ) and sum over ? mod
q, ? 6= ?0 to obtain
X
?0 6=? mod q
?(p?1 ) L(1, ?) = ?
q
a
1X
?
q a=1
q
X
?(ap?1 )
(8.43)
?0 6=? mod q
= S1 + S2 ,
say, where
S1 = ?
X
q
1 X
a
?
?(ap?1 )
q a=1
q
? mod q
and
S2 =
q
1 X
a
?
?0 (ap?1 ).
q a=1
q
By the orthogonality (8.38) of characters,
p
?(q)
?
S1 = ?
.
q
q
(8.44)
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The sum S2 is
q?1
1 X?
a
S2 =
,
?
q a=1
q
the star on the summation sign indicating the sum over all a?s, relatively
prime X
to q, (a, q) = 1, which condition may be replaced by introducing the
х(d). Writing the condition d|(a, q) as d|q, a = a0 d ? q ? 1, we
sum
d|(a,q)
have
0
d ?1
X
1 X
a
S2 =
х(d)
? q
q
0
d
q
a =1
d|q
q
d
X
a+1
q
q
q
?
whose inner sum is
= ? log ? ? + ? by Lemma 8.2.
q
d
d
d
d
a=0
Hence
S2 = ?
1
?(q)
log q
?(q) ? log Nq ?
?,
q
q
q
(8.45)
where log Nq is defined by (8.33).
Substituting (8.44) and (8.45) in (8.43), we conclude that
X
?0 6=? mod q
?(p?1 ) L(1, ?) =
?(q)
q
??
p
1
? log q ?
log Nq ? ? .
q
?(q)
(8.46)
It remains to calculate the left-hand side of (8.46) by dividing the sum
into two parts:
X
?0 6=? even
and
X
? odd
substituting therewith (8.31) and (8.30), respectively.
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Around Dirichlet?s L-functions
First, by (8.31),
X
?(p?1 ) L(1, ?)
(8.47)
?0 6=? mod q
q?1
a
1 X
log 2 sin ?
= ??
q a=1
q
X
?0 6=? even
?(p) ?
b(a)
X
q?1
a
1 X
??0 (p) ?
log 2 sin ?
= ??
c0 (a)
q a=1
q ? even
= T1 + T2 ,
say, where
X
q?1
1 X
1
a
?(p) ?
T1 = ? ?
log 2 sin ?
q a=1
q
q
? even
X
k
?(k) e?2?i q a
(8.48)
k mod q
and
q?1
a
1 X
1
log 2 sin ? ?
T2 = ?
q a=1
q
q
X
k
?0 (k) e?2?i q a .
(8.49)
k mod q
The inner double sum of T1 is
1 X ?2?i kq a X
e
?(kp?1 )
?
q
? even
k mod q
?(q)
p
1 ?(q) ?2?i pq a
p
+ e2?i q a = ? cos 2? a
e
=?
q 2
q
q
by Lemma 8.3, and so
T1 = ?
q?1
?(q) X
a
p
cos 2 a? log 2 sin ? ,
q a=1
q
q
while the inner sum for T2 is
X
? ?2?i kq a
e
k mod q
which is equal to
х
?(q)
?
q
(a,q)
q
(a,q)
,
(8.50)
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by Ho?lder?s result (cf. [Leh, p.133]).
Hence
q
х
q?1
X
(a,q)
?(q)
a
T2 =
log 2 sin ?
q
q a=1
q
?
(a,q)
but this is ? q1 log Nq by (8.34), i.e.
T2 = ?
1
log Nq .
q
(8.51)
Substituting (8.50) and (8.51) in (8.47), we obtain
q?1
X
?(q) X
a
1
p
?1
?(p ) L(1, ?) = ?
cos 2? a? log 2 sin ? ? log Nq .
q a=1
q
q
q
?0 6=? even
(8.52)
On the other hand, by (8.30) and Lemma 8.3,
X
q?1
a X
? X
cot ?
?(ap?1 )
2q a=1
q
? odd
? ?(q)
p
?(q)
?p
=
cot ? ?
cot
?
2q
2
q
2
q
??(q)
p
=
cot ?.
2q
q
?(p?1 ) L(1, ?) =
(8.53)
Combining (8.52) and (8.53) implies
X
?(p?1 ) L(1, ?)
(8.54)
? odd
?0 6=? mod q
q?1
p
?(q) X
a
1
?(q)
p
cos 2 a? log 2 sin ? ? log Nq +
=?
? cot ?.
q a=1
q
q
q
2q
q
Comparing (8.46) and (8.54), we see that the terms involving log Nq
cancel each other and (8.29) follows. This completes the proof.
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Appendix A
Complex functions
A.1
Function series
Fig. A.1
D?alembert
Theorem A.1 A uniformly convergent series of analytic functions may
be integrated term by term along any curve inside the region of uniform
convergence. Namely, if the functions
f1 (z), f2 (z), ...
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are analytic in D and the series
?
X
fn (z) = f (z)
n=1
is uniformly convergent in D, then for any curve C ? D, we have
Z
f (z) dz =
Z X
?
fn (z) dz =
C n=1
C
? Z
X
n=1
fn (z) dz.
C
Proof. fn (z) need not be analytic but enough to be continuous in D
(since analyticity ? continuity, the assumption
R is excessive). Since f (z) is
continuous in D, it follows that the integral C fn (z) dz, n ? N exists. So
n
R
P
fi (z). Since sn (z) converges to
does the integral C sn (z) dz for sn (z) =
i=1
f (z) uniformly on D, we have
?? > 0, ?n0 = n0 (?) ? N s.t. n > n0 ? |sn (z) ? f (z)| < ?,
?z ? D.
Hence for n ? n0 , we have
Z
(sn (z) ? f (z)) dz < ? ?(C),
C
where ?(C) is the length of C so that
Z
Z
lim
sn (z) dz =
f (z) dz
n??
C
C
whose left-hand side is nothing other than the definition of
? Z
X
n=1
fn (z) dz.
C
Theorem A.2 The limit of the uniformly convergent series of analytic
functions is interchangeable with integration along any curve lying in the
region of its uniform convergence. Namely, if
f1 (z), f2 (z), ...
are analytic in D and
lim fn (z) = f (z),
n??
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Complex functions
uniformly in D, then for any Jordan curve C ? D, we have
Z
Z
Z
f (z) dz = lim
fn (z) dz =
lim fn (z) dz.
C
n??
C n??
C
Definition A.1 If a sequence (respectively, series) of functions defined
on D are uniformly convergent on any bounded closed subset of D (i.e.,
on any compact subset D 0 such that D0 ? D), we say that the sequence
(respectively, series) is uniformly convergent on D in the wide sense.
Theorem A.3
If the functions
f1 (z), f2 (z), ...
are (i) analytic in D and (ii) the series
?
X
fn (z)
n=1
is uniformly convergent in D in the wide sense, then its sum
?
X
f (z) =
fn (z)
n=1
is analytic in D and its derivative may be obtained by termwise differentiation:
0
f (z) =
?
X
fn0 (z)
n=1
Also, the termwise differentiated series is uniformly convergent in the wide
sense in D.
Corollary A.1
Any function series
?
X
fn (z) := f (z)
n=1
that is uniformly convergent in the wide sense in D is termwise differentiable infinitely many times:
f (k) (z) =
?
X
n=1
fn(k) (z),
k ? N.
(and the k-times differentiated series is also uniformly convergent in the
wide sense in D.)
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Proof. We shall prove both Theorem A.3 and its Corollary A.1 at the
same time. By the Cauchy integral formula in Theorem A.7, we have for
any rectifiable simple curve C in D and any point z in C,
?
? Z
X
fn (w)
1 X
dw.
f (z) =
fn (z) =
2?i
w?z
n=1
n=1 C
But
?
P
fn (w) is uniformly convergent on C, and so Theorem A.1 allows
n=1
us to integrate term by term after multiplying (w ? z)?1 :
f (z) =
1
2?i
Z X
Z
?
1
fn (w)
f (w)
dw =
dw.
2?i C w ? z
C n=1 w ? z
Hence the Cauchy integral formula holds for f (z) and so it follows that
f (z) is analytic in C and that
Z
f (w)
k!
dw, k ? N ? {0}.
f (k) (z) =
2?i C (w ? z)k+1
Let
Sn (z) =
n
X
fk (z)
k=1
be the n-th partial sum of
?
X
fn (z)
n=1
and take any bounded closed subset D 0 in D. Then take any simple closed
contour C ? D of finite length containing D 0 and suppose dist (D 0 , C) =
? > 0. The we have
Z
Sn (w)
k!
(k)
dw, k ? N ? {0}.
Sn (z) =
2?i C (w ? z)k+1
Hence, it follows that
k!
?(C) max |f (w) ? Sn (w)| ,
w?C
2?? k+1
whence we have
?? > 0,
?n0 = n0 (?) ? N s.t. n > n0 ?
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Complex functions
|f (w) ? Sn (w)| < ?
on C. Hence,
lim Sn(k) (z) = f (k) (z).
n??
uniformly on D0 .
Corollary A.2 (The Weierstrass double series theorem) Suppose
{fn (z)} are analytic in |z ? z0 | < r and has the Taylor expansion
fn (z) =
?
X
k=0
(n)
ak (z ? z0 )k .
Then if
?
X
fn (z) = f (z)
n=1
uniformly in the wide sense on |z ? z0 | < r, then f (z) is analytic on
|z ? z0 | < r and its Taylor expansion is given by
f (z) =
?
X
k=0
ak (z ? z0 )k
(|z ? z0 | < r) ,
ak =
?
X
(n)
ak .
n=1
That is, the iterates of the double series coincide ? the order of summation
being interchangeable ?
!
? X
?
?
?
X
X
X
(n)
k
k
ak (z ? z0 ) =
fn (z) = f (z) =
ak (z ? z0 )
n=1 k=0
n=1
k=0
Proof. This is a special case of Theorem A.2. The relation between
coefficients follows from the Theorem A.8:
ak =
Theorem A.4
?
?
X
1 (k)
1 X (k)
(n)
f (z0 ) =
fn (z0 ) =
ak .
k!
k! n=1
n=1
If {fn (z)} are analytic in D and
lim fn (z) = f (z)
n??
uniformly in the wide sense on D, then f (z) is again analytic on D and
f (k) (z) = lim fn(k) (z)
n??
uniformly in D in the wide sense.
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Most of the above results on infinite series apply to infinite integrals
in spite of the fact that in the case of partial sums SnR(z), n goes to ?
b
discretely while in the case of partial integral I(b) = 0 f (z) dz, b goes
to ? continuously.
To assure the uniform convergence in case the series (integrals) are absolutely convergent, the main tool is Weierstrass? M-test (Majorant series
test) which in the case of integrals asserts that given a (complex-valued)
function f (x, y), x ? [a, b] (resp (a, b) as the case may be), y ? Y , if there
is a positive (-valued) function M (x) such that for any y ? Y ,
|f (x, y)| ? M (x)
and
Z
?
a
M (x) dx < ? (resp.
Z
b
a
M (x) dx < ?),
R?
Rb
then a f (x, y) dx (resp. a M (x) dx < ?) is absolutely and uniformly
convergent on Y .
Following Titchmarsh [Tit], we often refer to this as ?by absolute convergence.?
If the series or integrals are convergent but not absolutely convergent,
i.e. conditionally convergent, we need to appeal to more delicate convergence tests such as Dirichlet?s (cf. ДB.2).
Exercise A.1 Noting that the principal branch of the natural logarithm
log z (often denoted by Log z) may be defined by the Condition
1
d
log z = , log 1 = 0,
dz
z
prove the integral representation (Re z > 0)
Z ? ?t
e ? e?zt
dt.
log z =
t
0
(A.1)
(A.2)
e?(z?1)t ? 1
Since the integrand f (t) = f (tz) = ?e?t
? z?1
t
Z ?
as t ? 0, the improper integral
f (t) dt is absolutely convergent. Hence
Solution
0
we may differentiate under the integral sign to get
Z ?
Z ?
d
1
f (t) dt =
e?zt dt = .
dz 0
z
0
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Complex functions
Since f (t, 1) = 0, we have log 1 =
satisfied.
Example A.1 (Power series)
polynomial
R?
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f (t, 1) dt = 0, and Condition (A.1) is
Series in the form of an infinite degree
f (z) =
?
X
n=0
an (z ? z0 )n ,
centered at z0 , is called a power series centered at z0 . By translation it
suffices to consider the power series at 0. We speak of absolute convergence of power series, and the region of (absolute) convergence of a power
series is a disc with the boundary called the circle of convergence whose
radius is called the radius of convergence and is most conveniently given by
D?Alembert?s formula
an r = lim n?? an+1 if the limit exists (including ? ).
We consider the case where 0 < r ? ? and denote the region of convergence
by D.
By above theorems we have
Theorem A.5 Inside D, the region of convergence, (i) a power series
may be integrated term by term along any path lying in D, (ii) a power
series is analytic in D and the derivatives may be obtained by term by term
P
P
differentiation and (iii) two power series
an z n ,
bn z n (in their common
region D of convergence) may be added, subtracted, multiplied and divided;
in particular, the multiplication is carried out by the Cauchy product
!
! ?
?
?
X
X
X
n
m
bn z
=
cl z l ,
am z
n=0
m=0
cl =
X
l=0
a m bn
m+n=l
(cf. Remark 1.1); the division is carried out exactly as we do with ordinary
polynomials: e.g. to check the numerical values of Bernoulli numbers in
1 2
1 3
Example 1.1, we may divide z by z + 2!
z + 3!
z + и и и to obtain
1
1
1
z
6 2
30 4
=
1
?
z
+
z
?
z +иии
ez ? 1
2
2!
4!
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Similarly, we may carry out the division in Exercise 5.4.
To sum up, Theorem A.5 says that we may treat power series as ordinary
polynomials (, which disposition is due to Euler).
Example A.2 (Dirichlet series) For an increasing sequence of positive
?s log ?n
reals ?n , the series of functions fn (s) = ??s
, with logarithm
n = e
indicating the principal branch,
?
X
??s
n
n=1
is called the Dirichlet series. Contrary to the case of power series, the region
of absolute convergence of Dirichlet series is a right half-plane and we may
speak about the abscissa of absolute convergence, often denote by ?a .
We have a counterpart of Theorem A.5 for Dirichlet series.
Theorem A.6
Let ?a denote the abscissa of absolute convergence of
?
P
??s
a Dirichlet series f (s) =
n . Then in the region ? > ?a , f (s) is
n=1
analytic, integrable and differentiable term by term. Two Dirichlet series
?
?
P
P
f (s) =
am m?s and g(s) =
bn n?s may be multiplied by the Dirichm=1
n=1
let convolution:
f (s) g(s) =
?
X
cl l?s ,
cl =
l=1
X
a m bn .
mn=l
The last sum is often expressed as
X
X
cl =
ad bl/d =
al/d bd ,
d|l
d|l
with d|l meaning that d runs though all positive divisors of l.
Example A.3 (i) The integral in (2.1) defining the gamma function is
(absolutely and) uniformly convergent in the wide sense in ? > 0. (ii) The
series (3.2) defining the Riemann zeta-function is (absolutely and) uniformly convergent in the wide sense in ? > 1.
Proof. (i) The integral is improper at both end points. We apply Weierstrass? M-test to e?x x??1 . Let s lie in the compact region 0 < ?0 ? ? ? R,
|t| ? R, R > 0. Then for 0 < x < 1, e?x x??1 < e?xRx?0 ?1 = O(x?0 ?1 ) and
1
x > 1, e?x x??1 < e?x xR?1 = O(x?2 ). Since 0 t?0 ?1 dt = O(1) and
Rfor
? ?2
dt = O(1), we conclude the assertion.
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(ii) Let 2m?1 ? N < 2m . Them
N
X
n=1
n?? < 1 + 2 и 2?? + 4 и 4?? + и и и + 2m?1 (2m?1 )??
= 1 + 21?? + (21?? )2 + и и и + (21?? )m?1
1
.
<
1 ? 21??
P?
Hence n=1 n?? < ?, and the series is absolutely convergent for ? > 1.
We may also apply (B.4) to obtain
Z ?
X
1
x1??
1
??
+ +
??
n =
B 1 (w)???1 du + O(x?? ), ? > 0
??1 2 1??
1
n?x
= O(1),
? > 1.
Theorem A.7 (Goursat) If f (z) is analytic in a domain D, them it
has all orders of derivatives f (k) (z), which are also analytic in D, given by
the Cauchy integral formula
Z
f (w)
1
f (k) (z)
=
dw
k!
2?i C (w ? z)k+1
Z
1
dk f (w)
=
dw,
2?i C dz k w ? z
where C is a closed Jordan curve contained in D.
This theorem of Goursat draws a clear line between analytic functions
and real differentiable functions. The requirement that a function is analytic at a point (in the neighborhood of a point) is such a stringent restriction that it already implies the existence of derivatives of all orders.
Theorem A.8 (Cauchy-Taylor) If f (z) is analytic at z0 , then it can
be expanded into the Taylor series in the maximal circle contained in the
domain D of analyticity:
f (z) =
?
X
n=0
an (z ? z0 )n ,
where an is given by (Theorem A.7)
1
f (n) (z0 )
=
an =
n!
2?i
Z
C
dn f (w)
dw,
dz0n w ? z0
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C being any closed contour contained in D.
Actual determination of Taylor coefficients may be done by the method
of undetermined coefficients.
Theorem A.9 (Consistency Theorem or the Principle of Analytic
Continuation) If two functions f (z) and g(z) are analytic in a domain
D and f (z) = g(z) on a subset of D containing an accumulation point,
then f (z) = g(z) all over D.
This theorem often applies when two functions f (z), g(z) coincide on a
segment ? R, in which case we may extend the domain of analytic functions.
Corollary A.3 If a function of the real variable x is real analytic at x0 ,
i.e. it has the power series expansion in the neighborhood |x ? x0 | < r,
r>0
f (x) =
?
X
n=0
an (x ? x0 )n ,
an =
f (n) (x0 )
,
n!
then there is a unique function f (z) analytic on |z ? x0 | < r and coinciding
with f (x) on |x ? x0 | < r, which is given by
?
X
f (n) (x0 )
(z ? x0 )n .
f (z) =
n!
n=0
This is called an analytic continuation of f (x). Most of elementary
functions have their analytic continuation as examples of Corollary A.3:
ez , sin z, cos z. In a word, given a real power series in x, we get its analytic
continuation by changing x by z, the complex variable.
Theorem A.10 (Laurent expansion) If f (z) is (one-valued and) analytic in the annulus (ring-shaped domain) D : r < |z ? z0 | < R (0 < r < R),
then we have the Laurent expansion of f (z):
f (z) =
?
X
n=??
an (z ? z0 )n ,
where the n-th Laurent coefficient an is given by (r < ? < R)
Z
1
f (z)
an =
dz
2?i |z?z0 |=? (z ? z0 )n+1
Z
dn
1
1
f (z) n (z ? z0 )?1 dz, n ? Z
=
2?i |z?z0 |=? n!
dz0
(A.3)
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The Laurent series (A.3) converges uniformly in any annulus contained
in D.
1 (n)
f (z0 ) are the Taylor coefficients. The negative
For n ? 0, an = n!
P?1
power part n=?? an (z ?z0 )n is called the principal part of f (z) at z = z0 .
P?1
If the principal part is finite, n=?m an (z ? z0 )n , a?m 6= 0, say, then
f (z) is said to have an m-th pole at z = z0 , and the coefficient a?1 is called
the residue of f (z) at z = z0 , denoted by
Z
1
f (z) dz.
(A.4)
a?1 = Resz=z0 f (z) =
2?i |z?z0 |=?
If z = z0 is a pole of order m of f (z), then the residue may be calculated
in a similar way as the Taylor coefficients, by the method of undetermined
coefficients:
Clearing the denominator, we have
(z ? z0 )m f (z) = a?m + a?m+1 (z ? z0 ) + и и и + a?1 (z ? z0 )m?1 + и и и .
Hence differentiating m ? 1 times, we get
dm?1
((z ? z0 )m f (z)) = (m ? 1)! a?1 + O(z ? z0 ).
dz m?1
Hence
Resz=z0 f (z) = a?1 =
1
dm?1
lim
((z ? z0 )m f (z))
(m ? 1)! z?z0 dz m?1
(A.5)
which is applicable to many similar settings. It is advisable to remember
the process rather than Formula (A.5).
A.2
Residue theorem and its applications
Theorem A.11 (The (Cauchy) Residue Theorem)
Let C be a
piecewise smooth (Jordan) curve. Suppose f (z) is analytic in a region D
containing C except for finitely many singularities (which we may suppose
are poles) z1 , z2 , и и и , zn (n = 0 inclusive). Then we have
Z
n
X
f (z) dz = 2?i
Resz=zi f (z).
(A.6)
C
i=1
Remark A.1 By Theorem A.11, the value of the integral may be determined by computing the residues, which, as stated above, amounts to
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clearing the denominator. The vacuous case n = 0 implies the most fundamental Cauchy Integral Theorem which asserts that the integral along a
closed curve contained in the region of analyticity is 0, which in turn originates from the fact that in this case the region can be made to shrink into
a point, a topological feature of analytic functions in a region.
Theorem A.12 Suppose f (z) is a rational function in z satisfying the
conditions (i) it has no poles on the positive real axis and (ii) for some
a ? R, a 6? Z, we have
lim z a+1 f (z) = 0,
z?0
lim z a+1 f (z) = 0.
z??
Then,
Z
?
xa f (x) dx =
0
2?i X
Res (z a f (z)) ,
1 ? e2?ia
(A.7)
z6=0
where the power function is defined by z a = exp (a Log z), Log z signifying
the principal branch.
Proof. First note that the improper integral (A.7) is absolutely convergent both at 0 and ? by the Weierstrass M-test. Since f (z) has only finitely
many poles, we may choose 0 < r < R such that all the poles of f (z) other
than the origin lies in the annulus r < |z| < R. Let D denote this annulus
with branch cut along the positive real axis, i.e. its boundary consisting
of the curves C1 : starting from r and moving along the upper edge of the
positive real axis to R, moving along the bigger circle CR and returning
back to the point R, then moving along the lower edge of the positive real
axis to r, moving along the smaller circle cr , and returning to the starting point r. C1 : z = x, 0 ? x ? R; CR : z = Rei? , 0 < ? < 2?;
C2 : z = xe2?i , x : R ? r; cr : z = re?i? , 0 < ? < 2?. Then we apply
the residue theorem to this cut region. Since the argument of z increases
from 0 to 2?, we have
Z r
Z
a
z f (z) dz =
xa e2?ia f (x) dx,
R
C2
whence it follows that
1?e
2?ia
Z
?
xa f (x) dx = 2?i
0
X
z6=0
Res (z a f (z)) .
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Complex functions
We shall assign a precise meaning to this procedure. We integrate the
branch
g(z) = z a f (z),
0 < arg z < 2?
(with branch cut along the positive real axis as above) along CR and cr :
Z
Z
g(z) dz,
g(z) dz.
CR
cr
By dividing the annulus by any ray starting from the origin and lying
inside the second and the third quadrants, we introduce two regions D1
and D2 with branch cut along the negative and positive imaginary axis,
respectively. For concreteness? sake, we choose the negative real axis (any
ray can do if there are no poles of the integrand on it):
L : z = xe?i , x : R ? r
and ?L. Now integrate the branch
g1 (z) = z a f (z),
3
1
z 6= 0, ? ? < z < ?
2
2
along ?D1 = C1 + CR,1 + L + cr,1 , to get
Z R
Z
Z
Z
a(Log x+i0)
e
f (x) dx +
g1 +
g1 +
r
CR,1
L
g1
cr.1
= 2?i Res ea(log|z|+i arg z) f (z),
z?D1
where CR,1 (resp. cr,1 ) signifies the upper half of CR , (resp. cr ,) . Also
integrating the branch
g2 (z) = z a f (z),
z 6= 0,
5
1
?<z< ?
2
2
along
?D2 = C2 + CR,2 + (?L) + cr,2 ,
to get
?
Z
R
ea(Log x+i0) f (x) dx +
r
Z
g2 +
CR,2
= 2?i Res ea(log|z|+i arg z) f (z),
z?D2
Z
g2 +
?L
Z
g2
cr.2
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where CR,2 (resp. cr,2 ) signifies the lower half of CR , (resp. cr , ). Now note
that except on the positive real axis,
g1 (z) = g(z),
D1 ? ?D1
g2 (z) = g(z),
D2 ? ?D2 ,
and
so that
Z
+
C
Z
+ 1 ? e2?ia
cr
Z
R
ea log x f (x) dx =
r
X
Res z a f (a),
z?D
where D is the union of D1 and D2 with L and ?L removed:
D = (D1 ? D2 ) ? L ? (?L).
Now
Z
and
=
CR4
Z
=
cr
Z
Z
2?
a ia?
R e
f Re
i?
0
0
2?
i?
iRe d? = O
ra eia? f rei? irei? d? = O
Z
Z
R
0
f Re d?
i?
Ra+1 f Rei? d? .
0
?
xa f (x) dx
0
thereby completing the proof.
Corollary A.4
a+1 2?
whence as r ? 0+ and R ? ?, we obtain
Z
Z
z a f (z) dz ? 1 ? e2?ia
?D
2?
For 0 < Re a < 1 we have
Z ? ?a
?
x
dx =
.
x+1
sin ?a
0
We note that Exercise 2.3 could be thought of as giving the value if we
assume those formulas appearing there.
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Appendix B
Summation formulas and convergence
theorems
B.1
Summation formula and its applications
Lemma B.1 (The General Newton-Leibniz Principle) Suppose f :
[a, b] ? R is differentiable on [a, b] except for a finite number of points and
?
+
that f 0 (x) = 0 (at endpoints we assume ddx f (a), ddx f (a) exist.) Further,
suppose f (x) is continuous on [a, b]. Then f is a constant on [a, b].
Proof. If f (x) is differentiable at each point of (c0 , c2 ) ? [a, b] except
for c1 , then by the standard Newton-Leibniz Principle, we have f (x) = C1
(c0 < x < c1 ), and f (x) = C2 (c1 < x < c2 ). By continuity, we obtain
C1 = lim f (x) = lim f (x) = C2 . Therefore, f is a constant on
x?c1 ?0
x?c1 +0
(c0 , c2 ). In case f is not differentiable at the endpoint a, by the continuity
at a, we have f (a) = lim f (x) = C1 .
x?a+0
Theorem B.1 (Abel Summation Formula)
P
a(n),
putting A(t) =
If f (t) ? C 1 [a, x], then
a<n?t
X
a<n?x
a(n) f (n) = A(x) f (x) ?
Z
x
A(t) f 0 (t) dt.
(B.1)
a
Proof. (The first proof due to Arhipov and Chubarikov.) For x > a(? 0),
put
F (x) =
X
a<n?x
Z x
G(x) = ?
a(n) f (n) ? A(x) f (x),
A(t) f 0 (t) dt.
a
197
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Then for x = m ? Z we have
lim F (x) =
x?m+0
X
a<n?m
f (n) a(n) ? f (m) A(m).
On the other hand,
lim F (x) =
x?m?0
X
a<n?m?1
=
X
a<n?m
=
X
a<n?m
=
X
a<n?m
f (n) a(n) ? f (m) A(m ? 1)
f (n) a(n) ? f (m) a(m) ? f (m) A(m ? 1)
f (n) a(n) ? f (m) (a(m) + A(m ? 1))
f (n) a(n) ? f (m) A(m)
whence it follows that F (x) and G(x) are continuous on [a, x] and piecewise
differentiable on [a, x]. Furthermore
F 0 (x) = G0 (x),
x?
/ Z.
Hence, Lemma B.1 applies, and we have F (x) = G(x) + C. And since
F (a) = G(a) = 0, it follows that F (x) = G(x), i.e. (B.1) ensues.
The second proof uses a special case of the formula for integration by
parts in the theory of Stieltjes integrals stated in the following theorem:
Theorem B.2 Suppose f (t) is continuous Ron [a, b] and that ?(t) is of
b
bounded variation. Then the Stieltjes integral a f (t) d?(t) exists. Further,
Rb
if f (t) is of bounded variation and ?(t) is continuous, then a ?(t) df (t)
exists and the formula for integration by parts
Z b
h
ib Z b
f (t) d?(t) = f (t)?(t) ? ?(t) df (t)
(B.2)
a
a
a
holds true.
Proof. (The second proof.) Putting ?(t) = A(t) =
P
a<n?t
P
a(n), ?(t) is
a step function and so of bounded variation. Since
a(n) f (n) =
a<n?x
Rx
a f (t) d?(t), it follows from (B.2) that
Z x
h
ix Z x
X
a(n) f (n) =
f (t) dA(t) = f (t) A(t) ?
A(t) df (t). (B.3)
a<n?x
a
a
a
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Since f ? C 1 , the last integral may be written as
leads to (B.1).
Rx
a
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199
A(t)f 0 (t) dt, and (B.3)
Rb
The Stieltjes integral a f (t) d?(t) exists under weaker conditions than
those stated in Theorem B.2.
Theorem B.3 If f (t) and ?(t) are both of bounded variation on [a, b]
and have no common discontinuity, then each one is integrable with respect
to the other from a to b.
Proof. (The third proof of Theorem B.1.) Substituting a(n) = A(n) ?
A(n ? 1), we see that
LHS =
X
A(n) f (n) ?
a<n?x
X
A(m) f (m + 1)
a?1<m?x?1
[x]?1
= ?A([a]) f ([a]) +
X
A(n) (f (n) ? f (n + 1)) + A(x) f ([x]) ,
n=[a]+1
where
R n+1as usual [x] is the integral part of x. Using f (n) ? f (n + 1) =
? n f 0 (t) dt and noting that A(t) = A(n), n < t < n + 1, we obtain
X Z
[x]?1
LHS = ?
=?
n=[a]+1
Z
n+1
A(t) f 0 (t) dt + A(x) f ([x])
n
[x]
A(t) f 0 (t) dt + A(x) f ([x]) .
[a]+1
We rewrite the RHS as
?
Z
x
A(t) f 0 (t) dt +
a
= A(x) f (x) ?
Z
x
Z
[a]+1
A(t) f 0 (t) dt +
a
Z
x
A(t) f 0 (t) dt + A(x) f ([x])
[x]
A(t) f 0 (t) dt.
a
Proof. (The fourth proof.) We transform the integral on the RHS of (B.1)
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by changing the order of summation and integration,
Z x
Z x X
? A(t) f 0 (t) dt = ?
a(n) f 0 (t) dt
a
a a<n?t
=?
=?
X
a(n)
a<n?x
X
Z
x
f 0 (t) dt
n
a(n) f (x) ? f (n)
a<n?x
= ?f (x)
X
a(n) +
a<n?x
X
a(n) f (n),
a<n?x
which leads to (B.1).
Theorem B.4 (Euler?s summation formula)
Let B1 (x) =
B1 (x ? [x]) = x ? [x] ? 21 denote the 1-st periodic Bernoulli polynomial
(cf. (7.9)). Then for f (t) ? C 1 ([a, x]), we have
Z x
h
ix Z x
X
f (t) dt ? B1 (t) f (t) +
f (n) =
B1 (t) f 0 (t) dt.
a<n?x
a
a
a
Putting a(n) = 1 in (B.1), we have
X
A(t) =
1 = [t] ? [a] = t ? B1 (t) ? a ? B1 (a) .
Proof.
a<n?t
Hence (B.1) reads
X
f (n)
a<n?x
Z
x
f (x) ?
t f 0 (t) dt
a
Z x
+ a ? B1 (a) (f (x) ? f (a)) +
B1 (t)f 0 (t) dt
a
Z x
ix Z x
h
ix h
f (t) dt +
= t ? B1 (t) f (t) ? t f (t) +
B1 (t) f 0 (t) dt,
= x ? B1 (x) ? a ? B1 (a)
a
a
a
a
which leads to the RHS.
Proof. (a la? Arhipov and Chubarikov) Putting
X
F (x) =
f (n) ? B1 (a) f (a) + B1 (x) f (x)
a<n?x
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and
G(x) =
Z
x
a
f (t) + B1 (t) f 0 (t) dt,
we shall show that F (x) = G(x). Clearly F (a) = G(a) = 0. G(x), being a
function of the upper limit of integration. is continuous and so is F (x) for
x?
/ Z by definition. And for x = m ? Z, we note that
X
lim F (x) =
x?m+0
f (n) ? B1 (a) f (a) ?
a<n?m
1
f (m) (= F (m)) = lim F (x),
x?m?0
2
i.e. F (x) is continuous for x = m ? Z. This is because the positive jump
P
f (n) at x = m cancels the negative jump of F2 (x) =
of F1 (x) =
a<n?x
B1 (a)f (a) ? B1 (x)f (x), and their sum F (x) is continuous at x = m. Also
it is clear that at x ?
/ Z, both G(x) and F (x) are differentiable and G0 (x) =
F 0 (x). Hence the Newton-Leibniz principle in Lemma B.1 applies.
Theorem B.5 (the Euler-Maclaurin summation formula [Wal]. A
general form of Theorem B.4)
Let B r (x) = Br (x ? [x]) denote the
r-th periodic Bernoulli polynomial. Then for f (t) ? C l [a, x], we have
X
f (n) =
a<n?x
Z
x
l
o
X
(?1)r n
Br (x) f (r?1) (x) ? Br (a) f (r?1) (a)
r!
r=1
Z x
Bl (t) f (l) (t) dt.
f (t) dt +
a
+
(?1)l+1
l!
a
Proof. (The first proof.) Apply integration by parts to Theorem B.4.
Proof. (The second proof.) (Similar to the second proof of Theorem B.4)
Putting
F (x) =
X
f (n) ?
a<n?x
l
o
X
(?1)r n
Br (x) f (r?1) (x) ? Br (a) f (r?1) (a)
r!
r=1
and
G(x) =
Z x
a
(?1)l
(l)
Bl (t) f (t) dt,
f (t) +
l!
we note that F (a) = G(a) = 0, and F 0 (x) = G0 (x), ? x ?
/ Z.
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B.2
Application to the Riemann zeta-function
We illustrate our theory in ДB.1 by specifying to the Riemann zeta-function.
Theorem B.4 with a = 1, x > 0, f (u) = u?s , s ? C gives, on noting
B 1 (1) = ? 12 , B1 (1) = 21 ,
X
n?s =
1<n?x
=
Z
ix Z
h
u?s du ? B1 (u) u?s +
x
1
1
x
1
B1 (u) ?s u?s?1 du
x1?s
1
1
+
? B 1 (x) x?s ? ? s
1?s s?1
2
Z
x
B 1 (u) u?s?1 du,
1
which may be expressed for ? > 0 as
X
Z x
1
x1?s
1
+ +
? B 1 (x) x?s ? s
B 1 (u) u?s?1 du
s?1 2 1?s
1
Z ?
1
x1?s
1
+ +
?s
=
B 1 (u) u?s?1 du + O x?? . (B.4)
s?1 2 1?s
1
n?s =
n?x
For ? > 1, letting x ? ?, we obtain
1
1
?(s) =
+ ?s
s?1 2
Z
?
B 1 (u) u?s?1 du.
(B.5)
1
We note that the RHS is meromorphic in ? > 0, providing a meromorphic
continuation of the LHS to ? > 0. We now showR that the integral is
?
analytic for ? > ?1. For this it suffices to show that 1 B 1 (u) u?s?1 du is
uniformly convergent. To this end we may apply the mean value theorem
or integration by parts:
Z
?
B 1 (u) u?s?1 du =
1
1
B 2 (u) u?s?1
2
?
1
+
s+1
2
Z
?
B 2 (u) u?s?2 du.
1
1
The first term is?
2 B2 and the second term is absolutely convergent for
R?
? > ?1, whence 1 B 1 (u) u?s?1 du is analytic for ? > ?1 and it follows
that (B.5) holds and ?(s) is analytic for ? > ?1 except for s = 1, where it
has a simple pole with residue 1.
Now if we restrict to ?1 < ? < 0, then
s
Z
1
B 1 (u) u
0
?s?1
du = s
Z 1
0
1
u?
2
u?s?1 du = ?
1
1
? .
s?1 2
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Hence (B.5) is transformed into
Z ?
?(s) = ?s
B 1 (u) u?s?1 du,
0
?1 < ? < 0,
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203
(B.6)
where the RHS of (B.6) is meromorphic in ?1 < ?. We substitute the
boundedly convergent Fourier series (7.9) for B 1 (x).
Lebesgue?s theorem allows us to integrate term by term to obtain:
Z
?
?X
sin 2?nu ?s?1
u
du
n
0
n=1
Z
?
s X 1 ? ?s?1
=
u
sin 2?nu du.
? n=1 n 0
?(s) =
s
?
We apply a formula in the theory of Mellin transforms (cf. Д7.4)
Z ?
1
z?1
?z
x
sin ax dx = a ?(z) sin ?z , ?1 < Rez < 1.
2
0
(B.7)
(B.8)
Substituting (B.8) in (B.7) we deduce that
?
? X
1
1 s
s
(2?) s ?(?s) sin ? s
n
?
2 n=1 n
? = 2 (2?)s?1 ?(1 ? s) sin s ?(1 ? s)
2
?(s) =
(B.9)
for ?1 < ? < 0, where we used (2.5). For ? < 0, ?(1 ? s) has the Dirichlet
series expression. Formula (B.9) is the asymmetric form of the functional
equation (5.54) for the Riemann zeta-function.
Example B.1
Solution
Find the value of ?(2).
Applying (2.15) to rewrite (B.9) as
?(s) = (2?)s?1
?
?(1 ? s)
?(s) cos ?2 s
(B.10)
Putting s = 2 in (B.10), we get
?(2) = 2?
?
?(?1) = 2? 2 (??(?1))
?(2) (?1)
and we are led to find the value of ?(?1), which we may find with the aid
of (4.3).
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Theorem B.6 (Abel?s continuity theorem) Suppose the power series
?
X
f (z) =
an z n converges at the point z0 on the circle of convergence.
n=0
?
Then as long as z approaches z0 in the angular region |arg(z ? z0 )| < ,
2
?
X
n
the value f (z) approaches the value
an z0 :
n=0
lim f (z) =
z?z0
?
X
an z0n .
n=0
Theorem B.7 (Generalization of Dirichlet?s theorem) Given two
sequences of functions {an (x)}, {bn (s)} defined on R ? R and D ? C,
respectively, suppose that
?
X
X
(i) the partial sums AN (x) =
an (x) of the series
an (x) are
n=1
n?N
bounded uniformly in x
(ii) bn (s) ? 0 uniformly on D
and that
(iii) there is a Majorant series
?
X
n=1
that
cn < ? of positive terms cn such
|bn (s) ? bn+1 (s)| ? cn
for every s ? D and for all n sufficiently large.
P?
Then the series n=1 an (x) bn (s) is uniformly convergent on R and D.
P?
If bn (s) are analytic in D, so is the sum function n=1 an (x) bn (s) for
each x ? R.
Proof. By the formula for partial summation (cf. the third proof of Theorem B.3.), we have for integers M , N , M < N ,
X
M <n?N
an (x) bn (s) =
X
M <n?N
An (x) (bn (s) ? bn+1 (s))
+ AN (x) bN (s) ? AM (x) bM +1 (s)
?
?
X
= O?
cn ? + O(|bN (s)| + |bM +1 (s)|)
M <n?N
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which can be made < ? for any ? > 0 uniformly in x and s provided that
M and N are sufficiently large. Hence the Cauchy criterion applies and
uniform convergence follows.
For fixed x ? R, analyticity of the sum function is a consequence of
Theorem A.3
Corollary B.1 (Dirichlet?s
N
test for uniform convergence) If bn ? 0
X
as n ? ?, and an (x) = O(1) uniformly in x ? R, then the series
n=1
?
X
bn an (x) is uniformly convergent on R.
n=1
Example B.2
If bn ? 0, then
?
X
bn sin 2?nx is uniformly convergent
n=1
in any interval not containing an integer. This follows from Exercise 7.8,
(7.14). In particular, the case bn = n1 establishes the uniform convergence of
?
1 X sin 2?nx
the Fourier series ?
for B1 (x) in any interval not containing
? n=1
n
an integer. Therefore, Lebesgue?s theorem allows to integrate it term by
term to obtain B2 (x) (and higher order periodic Bernoulli polynomials).
This example is a special case of the following.
Proposition B.1 The series for the polylogarithm function ls (x) defined
in the first instance for ? > 1 by
ls (x) =
?
X
e2?inx
ns
n=1
is uniformly convergent in any interval of x free from an integer (0 < x < 1)
for ? > 0.
Proof.
By Exercise 7.8, the partial sums satisfy
n
X
1
1
2?ikx ?i(n+1)x sin ?n =O
?
,
e
= e
sin ?x | sin ?x|
kxk
k=1
kxk indicating the distance to the nearest integer.
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Sci. USA 35 (1949), 371?374; Collected Papers of Atle Selberg I, Springer
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[GR] I. S. Gradshteyn and I. M. Ryzhik, Table of integral series and products,
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[Hashimoto] M. Hashimoto, Examples of the Hurwitz transform, to appear.
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[H] A.PHurwitz,
1 Einige Eigenschaften der Dirichlet?schen Funktionen F (s) =
D
и ns , die bei der Bestimmung der Classenanzahlen bina?rer quadratisn
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[Is] M. Ishibashi, An elementary proof of the generalized Eisenstein formula,
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aspects of the modular relation. Number Theory: tradition and modernization, ed. by W. Zhang and Y. Tanigawa, (Developments in Mathematics,
Vol. 15) Springer, (Febuary 2006) 103-118
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Nombres, J.-M. De Koninck and C. Levesque (e?ds.), 1989, 459?474, Walterde Gruyter 1989.
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of the mean square of the Hurwitz zeta-function, J. Number Theory 120
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lattice sum limits, J. Math. Anal. Appl. 294 (2004), 7?14.
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[KTZ2] S. Kanemitsu, Y. Tanigawa and J.-H. Zhang, Evaluation of the Spannenintegrals of the product of two zeta-functions, to appear.
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[Kat] M. Katsurada, An application of Mellin-Barnes? type integrals to mean
square of Lerch zeta-functions, Collect. Math. 48 (1997), 137?153.
[Kat1] M. Katsurada, Power series and asymptotic series associated with the
Lerch zeta-function, Proc. Japan Acad. Ser. A Math. Sci. 74 (1998), 167?
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[Ko] N, S. Koshlyakov Investigation of some questions of analytic theory of the
rational and quadratic fields, I-III (Russian), Izv. Akad. Nauk SSSR, Ser.
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[LG] A. Laurinchikas and R. Garunkstis, The Lerch zeta-function, Kluwer Academic Publ., Dordrecht-Boston-London 2002.
[Lan] E.Landau, Vorlesungen uber Zahlentheorie Bd II, Leipzig1927=Chesea ??
[Leb] N. N. Lebedev, Special functions and their applications, Dover 1972.
P e2?kiw
[Le] M. Lerch, Note sur la function K(w, x, s) =
Acta Math. 11 (1887),
(w+k)s
19?24
[Leh] D. H. Lehmer, Euler constants for arithmetic progressions, Acta Arith. 27
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Babbage Res. Center, Manitoba, 1981.
[Leh2] D. H. Lehmer, A new approach to Bernoulli polynomials, Amer. Math.
Monthly 95 (1988), 905?911 =Selected papers of D. H. Lehmer,
[LDH] H .-L. Li, L.-P. Ding and M. Hashimoto, Structural elucidation of Eisenstein?s formula, to appear.
[Li] R. Lipschitz, Untersuchungen der Eigenschaften einter Gattung von unendlichen Reihen, J. Reine Angew. Math. 105 (1889), 127?156.
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J. Anal. Appl. 25 (2006), 517?533.
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J. Reine Angew. Math. 38 (1849), 1?39.
[Matsumoto] K. Matsumoto, Recent developments in the mean square theory
of the Riemann zeta and other zeta-functions, Number Theory ed. by
R. P. Bambah et al, Hindustan Books Agency, 2000, 241?286.
[Me] Hj. Mellin, Die Dirichletschen Reihen, die zahlentheoretischen Funktionen
und die unendlichen Produkte von endlichem Geschlecht, Acta Soc. Fenn.
31 (1902), 1?48.
[M1] M. Mikola?s, Mellinsche Transformation und Orthogonalita?t bei ?(s, u); Verallgemeinerung der Riemannschen Funkutionalgleichung von ?(s), Acta Sci.
Math. (Szeged) 17 (1956), 143?164.
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261?263.
[M4] M. Mikola?s, New proof and extension of the functional equation for Lerch?s
zeta-function, Ann. Univ. Sci. Budapest 14 (1971), 111?116.
[Mi] J. Milnor, On polylogarithms, Hurwitz zeta functions, and the Kubert identities, Enseign. Math. (2) 29 (1983), 281?322.
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[PK] R. B. Paris and D. Kaminski, Asymptotics and Mellin-Barnes Integrals,
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[Sch] O. Schlo?milch, Uebungsaufgaben fu?r Schuler, Lehrsatz von dem Herrn Prof.
Dr. SCHLO?MILCH, Arch. Math. u. Phys. (Grunert?s Archiv) 12 (1849),
415.
[Sla] L. J. Slater, Confluent Hypergeometric Functions, Cambridge University
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Index
Abel convolution, 5
Abel summation formula, 197
Abel?s continuity theorem, 204
Abelian character, 167
abscissa of absolute convergence, 190
addition formula, 3
analytic continuation, 192
Appell sequence, 1
arithmetic functions, 165
Ars Conjectandi, 4
asymmetric form of the functional
equation, 98
asymptotic formula, 56
complex Fourier transform, 153
confluent hypergeometric function, 73
cotangent function, 100
D?Alembert?s formula, 189
delta function, 72, 155
difference equation, 3
difference operator, 3, 62
difference relation, 29
Dirichlet convolution, 190
Dirichlet kernel, 140
Dirichlet series, 190
distribution property, 4, 97, 170
dual lattice, 115
Dufresnoy-Pisot type uniqueness
theorem, 94
Bernoulli number, 2
Bernoulli polynomial, 1, 3, 22
Bessel function, 106
Bessel?s inequality, 134
beta function, 32
block decomposition, 128
body-centered cubic, 118
Bohr-Mollerup theorem, 96
by absolute convergence, 188
Epstein zeta-function, 109
Euler digamma function, 54
Euler function, 149, 173
Euler?s constant, 86
Euler?s identity, 102
Euler?s interpolation formula, 40, 89
Euler?s product formula, 88
Euler?s summation formula, 200
Euler-Maclaurin summation formula,
201
Eulerian integral of the first kind, 32
Eulerian integral of the second kind,
29
exponential Fourier transform, 152
Cauchy integral formula, 191
Cauchy Integral Theorem, 194
Cauchy product, 5, 189
causal function, 153
characteristic function, 167
Chebyshe?v?s inequalities, 38
circle of convergence, 189
complete orthonormal system, 133
213
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face-centered cubic, 117
falling factorial, 12, 31
finite expressions, 174
formula of Gauss, 174
Fourier coefficient, 132
Fourier cosine coefficient, 135
Fourier cosine series, 136
Fourier cosine transform, 152
Fourier expansion, 168
Fourier Integral Theorem, 154
Fourier series, 4, 132
Fourier sine coefficient, 135
Fourier sine series, 136
Fourier sine transform, 152
Fourier transform, 153, 168
functional equation, 99, 100
gamma function, 29
Gauss? integral representation, 44
Gaussian representation, 86, 93
generalized Chowla-Selberg type
formula, 128
generalized Gauss sum, 178
generating function, 3
Generating functionology, 3
generic definition, 86
Gram matrix, 115
harmonic number, 54
Hurwitz formula, 69, 75, 99
Hurwitz transform, 69
Hurwitz zeta-function, 53
hyperbolic cotangent function, 100
incomplete gamma function of the
first kind, 75
incomplete gamma function of the
second kind, 73
incomplete gamma functions, 53
inner product, 166
integral part, 4
integral representation, 55, 56, 61
integral representation for Euler?s
constant, 44
integral transform, 151
inverse Fourier transform, 154
inverse Mellin transform, 162
Kubert identity, 3, 97, 170
Lagrange Interpolation Method, 26
Lagrange interpolation polynomial,
26
Laplace transform, 154
lattice zeta-function, 115
Laurent coefficient, 60, 192
Laurent expansion], 192
Legendre?s integral representation, 45
Lerch zeta-function, 53
Lerch?s formula, 83, 84
Mellin transform, 29, 152, 161
Mellin-Barnes integral, 126
modified Bessel function of the first
kind, 108
modified Bessel function of the
second kind, 108, 126
modified Kubert identity, 76
multiplication formula, 4
multiplicative group of reduced
residue classes, 167
one-sided Laplace transform, 152
orthogonal system, 133
orthogonality, 168
orthonormal system, 133
pair of its Mellin transforms, 162
pair of the Fourier transform, 156
partial fraction expansion, 47, 100
partial sum, 55, 132
Pochhammer symbol, 31
pole, 193
polylogarithm function, 53, 171
positive semidefinite Hermitian form,
133
power series centered at, 189
principal character, 168
principal part, 193
principle of analytic continuation, 192
probability integral, 32, 34
product representation, 99
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215
Index
Prym?s decomposition, 32
psi function, 54
radius of convergence, 189
Ramanujan, 62
real analytic, 192
reciprocal relation, 8, 98
reciprocity relation, 34
residue, 193
residue theorem, 193
Riemann Hypothesis, 149
Riemann zeta-function, 53
shift operator, 12
shifted factorial, 31
simple cubic, 116
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Stirling number of the second kind, 11
Stirling?s formula, 37, 93
sums of powers, 4
Taylor series, 191
the space of (formal) Dirichlet series,
170
two-sided Laplace transform, 152
uniformly convergent in the wide
sense, 185
Wallis? formula, 35
Weierstrass? canonical product of
genus 1, 88
Weierstrass? M-test, 188
f ? C(q), ? 1 .
D(q) =
ns n=1
If we impose some growth condition on f , the Dirichlet series is convergent
in some half-plane, ? 1. The spaces C(q) and D(q) are isomorphic
(cf. Yamamoto [Ya]).
We may express (8.8) in terms of the Dirichlet series
?
?
X
X
X
f (n)
?a (n)
f
(a)
=
s
n
ns
n=1
n=1
a mod q
X
=
f (a) ?(s, a mod q),
(8.14)
a mod q
where ?(s, a mod q) signifies the partial zeta-function
X
1
,
?(s, a mod q) =
ns
n?a mod q
(8.15)
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Around Dirichlet?s L-functions
which can be expressed by the Hurwitz zeta-function
a
?s
?(s, a mod q) = q ? s,
,
q
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171
(8.16)
so that it is meromorphic over the whole plane. Hence (8.12) is stated as
?
X
X
f (n)
a
?s
,
(8.17)
=q
f (a) ? s,
ns
q
n=1
a mod q
which continues the left-hand side meromorphically over the whole plane.
Example 8.4
we get
If we trace the above argument in the reverse direction,
X
?
a
?a (n)
=
q ?s ? s,
.
q
ns
n=1
Substituting (8.4) and (8.3) for ?a , we obtain
X
?
X
1
a
=
??j (a) ?j (n) s
q ?s ? s,
q
n
n=1
j mod q
?
X
1 X
j (n)
=
,
?a (j)
q
ns
n=1
j mod q
whence we have
q
?s
a
? s,
q
j
1 X
?a (j) ls
=
,
q
q
(8.18)
j mod q
being valid for all s 6= 1, whose ls (x) indicates the polylogarithm function
defined by (3.3) (cf. Ishibashi [Is, p.447]).
But, if we state (8.18) in the form (the genuine generalization of the
Eisenstein formula)
q?1
X
a
k
a
e?2?i q k ls
= q 1?s ? s,
? ?(s),
(8.19)
q
q
k=1
and interpret the case s = 1 as the limit as s ? 1, then (8.19) is valid for all
s ? C. The limit interpretation of the right-hand side of (8.19) corresponds
to the normalization ls (0) = ?(s) in Milnor [Mi], but even under this, (8.16)
is not valid and only (8.19) stands. We state the limiting case of (8.19) as
Theorem 8.1
The limiting case of (8.19) implies Gauss? formula (8.29).
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For details, cf. [LDH].
Exercise 8.2
Prove that the formula
q?1
r
1 X ?2?i qj r
a+j
? s, a, ? +
, q? =
e
e2?ij? q ?s ? s,
q
q r=0
q
(8.20)
is the two different expressions for the j-component
?
X
e2?im?
?j (s, a, ?) =
?j (m)
(m + a)s
m=0
of
?(s, a, ?) =
q?1
X
?j (s, a, ?),
(8.21)
j=0
where ?(s, a, ?) = ?(s, a, e2?i? ) (cf. (8.12)). Note that one expression is a
consequence of (8.4),
?a (m) =
1 X
1 X 2?i m?a
q j.
a (j) j (m) =
e
q
q
j mod q
Solution
Since ?j (s, a, ?) =
?
X
m=0
m?j ( mod q)
e2?im?
, we may express it, on
(m + a)s
writing m = j + lq, l = 0, 1, и и и , as
?
X
e2?i(lq?+j?)
s
a+j
l=0 l + q
a+j
2?ij? ?s
=e
q ? s,
, q? .
q
?j (s, a, ?) = q ?s
On the other hand, substituting (8.3), we derive that
?j (s, a, ?) =
=
(8.22)
j mod q
q?1
?
X
e2?im? 1 X 2?i m?j
q r
e
s q
(m
+
a)
m=0
r=0
r
q?1
?
1 X ?2?i qj r X e2?i(?+ q )m
e
,
q r=0
(m + a)s
m=0
which is the right-hand side of (8.21).
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173
Around Dirichlet?s L-functions
We confirm that adding the right-hand side of (8.21) over j mod q, we
obtain ?(s, a, ?), i.e. (8.22) in terms of the right-hand side of (8.21).
As suggested by (5.56)0 , there is a counterpart of (8.11), which is a
decomposition into residue classes mod q:
q?1
1 X
a
L(s, ?) = s
?(a) ? s,
q a=1
q
(8.23)
being valid for any Dirichlet character mod q, not necessarily primitive.
Recall the Laurent expansion for ?(s, x),
?(s, x) =
1
? ?(x) + O(s ? 1),
s?1
s ? 1,
(8.24)
where ?(x) signifies the Euler digamma function (cf. Д5.1)
?(x) =
?0
0
(x) = (log ?(x)) .
?
(8.25)
Also recall the orthogonality of characters
q?1
X
a=1
?(a) =
(
0,
?(q),
if ? 6= ?0 ,
(8.26)
if ? = ?0 ,
where ?0 and ?(q) stand for the principal character mod q and the Euler
function defined by
?(q) =
X
1,
(8.27)
1?a?q
(a,q)=1
respectively.
From (8.23), (8.24) and (8.26) we obtain
q?1
1X
a
L(s, ?) = ?
?(a) ?
+ O(s ? 1),
q a=1
q
s?1
and a fortiori
q?1
a
1X
?(a) ?
L(1, ?) = ?
.
q a=1
q
(8.28)
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For the values of ? pq , we have a formula of Gauss (2.58) which may be
stated by Lemma 8.1 below as
X
p
?
2pk
p
k
= ?? ? log q ? cot ? + 2
cos
?
? log 2 sin ? , (8.29)
q
2
q
q
q
q
k? 2
where ? is the Euler constant (?(1) = ??) ([Bo?h], [Ca], [GR]).
It was D. H. Lehmer [Leh] who first used (8.29) in his study of generalized Euler constants ?(p, q) for an arithmetic progression p mod q. He
deduced (8.29)
[Leh, (11)], and the relation [Leh, Theorem 7], between
from
p
?(p, q) and ? q , and stated ([Leh, p.135]) ?Our proof via finite Fourier
series indicates that Gauss? remarkable result has a completely elementary
basis.?
Our main purpose is to elaborate on this statement of Lehmer and,
on streamlining the argument, to show that (8.29) has a purely numbertheoretic basis and ? is a number-theoretic function. As a converse to this,
we shall also put into practice the statement of Deninger [D, p.180], to the
effect that (8.29) can be used to evaluate L(1, ?). Indeed, Funakura was on
these lines (cf. [Fu, (1)]) but he appealed to the integral representation of
Legendre and applied Lehmer?s argument of using ? log(1 ? e2?ix ), 0 <
x < 1.
8.2
The Dirichlet class number formula
We may now state our main theorem in this chapter.
Theorem 8.2
for L(1, ?).
Gauss? formula (8.29) is equivalent to finite expressions
L(1, ?) =
q?1
a
? X
?(a) cot ?
2q a=1
q
(8.30)
for ? odd and
q?1
a
1 X
?
b(a) log 2 sin ?
L(1, ?) = ? ?
q a=1
q
(8.31)
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for ? even, where
?
b(a) =
k
1 X
?(k) e?2?i q a
q
(8.32)
k mod q
is the finite Fourier transform of ? (intimately related to the generalized
Gauss sum G(a, ?), see (8.39) below).
Corollary 8.1
For primitive ?, (8.30) and (8.31) reduce, respectively, to
q?1
?i X
a
L(1, ?odd ) = ?
?(a) B1
G(?) a=1
q
(8.30)0
q?1
1 X
a
L(1, ?even) = ?
?(a) log 2 sin ? ,
G(?) a=1
q
(8.31)0
and
where G(?) = G(1, ?) is the normalized Gauss sum.
Remark 8.1
(i) On symmetry grounds, (8.30) may be stated as
q?1
?i X
a
?
b(a) B1
,
L(1, ?odd ) = ? ?
q a=1
q
(8.30)00
which can be explicitly computed to be (8.30) (cf, e.g. [Fu]).
Although both Funakura [Fu] and Ishibashi-Kanemitsu [IK] treated the
case of periodic functions f (n) of period q, the formulas (8.30)00 and (8.31)
are implicit in Yamamoto?s work [Ya], depending on (8.11) and (7.18) &
(7.19).
(ii) The last statement
of Corollary 8.1 follows, on recalling that the
Kronecker characters |d|
(d < 0) and dи (d > 0) are primitive odd and
и
even characters mod |d|, respectively.
In the course of proof of Theorem 8.2, we shall encounter an interesting
P
number-theoretic function log Nq = ? d|q (х(d) log d) dq which eventually
cancels out in view of the following Theorem 8.3. We believe this function
deserves wider attention and we state
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Theorem 8.3 For q > 1, the number-theoretic function log N (q) =
log Nq admits the following expressions.
log Nq = ?q
X х(d)
d
d|q
log d
?
q?1
X
(8.33)
?
х
q
(a,q)
??
a
??
log?2 sin? ? q
q ?
a=1
(a,q)
q X
?(d) ?
=
d
= ??(q)
(8.34)
(8.35)
d|q
= ?(q)
X log p
,
p?1
(8.36)
p|q
the product extending over all prime divisors of q, where х and ? signify
the Mo?bius function and the von Mangoldt function, respectively.
For a proof cf. [HKT].
8.3
Proof of the theorems
Let f (n) be an arithmetic periodic function of period q:
f : Z ? C;
f (n + q) = f (n),
n ? Z.
We define the parity of f as follows: f is called even if f (?n) = f (n) and
odd if f (?n) = ?f (n).
We prepare some lemmas, of which Lemma 8.1 is repeatedly used in
what follows, without notice.
Lemma 8.1
If f is odd, then
q?1
X
f (a) = 0
a=1
and if f is even, then
q?1
X
a=1
f (a) = 2
X
a? 2q
f (a) = 2
X
a< 2q
f (a) +
1 + (?1)q q .
f
2
2
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Around Dirichlet?s L-functions
In particular, if f and ? mod q are of opposite parity, then
q?1
X
?(a) f (a) = 0
a=1
while if f and ? are of the same parity and q > 2, then
q?1
X
?(a) f (a) = 2
a=1
X
?(a) f (a) = 2
X
?(a) f (a).
a< 2q
a? q2
Lemma 8.2 The ? function satisfies Gauss? multiplicative formula or
the modified Kubert identity
?(x) = log q +
Lemma 8.3
and
q?1 x+a
1 X
?
.
q a=0
q
(8.37)
Let ? denote a Dirichlet character mod q, q ? 3. Then
?
?0
if n 6? ▒1 (mod q),
X
?(n) = ?(q)
?
if n ? ▒1 (mod q).
? even
2
?
?
0
?
?
? ?(q)
X
?(n) =
2
?
?
? odd
?
?? ?(q)
2
if
n 6? ▒1
if
n?1
if
n ? ?1
(mod q),
(mod q),
(mod q),
where the sum is extended over all even and odd characters, respectively.
Proof. For q ? 3, the set {▒1} forms a subgroup of the reduced residue
О
class group G = (Z/qZ) of index 2. Hence the factor group G/{▒1} has
?(q)
order
. Since the set of all even characters coincides with the character
2
group of G/{▒1}, it follows, from the orthogonality of characters, that
X
X
?(a) =
?(n)
? even
\
??G/{▒1}
?
?0
= ?(q)
?
2
if n 6= 1 in G/{▒1},
if n = 1 in G/{▒1},
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which proves the first assertion. The second assertion follows from the first
and the orthogonality relation
(
X
0
if n 6? 1 (mod q)
?(n) =
(8.38)
?(q) if n ? 1 (mod q).
b
? ?G
This completes the proof.
It is instructive to give a proof of Corollary 8.1 first. We introduce the
generalized Gauss sum
G(k, ?) =
q?1
X
a
?(a) e2?i q k
(8.39)
a=1
= q ?(?1) ?
b(k)
(cf. (8.32)) and note that it decomposes into
G(k, ?) = ?(k) G(?)
(8.40)
if and only if ? is primitive ([Ap4], [Da]). We derive (8.30)0 by appealing
to Eisenstein?s formula
q?1 X
p
k
k
?2?i p
q
= B1 ? q B 1
l0
e
q
q
k=1
or rather its converse (cf. [LDH], [Is], [Wa])
q?1
X
k=1
B1
p
p
k
e2?i q k = ?l0
? 1 ? B1
q
q
p
i
= ? cot ?.
2
q
Substituting (8.41) into (8.30), we find that
L(1, ?) =
q?1
?i X
k
B1
G(k, ?).
q
q
k=1
Using (8.38) and other known facts
G(?) = ?(?1) G(?),
we conclude (8.30)0 .
|G(?)|2 = q,
(8.41)
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Around Dirichlet?s L-functions
We may deduce (8.31)0 from (8.31) in a similar way. Substituting (8.32)
into (8.31), we obtain
L(1, ?) = ?
X
q?1
q?1
1X
2?k
k
a
?(a) cos
log 2 sin ?
q
q
q
a=1
(8.42)
k=1
whose inner sum is again G(k, ?). Therefore for ? primitive, we have
q?1
k
G(?) X
?(k) log 2 sin ? ,
L(1, ?) = ?
q
q
k=1
whence (8.31)0 follows in the same way.
We now turn to the proof of Theorem 8.2.
Proof. That (8.29) implies (8.30) and (8.31) is immediate. Indeed, substituting (8.29) in (8.28) and using Lemma 8.1, we obtain (8.30) for ? odd
and (8.42) for ? even, which is the same as (8.31).
Now we are to prove the converse, i.e. we are to deduce (8.29) from
(8.30) and (8.31).
With p, (p, q) = 1, we multiply (8.28) by ?(p?1 ) and sum over ? mod
q, ? 6= ?0 to obtain
X
?0 6=? mod q
?(p?1 ) L(1, ?) = ?
q
a
1X
?
q a=1
q
X
?(ap?1 )
(8.43)
?0 6=? mod q
= S1 + S2 ,
say, where
S1 = ?
X
q
1 X
a
?
?(ap?1 )
q a=1
q
? mod q
and
S2 =
q
1 X
a
?
?0 (ap?1 ).
q a=1
q
By the orthogonality (8.38) of characters,
p
?(q)
?
S1 = ?
.
q
q
(8.44)
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The sum S2 is
q?1
1 X?
a
S2 =
,
?
q a=1
q
the star on the summation sign indicating the sum over all a?s, relatively
prime X
to q, (a, q) = 1, which condition may be replaced by introducing the
х(d). Writing the condition d|(a, q) as d|q, a = a0 d ? q ? 1, we
sum
d|(a,q)
have
0
d ?1
X
1 X
a
S2 =
х(d)
? q
q
0
d
q
a =1
d|q
q
d
X
a+1
q
q
q
?
whose inner sum is
= ? log ? ? + ? by Lemma 8.2.
q
d
d
d
d
a=0
Hence
S2 = ?
1
?(q)
log q
?(q) ? log Nq ?
?,
q
q
q
(8.45)
where log Nq is defined by (8.33).
Substituting (8.44) and (8.45) in (8.43), we conclude that
X
?0 6=? mod q
?(p?1 ) L(1, ?) =
?(q)
q
??
p
1
? log q ?
log Nq ? ? .
q
?(q)
(8.46)
It remains to calculate the left-hand side of (8.46) by dividing the sum
into two parts:
X
?0 6=? even
and
X
? odd
substituting therewith (8.31) and (8.30), respectively.
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Around Dirichlet?s L-functions
First, by (8.31),
X
?(p?1 ) L(1, ?)
(8.47)
?0 6=? mod q
q?1
a
1 X
log 2 sin ?
= ??
q a=1
q
X
?0 6=? even
?(p) ?
b(a)
X
q?1
a
1 X
??0 (p) ?
log 2 sin ?
= ??
c0 (a)
q a=1
q ? even
= T1 + T2 ,
say, where
X
q?1
1 X
1
a
?(p) ?
T1 = ? ?
log 2 sin ?
q a=1
q
q
? even
X
k
?(k) e?2?i q a
(8.48)
k mod q
and
q?1
a
1 X
1
log 2 sin ? ?
T2 = ?
q a=1
q
q
X
k
?0 (k) e?2?i q a .
(8.49)
k mod q
The inner double sum of T1 is
1 X ?2?i kq a X
e
?(kp?1 )
?
q
? even
k mod q
?(q)
p
1 ?(q) ?2?i pq a
p
+ e2?i q a = ? cos 2? a
e
=?
q 2
q
q
by Lemma 8.3, and so
T1 = ?
q?1
?(q) X
a
p
cos 2 a? log 2 sin ? ,
q a=1
q
q
while the inner sum for T2 is
X
? ?2?i kq a
e
k mod q
which is equal to
х
?(q)
?
q
(a,q)
q
(a,q)
,
(8.50)
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by Ho?lder?s result (cf. [Leh, p.133]).
Hence
q
х
q?1
X
(a,q)
?(q)
a
T2 =
log 2 sin ?
q
q a=1
q
?
(a,q)
but this is ? q1 log Nq by (8.34), i.e.
T2 = ?
1
log Nq .
q
(8.51)
Substituting (8.50) and (8.51) in (8.47), we obtain
q?1
X
?(q) X
a
1
p
?1
?(p ) L(1, ?) = ?
cos 2? a? log 2 sin ? ? log Nq .
q a=1
q
q
q
?0 6=? even
(8.52)
On the other hand, by (8.30) and Lemma 8.3,
X
q?1
a X
? X
cot ?
?(ap?1 )
2q a=1
q
? odd
? ?(q)
p
?(q)
?p
=
cot ? ?
cot
?
2q
2
q
2
q
??(q)
p
=
cot ?.
2q
q
?(p?1 ) L(1, ?) =
(8.53)
Combining (8.52) and (8.53) implies
X
?(p?1 ) L(1, ?)
(8.54)
? odd
?0 6=? mod q
q?1
p
?(q) X
a
1
?(q)
p
cos 2 a? log 2 sin ? ? log Nq +
=?
? cot ?.
q a=1
q
q
q
2q
q
Comparing (8.46) and (8.54), we see that the terms involving log Nq
cancel each other and (8.29) follows. This completes the proof.
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Appendix A
Complex functions
A.1
Function series
Fig. A.1
D?alembert
Theorem A.1 A uniformly convergent series of analytic functions may
be integrated term by term along any curve inside the region of uniform
convergence. Namely, if the functions
f1 (z), f2 (z), ...
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are analytic in D and the series
?
X
fn (z) = f (z)
n=1
is uniformly convergent in D, then for any curve C ? D, we have
Z
f (z) dz =
Z X
?
fn (z) dz =
C n=1
C
? Z
X
n=1
fn (z) dz.
C
Proof. fn (z) need not be analytic but enough to be continuous in D
(since analyticity ? continuity, the assumption
R is excessive). Since f (z) is
continuous in D, it follows that the integral C fn (z) dz, n ? N exists. So
n
R
P
fi (z). Since sn (z) converges to
does the integral C sn (z) dz for sn (z) =
i=1
f (z) uniformly on D, we have
?? > 0, ?n0 = n0 (?) ? N s.t. n > n0 ? |sn (z) ? f (z)| < ?,
?z ? D.
Hence for n ? n0 , we have
Z
(sn (z) ? f (z)) dz < ? ?(C),
C
where ?(C) is the length of C so that
Z
Z
lim
sn (z) dz =
f (z) dz
n??
C
C
whose left-hand side is nothing other than the definition of
? Z
X
n=1
fn (z) dz.
C
Theorem A.2 The limit of the uniformly convergent series of analytic
functions is interchangeable with integration along any curve lying in the
region of its uniform convergence. Namely, if
f1 (z), f2 (z), ...
are analytic in D and
lim fn (z) = f (z),
n??
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Complex functions
uniformly in D, then for any Jordan curve C ? D, we have
Z
Z
Z
f (z) dz = lim
fn (z) dz =
lim fn (z) dz.
C
n??
C n??
C
Definition A.1 If a sequence (respectively, series) of functions defined
on D are uniformly convergent on any bounded closed subset of D (i.e.,
on any compact subset D 0 such that D0 ? D), we say that the sequence
(respectively, series) is uniformly convergent on D in the wide sense.
Theorem A.3
If the functions
f1 (z), f2 (z), ...
are (i) analytic in D and (ii) the series
?
X
fn (z)
n=1
is uniformly convergent in D in the wide sense, then its sum
?
X
f (z) =
fn (z)
n=1
is analytic in D and its derivative may be obtained by termwise differentiation:
0
f (z) =
?
X
fn0 (z)
n=1
Also, the termwise differentiated series is uniformly convergent in the wide
sense in D.
Corollary A.1
Any function series
?
X
fn (z) := f (z)
n=1
that is uniformly convergent in the wide sense in D is termwise differentiable infinitely many times:
f (k) (z) =
?
X
n=1
fn(k) (z),
k ? N.
(and the k-times differentiated series is also uniformly convergent in the
wide sense in D.)
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Proof. We shall prove both Theorem A.3 and its Corollary A.1 at the
same time. By the Cauchy integral formula in Theorem A.7, we have for
any rectifiable simple curve C in D and any point z in C,
?
? Z
X
fn (w)
1 X
dw.
f (z) =
fn (z) =
2?i
w?z
n=1
n=1 C
But
?
P
fn (w) is uniformly convergent on C, and so Theorem A.1 allows
n=1
us to integrate term by term after multiplying (w ? z)?1 :
f (z) =
1
2?i
Z X
Z
?
1
fn (w)
f (w)
dw =
dw.
2?i C w ? z
C n=1 w ? z
Hence the Cauchy integral formula holds for f (z) and so it follows that
f (z) is analytic in C and that
Z
f (w)
k!
dw, k ? N ? {0}.
f (k) (z) =
2?i C (w ? z)k+1
Let
Sn (z) =
n
X
fk (z)
k=1
be the n-th partial sum of
?
X
fn (z)
n=1
and take any bounded closed subset D 0 in D. Then take any simple closed
contour C ? D of finite length containing D 0 and suppose dist (D 0 , C) =
? > 0. The we have
Z
Sn (w)
k!
(k)
dw, k ? N ? {0}.
Sn (z) =
2?i C (w ? z)k+1
Hence, it follows that
k!
?(C) max |f (w) ? Sn (w)| ,
w?C
2?? k+1
whence we have
?? > 0,
?n0 = n0 (?) ? N s.t. n > n0 ?
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Complex functions
|f (w) ? Sn (w)| < ?
on C. Hence,
lim Sn(k) (z) = f (k) (z).
n??
uniformly on D0 .
Corollary A.2 (The Weierstrass double series theorem) Suppose
{fn (z)} are analytic in |z ? z0 | < r and has the Taylor expansion
fn (z) =
?
X
k=0
(n)
ak (z ? z0 )k .
Then if
?
X
fn (z) = f (z)
n=1
uniformly in the wide sense on |z ? z0 | < r, then f (z) is analytic on
|z ? z0 | < r and its Taylor expansion is given by
f (z) =
?
X
k=0
ak (z ? z0 )k
(|z ? z0 | < r) ,
ak =
?
X
(n)
ak .
n=1
That is, the iterates of the double series coincide ? the order of summation
being interchangeable ?
!
? X
?
?
?
X
X
X
(n)
k
k
ak (z ? z0 ) =
fn (z) = f (z) =
ak (z ? z0 )
n=1 k=0
n=1
k=0
Proof. This is a special case of Theorem A.2. The relation between
coefficients follows from the Theorem A.8:
ak =
Theorem A.4
?
?
X
1 (k)
1 X (k)
(n)
f (z0 ) =
fn (z0 ) =
ak .
k!
k! n=1
n=1
If {fn (z)} are analytic in D and
lim fn (z) = f (z)
n??
uniformly in the wide sense on D, then f (z) is again analytic on D and
f (k) (z) = lim fn(k) (z)
n??
uniformly in D in the wide sense.
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Most of the above results on infinite series apply to infinite integrals
in spite of the fact that in the case of partial sums SnR(z), n goes to ?
b
discretely while in the case of partial integral I(b) = 0 f (z) dz, b goes
to ? continuously.
To assure the uniform convergence in case the series (integrals) are absolutely convergent, the main tool is Weierstrass? M-test (Majorant series
test) which in the case of integrals asserts that given a (complex-valued)
function f (x, y), x ? [a, b] (resp (a, b) as the case may be), y ? Y , if there
is a positive (-valued) function M (x) such that for any y ? Y ,
|f (x, y)| ? M (x)
and
Z
?
a
M (x) dx < ? (resp.
Z
b
a
M (x) dx < ?),
R?
Rb
then a f (x, y) dx (resp. a M (x) dx < ?) is absolutely and uniformly
convergent on Y .
Following Titchmarsh [Tit], we often refer to this as ?by absolute convergence.?
If the series or integrals are convergent but not absolutely convergent,
i.e. conditionally convergent, we need to appeal to more delicate convergence tests such as Dirichlet?s (cf. ДB.2).
Exercise A.1 Noting that the principal branch of the natural logarithm
log z (often denoted by Log z) may be defined by the Condition
1
d
log z = , log 1 = 0,
dz
z
prove the integral representation (Re z > 0)
Z ? ?t
e ? e?zt
dt.
log z =
t
0
(A.1)
(A.2)
e?(z?1)t ? 1
Since the integrand f (t) = f (tz) = ?e?t
? z?1
t
Z ?
as t ? 0, the improper integral
f (t) dt is absolutely convergent. Hence
Solution
0
we may differentiate under the integral sign to get
Z ?
Z ?
d
1
f (t) dt =
e?zt dt = .
dz 0
z
0
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Complex functions
Since f (t, 1) = 0, we have log 1 =
satisfied.
Example A.1 (Power series)
polynomial
R?
0
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f (t, 1) dt = 0, and Condition (A.1) is
Series in the form of an infinite degree
f (z) =
?
X
n=0
an (z ? z0 )n ,
centered at z0 , is called a power series centered at z0 . By translation it
suffices to consider the power series at 0. We speak of absolute convergence of power series, and the region of (absolute) convergence of a power
series is a disc with the boundary called the circle of convergence whose
radius is called the radius of convergence and is most conveniently given by
D?Alembert?s formula
an r = lim n?? an+1 if the limit exists (including ? ).
We consider the case where 0 < r ? ? and denote the region of convergence
by D.
By above theorems we have
Theorem A.5 Inside D, the region of convergence, (i) a power series
may be integrated term by term along any path lying in D, (ii) a power
series is analytic in D and the derivatives may be obtained by term by term
P
P
differentiation and (iii) two power series
an z n ,
bn z n (in their common
region D of convergence) may be added, subtracted, multiplied and divided;
in particular, the multiplication is carried out by the Cauchy product
!
! ?
?
?
X
X
X
n
m
bn z
=
cl z l ,
am z
n=0
m=0
cl =
X
l=0
a m bn
m+n=l
(cf. Remark 1.1); the division is carried out exactly as we do with ordinary
polynomials: e.g. to check the numerical values of Bernoulli numbers in
1 2
1 3
Example 1.1, we may divide z by z + 2!
z + 3!
z + и и и to obtain
1
1
1
z
6 2
30 4
=
1
?
z
+
z
?
z +иии
ez ? 1
2
2!
4!
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Similarly, we may carry out the division in Exercise 5.4.
To sum up, Theorem A.5 says that we may treat power series as ordinary
polynomials (, which disposition is due to Euler).
Example A.2 (Dirichlet series) For an increasing sequence of positive
?s log ?n
reals ?n , the series of functions fn (s) = ??s
, with logarithm
n = e
indicating the principal branch,
?
X
??s
n
n=1
is called the Dirichlet series. Contrary to the case of power series, the region
of absolute convergence of Dirichlet series is a right half-plane and we may
speak about the abscissa of absolute convergence, often denote by ?a .
We have a counterpart of Theorem A.5 for Dirichlet series.
Theorem A.6
Let ?a denote the abscissa of absolute convergence of
?
P
??s
a Dirichlet series f (s) =
n . Then in the region ? > ?a , f (s) is
n=1
analytic, integrable and differentiable term by term. Two Dirichlet series
?
?
P
P
f (s) =
am m?s and g(s) =
bn n?s may be multiplied by the Dirichm=1
n=1
let convolution:
f (s) g(s) =
?
X
cl l?s ,
cl =
l=1
X
a m bn .
mn=l
The last sum is often expressed as
X
X
cl =
ad bl/d =
al/d bd ,
d|l
d|l
with d|l meaning that d runs though all positive divisors of l.
Example A.3 (i) The integral in (2.1) defining the gamma function is
(absolutely and) uniformly convergent in the wide sense in ? > 0. (ii) The
series (3.2) defining the Riemann zeta-function is (absolutely and) uniformly convergent in the wide sense in ? > 1.
Proof. (i) The integral is improper at both end points. We apply Weierstrass? M-test to e?x x??1 . Let s lie in the compact region 0 < ?0 ? ? ? R,
|t| ? R, R > 0. Then for 0 < x < 1, e?x x??1 < e?xRx?0 ?1 = O(x?0 ?1 ) and
1
x > 1, e?x x??1 < e?x xR?1 = O(x?2 ). Since 0 t?0 ?1 dt = O(1) and
Rfor
? ?2
dt = O(1), we conclude the assertion.
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(ii) Let 2m?1 ? N < 2m . Them
N
X
n=1
n?? < 1 + 2 и 2?? + 4 и 4?? + и и и + 2m?1 (2m?1 )??
= 1 + 21?? + (21?? )2 + и и и + (21?? )m?1
1
.
<
1 ? 21??
P?
Hence n=1 n?? < ?, and the series is absolutely convergent for ? > 1.
We may also apply (B.4) to obtain
Z ?
X
1
x1??
1
??
+ +
??
n =
B 1 (w)???1 du + O(x?? ), ? > 0
??1 2 1??
1
n?x
= O(1),
? > 1.
Theorem A.7 (Goursat) If f (z) is analytic in a domain D, them it
has all orders of derivatives f (k) (z), which are also analytic in D, given by
the Cauchy integral formula
Z
f (w)
1
f (k) (z)
=
dw
k!
2?i C (w ? z)k+1
Z
1
dk f (w)
=
dw,
2?i C dz k w ? z
where C is a closed Jordan curve contained in D.
This theorem of Goursat draws a clear line between analytic functions
and real differentiable functions. The requirement that a function is analytic at a point (in the neighborhood of a point) is such a stringent restriction that it already implies the existence of derivatives of all orders.
Theorem A.8 (Cauchy-Taylor) If f (z) is analytic at z0 , then it can
be expanded into the Taylor series in the maximal circle contained in the
domain D of analyticity:
f (z) =
?
X
n=0
an (z ? z0 )n ,
where an is given by (Theorem A.7)
1
f (n) (z0 )
=
an =
n!
2?i
Z
C
dn f (w)
dw,
dz0n w ? z0
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C being any closed contour contained in D.
Actual determination of Taylor coefficients may be done by the method
of undetermined coefficients.
Theorem A.9 (Consistency Theorem or the Principle of Analytic
Continuation) If two functions f (z) and g(z) are analytic in a domain
D and f (z) = g(z) on a subset of D containing an accumulation point,
then f (z) = g(z) all over D.
This theorem often applies when two functions f (z), g(z) coincide on a
segment ? R, in which case we may extend the domain of analytic functions.
Corollary A.3 If a function of the real variable x is real analytic at x0 ,
i.e. it has the power series expansion in the neighborhood |x ? x0 | < r,
r>0
f (x) =
?
X
n=0
an (x ? x0 )n ,
an =
f (n) (x0 )
,
n!
then there is a unique function f (z) analytic on |z ? x0 | < r and coinciding
with f (x) on |x ? x0 | < r, which is given by
?
X
f (n) (x0 )
(z ? x0 )n .
f (z) =
n!
n=0
This is called an analytic continuation of f (x). Most of elementary
functions have their analytic continuation as examples of Corollary A.3:
ez , sin z, cos z. In a word, given a real power series in x, we get its analytic
continuation by changing x by z, the complex variable.
Theorem A.10 (Laurent expansion) If f (z) is (one-valued and) analytic in the annulus (ring-shaped domain) D : r < |z ? z0 | < R (0 < r < R),
then we have the Laurent expansion of f (z):
f (z) =
?
X
n=??
an (z ? z0 )n ,
where the n-th Laurent coefficient an is given by (r < ? < R)
Z
1
f (z)
an =
dz
2?i |z?z0 |=? (z ? z0 )n+1
Z
dn
1
1
f (z) n (z ? z0 )?1 dz, n ? Z
=
2?i |z?z0 |=? n!
dz0
(A.3)
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The Laurent series (A.3) converges uniformly in any annulus contained
in D.
1 (n)
f (z0 ) are the Taylor coefficients. The negative
For n ? 0, an = n!
P?1
power part n=?? an (z ?z0 )n is called the principal part of f (z) at z = z0 .
P?1
If the principal part is finite, n=?m an (z ? z0 )n , a?m 6= 0, say, then
f (z) is said to have an m-th pole at z = z0 , and the coefficient a?1 is called
the residue of f (z) at z = z0 , denoted by
Z
1
f (z) dz.
(A.4)
a?1 = Resz=z0 f (z) =
2?i |z?z0 |=?
If z = z0 is a pole of order m of f (z), then the residue may be calculated
in a similar way as the Taylor coefficients, by the method of undetermined
coefficients:
Clearing the denominator, we have
(z ? z0 )m f (z) = a?m + a?m+1 (z ? z0 ) + и и и + a?1 (z ? z0 )m?1 + и и и .
Hence differentiating m ? 1 times, we get
dm?1
((z ? z0 )m f (z)) = (m ? 1)! a?1 + O(z ? z0 ).
dz m?1
Hence
Resz=z0 f (z) = a?1 =
1
dm?1
lim
((z ? z0 )m f (z))
(m ? 1)! z?z0 dz m?1
(A.5)
which is applicable to many similar settings. It is advisable to remember
the process rather than Formula (A.5).
A.2
Residue theorem and its applications
Theorem A.11 (The (Cauchy) Residue Theorem)
Let C be a
piecewise smooth (Jordan) curve. Suppose f (z) is analytic in a region D
containing C except for finitely many singularities (which we may suppose
are poles) z1 , z2 , и и и , zn (n = 0 inclusive). Then we have
Z
n
X
f (z) dz = 2?i
Resz=zi f (z).
(A.6)
C
i=1
Remark A.1 By Theorem A.11, the value of the integral may be determined by computing the residues, which, as stated above, amounts to
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clearing the denominator. The vacuous case n = 0 implies the most fundamental Cauchy Integral Theorem which asserts that the integral along a
closed curve contained in the region of analyticity is 0, which in turn originates from the fact that in this case the region can be made to shrink into
a point, a topological feature of analytic functions in a region.
Theorem A.12 Suppose f (z) is a rational function in z satisfying the
conditions (i) it has no poles on the positive real axis and (ii) for some
a ? R, a 6? Z, we have
lim z a+1 f (z) = 0,
z?0
lim z a+1 f (z) = 0.
z??
Then,
Z
?
xa f (x) dx =
0
2?i X
Res (z a f (z)) ,
1 ? e2?ia
(A.7)
z6=0
where the power function is defined by z a = exp (a Log z), Log z signifying
the principal branch.
Proof. First note that the improper integral (A.7) is absolutely convergent both at 0 and ? by the Weierstrass M-test. Since f (z) has only finitely
many poles, we may choose 0 < r < R such that all the poles of f (z) other
than the origin lies in the annulus r < |z| < R. Let D denote this annulus
with branch cut along the positive real axis, i.e. its boundary consisting
of the curves C1 : starting from r and moving along the upper edge of the
positive real axis to R, moving along the bigger circle CR and returning
back to the point R, then moving along the lower edge of the positive real
axis to r, moving along the smaller circle cr , and returning to the starting point r. C1 : z = x, 0 ? x ? R; CR : z = Rei? , 0 < ? < 2?;
C2 : z = xe2?i , x : R ? r; cr : z = re?i? , 0 < ? < 2?. Then we apply
the residue theorem to this cut region. Since the argument of z increases
from 0 to 2?, we have
Z r
Z
a
z f (z) dz =
xa e2?ia f (x) dx,
R
C2
whence it follows that
1?e
2?ia
Z
?
xa f (x) dx = 2?i
0
X
z6=0
Res (z a f (z)) .
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Complex functions
We shall assign a precise meaning to this procedure. We integrate the
branch
g(z) = z a f (z),
0 < arg z < 2?
(with branch cut along the positive real axis as above) along CR and cr :
Z
Z
g(z) dz,
g(z) dz.
CR
cr
By dividing the annulus by any ray starting from the origin and lying
inside the second and the third quadrants, we introduce two regions D1
and D2 with branch cut along the negative and positive imaginary axis,
respectively. For concreteness? sake, we choose the negative real axis (any
ray can do if there are no poles of the integrand on it):
L : z = xe?i , x : R ? r
and ?L. Now integrate the branch
g1 (z) = z a f (z),
3
1
z 6= 0, ? ? < z < ?
2
2
along ?D1 = C1 + CR,1 + L + cr,1 , to get
Z R
Z
Z
Z
a(Log x+i0)
e
f (x) dx +
g1 +
g1 +
r
CR,1
L
g1
cr.1
= 2?i Res ea(log|z|+i arg z) f (z),
z?D1
where CR,1 (resp. cr,1 ) signifies the upper half of CR , (resp. cr ,) . Also
integrating the branch
g2 (z) = z a f (z),
z 6= 0,
5
1
?<z< ?
2
2
along
?D2 = C2 + CR,2 + (?L) + cr,2 ,
to get
?
Z
R
ea(Log x+i0) f (x) dx +
r
Z
g2 +
CR,2
= 2?i Res ea(log|z|+i arg z) f (z),
z?D2
Z
g2 +
?L
Z
g2
cr.2
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where CR,2 (resp. cr,2 ) signifies the lower half of CR , (resp. cr , ). Now note
that except on the positive real axis,
g1 (z) = g(z),
D1 ? ?D1
g2 (z) = g(z),
D2 ? ?D2 ,
and
so that
Z
+
C
Z
+ 1 ? e2?ia
cr
Z
R
ea log x f (x) dx =
r
X
Res z a f (a),
z?D
where D is the union of D1 and D2 with L and ?L removed:
D = (D1 ? D2 ) ? L ? (?L).
Now
Z
and
=
CR4
Z
=
cr
Z
Z
2?
a ia?
R e
f Re
i?
0
0
2?
i?
iRe d? = O
ra eia? f rei? irei? d? = O
Z
Z
R
0
f Re d?
i?
Ra+1 f Rei? d? .
0
?
xa f (x) dx
0
thereby completing the proof.
Corollary A.4
a+1 2?
whence as r ? 0+ and R ? ?, we obtain
Z
Z
z a f (z) dz ? 1 ? e2?ia
?D
2?
For 0 < Re a < 1 we have
Z ? ?a
?
x
dx =
.
x+1
sin ?a
0
We note that Exercise 2.3 could be thought of as giving the value if we
assume those formulas appearing there.
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Appendix B
Summation formulas and convergence
theorems
B.1
Summation formula and its applications
Lemma B.1 (The General Newton-Leibniz Principle) Suppose f :
[a, b] ? R is differentiable on [a, b] except for a finite number of points and
?
+
that f 0 (x) = 0 (at endpoints we assume ddx f (a), ddx f (a) exist.) Further,
suppose f (x) is continuous on [a, b]. Then f is a constant on [a, b].
Proof. If f (x) is differentiable at each point of (c0 , c2 ) ? [a, b] except
for c1 , then by the standard Newton-Leibniz Principle, we have f (x) = C1
(c0 < x < c1 ), and f (x) = C2 (c1 < x < c2 ). By continuity, we obtain
C1 = lim f (x) = lim f (x) = C2 . Therefore, f is a constant on
x?c1 ?0
x?c1 +0
(c0 , c2 ). In case f is not differentiable at the endpoint a, by the continuity
at a, we have f (a) = lim f (x) = C1 .
x?a+0
Theorem B.1 (Abel Summation Formula)
P
a(n),
putting A(t) =
If f (t) ? C 1 [a, x], then
a<n?t
X
a<n?x
a(n) f (n) = A(x) f (x) ?
Z
x
A(t) f 0 (t) dt.
(B.1)
a
Proof. (The first proof due to Arhipov and Chubarikov.) For x > a(? 0),
put
F (x) =
X
a<n?x
Z x
G(x) = ?
a(n) f (n) ? A(x) f (x),
A(t) f 0 (t) dt.
a
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Then for x = m ? Z we have
lim F (x) =
x?m+0
X
a<n?m
f (n) a(n) ? f (m) A(m).
On the other hand,
lim F (x) =
x?m?0
X
a<n?m?1
=
X
a<n?m
=
X
a<n?m
=
X
a<n?m
f (n) a(n) ? f (m) A(m ? 1)
f (n) a(n) ? f (m) a(m) ? f (m) A(m ? 1)
f (n) a(n) ? f (m) (a(m) + A(m ? 1))
f (n) a(n) ? f (m) A(m)
whence it follows that F (x) and G(x) are continuous on [a, x] and piecewise
differentiable on [a, x]. Furthermore
F 0 (x) = G0 (x),
x?
/ Z.
Hence, Lemma B.1 applies, and we have F (x) = G(x) + C. And since
F (a) = G(a) = 0, it follows that F (x) = G(x), i.e. (B.1) ensues.
The second proof uses a special case of the formula for integration by
parts in the theory of Stieltjes integrals stated in the following theorem:
Theorem B.2 Suppose f (t) is continuous Ron [a, b] and that ?(t) is of
b
bounded variation. Then the Stieltjes integral a f (t) d?(t) exists. Further,
Rb
if f (t) is of bounded variation and ?(t) is continuous, then a ?(t) df (t)
exists and the formula for integration by parts
Z b
h
ib Z b
f (t) d?(t) = f (t)?(t) ? ?(t) df (t)
(B.2)
a
a
a
holds true.
Proof. (The second proof.) Putting ?(t) = A(t) =
P
a<n?t
P
a(n), ?(t) is
a step function and so of bounded variation. Since
a(n) f (n) =
a<n?x
Rx
a f (t) d?(t), it follows from (B.2) that
Z x
h
ix Z x
X
a(n) f (n) =
f (t) dA(t) = f (t) A(t) ?
A(t) df (t). (B.3)
a<n?x
a
a
a
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Since f ? C 1 , the last integral may be written as
leads to (B.1).
Rx
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199
A(t)f 0 (t) dt, and (B.3)
Rb
The Stieltjes integral a f (t) d?(t) exists under weaker conditions than
those stated in Theorem B.2.
Theorem B.3 If f (t) and ?(t) are both of bounded variation on [a, b]
and have no common discontinuity, then each one is integrable with respect
to the other from a to b.
Proof. (The third proof of Theorem B.1.) Substituting a(n) = A(n) ?
A(n ? 1), we see that
LHS =
X
A(n) f (n) ?
a<n?x
X
A(m) f (m + 1)
a?1<m?x?1
[x]?1
= ?A([a]) f ([a]) +
X
A(n) (f (n) ? f (n + 1)) + A(x) f ([x]) ,
n=[a]+1
where
R n+1as usual [x] is the integral part of x. Using f (n) ? f (n + 1) =
? n f 0 (t) dt and noting that A(t) = A(n), n < t < n + 1, we obtain
X Z
[x]?1
LHS = ?
=?
n=[a]+1
Z
n+1
A(t) f 0 (t) dt + A(x) f ([x])
n
[x]
A(t) f 0 (t) dt + A(x) f ([x]) .
[a]+1
We rewrite the RHS as
?
Z
x
A(t) f 0 (t) dt +
a
= A(x) f (x) ?
Z
x
Z
[a]+1
A(t) f 0 (t) dt +
a
Z
x
A(t) f 0 (t) dt + A(x) f ([x])
[x]
A(t) f 0 (t) dt.
a
Proof. (The fourth proof.) We transform the integral on the RHS of (B.1)
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by changing the order of summation and integration,
Z x
Z x X
? A(t) f 0 (t) dt = ?
a(n) f 0 (t) dt
a
a a<n?t
=?
=?
X
a(n)
a<n?x
X
Z
x
f 0 (t) dt
n
a(n) f (x) ? f (n)
a<n?x
= ?f (x)
X
a(n) +
a<n?x
X
a(n) f (n),
a<n?x
which leads to (B.1).
Theorem B.4 (Euler?s summation formula)
Let B1 (x) =
B1 (x ? [x]) = x ? [x] ? 21 denote the 1-st periodic Bernoulli polynomial
(cf. (7.9)). Then for f (t) ? C 1 ([a, x]), we have
Z x
h
ix Z x
X
f (t) dt ? B1 (t) f (t) +
f (n) =
B1 (t) f 0 (t) dt.
a<n?x
a
a
a
Putting a(n) = 1 in (B.1), we have
X
A(t) =
1 = [t] ? [a] = t ? B1 (t) ? a ? B1 (a) .
Proof.
a<n?t
Hence (B.1) reads
X
f (n)
a<n?x
Z
x
f (x) ?
t f 0 (t) dt
a
Z x
+ a ? B1 (a) (f (x) ? f (a)) +
B1 (t)f 0 (t) dt
a
Z x
ix Z x
h
ix h
f (t) dt +
= t ? B1 (t) f (t) ? t f (t) +
B1 (t) f 0 (t) dt,
= x ? B1 (x) ? a ? B1 (a)
a
a
a
a
which leads to the RHS.
Proof. (a la? Arhipov and Chubarikov) Putting
X
F (x) =
f (n) ? B1 (a) f (a) + B1 (x) f (x)
a<n?x
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and
G(x) =
Z
x
a
f (t) + B1 (t) f 0 (t) dt,
we shall show that F (x) = G(x). Clearly F (a) = G(a) = 0. G(x), being a
function of the upper limit of integration. is continuous and so is F (x) for
x?
/ Z by definition. And for x = m ? Z, we note that
X
lim F (x) =
x?m+0
f (n) ? B1 (a) f (a) ?
a<n?m
1
f (m) (= F (m)) = lim F (x),
x?m?0
2
i.e. F (x) is continuous for x = m ? Z. This is because the positive jump
P
f (n) at x = m cancels the negative jump of F2 (x) =
of F1 (x) =
a<n?x
B1 (a)f (a) ? B1 (x)f (x), and their sum F (x) is continuous at x = m. Also
it is clear that at x ?
/ Z, both G(x) and F (x) are differentiable and G0 (x) =
F 0 (x). Hence the Newton-Leibniz principle in Lemma B.1 applies.
Theorem B.5 (the Euler-Maclaurin summation formula [Wal]. A
general form of Theorem B.4)
Let B r (x) = Br (x ? [x]) denote the
r-th periodic Bernoulli polynomial. Then for f (t) ? C l [a, x], we have
X
f (n) =
a<n?x
Z
x
l
o
X
(?1)r n
Br (x) f (r?1) (x) ? Br (a) f (r?1) (a)
r!
r=1
Z x
Bl (t) f (l) (t) dt.
f (t) dt +
a
+
(?1)l+1
l!
a
Proof. (The first proof.) Apply integration by parts to Theorem B.4.
Proof. (The second proof.) (Similar to the second proof of Theorem B.4)
Putting
F (x) =
X
f (n) ?
a<n?x
l
o
X
(?1)r n
Br (x) f (r?1) (x) ? Br (a) f (r?1) (a)
r!
r=1
and
G(x) =
Z x
a
(?1)l
(l)
Bl (t) f (t) dt,
f (t) +
l!
we note that F (a) = G(a) = 0, and F 0 (x) = G0 (x), ? x ?
/ Z.
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B.2
Application to the Riemann zeta-function
We illustrate our theory in ДB.1 by specifying to the Riemann zeta-function.
Theorem B.4 with a = 1, x > 0, f (u) = u?s , s ? C gives, on noting
B 1 (1) = ? 12 , B1 (1) = 21 ,
X
n?s =
1<n?x
=
Z
ix Z
h
u?s du ? B1 (u) u?s +
x
1
1
x
1
B1 (u) ?s u?s?1 du
x1?s
1
1
+
? B 1 (x) x?s ? ? s
1?s s?1
2
Z
x
B 1 (u) u?s?1 du,
1
which may be expressed for ? > 0 as
X
Z x
1
x1?s
1
+ +
? B 1 (x) x?s ? s
B 1 (u) u?s?1 du
s?1 2 1?s
1
Z ?
1
x1?s
1
+ +
?s
=
B 1 (u) u?s?1 du + O x?? . (B.4)
s?1 2 1?s
1
n?s =
n?x
For ? > 1, letting x ? ?, we obtain
1
1
?(s) =
+ ?s
s?1 2
Z
?
B 1 (u) u?s?1 du.
(B.5)
1
We note that the RHS is meromorphic in ? > 0, providing a meromorphic
continuation of the LHS to ? > 0. We now showR that the integral is
?
analytic for ? > ?1. For this it suffices to show that 1 B 1 (u) u?s?1 du is
uniformly convergent. To this end we may apply the mean value theorem
or integration by parts:
Z
?
B 1 (u) u?s?1 du =
1
1
B 2 (u) u?s?1
2
?
1
+
s+1
2
Z
?
B 2 (u) u?s?2 du.
1
1
The first term is?
2 B2 and the second term is absolutely convergent for
R?
? > ?1, whence 1 B 1 (u) u?s?1 du is analytic for ? > ?1 and it follows
that (B.5) holds and ?(s) is analytic for ? > ?1 except for s = 1, where it
has a simple pole with residue 1.
Now if we restrict to ?1 < ? < 0, then
s
Z
1
B 1 (u) u
0
?s?1
du = s
Z 1
0
1
u?
2
u?s?1 du = ?
1
1
? .
s?1 2
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Summation formulas and convergence theorems
Hence (B.5) is transformed into
Z ?
?(s) = ?s
B 1 (u) u?s?1 du,
0
?1 < ? < 0,
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203
(B.6)
where the RHS of (B.6) is meromorphic in ?1 < ?. We substitute the
boundedly convergent Fourier series (7.9) for B 1 (x).
Lebesgue?s theorem allows us to integrate term by term to obtain:
Z
?
?X
sin 2?nu ?s?1
u
du
n
0
n=1
Z
?
s X 1 ? ?s?1
=
u
sin 2?nu du.
? n=1 n 0
?(s) =
s
?
We apply a formula in the theory of Mellin transforms (cf. Д7.4)
Z ?
1
z?1
?z
x
sin ax dx = a ?(z) sin ?z , ?1 < Rez < 1.
2
0
(B.7)
(B.8)
Substituting (B.8) in (B.7) we deduce that
?
? X
1
1 s
s
(2?) s ?(?s) sin ? s
n
?
2 n=1 n
? = 2 (2?)s?1 ?(1 ? s) sin s ?(1 ? s)
2
?(s) =
(B.9)
for ?1 < ? < 0, where we used (2.5). For ? < 0, ?(1 ? s) has the Dirichlet
series expression. Formula (B.9) is the asymmetric form of the functional
equation (5.54) for the Riemann zeta-function.
Example B.1
Solution
Find the value of ?(2).
Applying (2.15) to rewrite (B.9) as
?(s) = (2?)s?1
?
?(1 ? s)
?(s) cos ?2 s
(B.10)
Putting s = 2 in (B.10), we get
?(2) = 2?
?
?(?1) = 2? 2 (??(?1))
?(2) (?1)
and we are led to find the value of ?(?1), which we may find with the aid
of (4.3).
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Theorem B.6 (Abel?s continuity theorem) Suppose the power series
?
X
f (z) =
an z n converges at the point z0 on the circle of convergence.
n=0
?
Then as long as z approaches z0 in the angular region |arg(z ? z0 )| < ,
2
?
X
n
the value f (z) approaches the value
an z0 :
n=0
lim f (z) =
z?z0
?
X
an z0n .
n=0
Theorem B.7 (Generalization of Dirichlet?s theorem) Given two
sequences of functions {an (x)}, {bn (s)} defined on R ? R and D ? C,
respectively, suppose that
?
X
X
(i) the partial sums AN (x) =
an (x) of the series
an (x) are
n=1
n?N
bounded uniformly in x
(ii) bn (s) ? 0 uniformly on D
and that
(iii) there is a Majorant series
?
X
n=1
that
cn < ? of positive terms cn such
|bn (s) ? bn+1 (s)| ? cn
for every s ? D and for all n sufficiently large.
P?
Then the series n=1 an (x) bn (s) is uniformly convergent on R and D.
P?
If bn (s) are analytic in D, so is the sum function n=1 an (x) bn (s) for
each x ? R.
Proof. By the formula for partial summation (cf. the third proof of Theorem B.3.), we have for integers M , N , M < N ,
X
M <n?N
an (x) bn (s) =
X
M <n?N
An (x) (bn (s) ? bn+1 (s))
+ AN (x) bN (s) ? AM (x) bM +1 (s)
?
?
X
= O?
cn ? + O(|bN (s)| + |bM +1 (s)|)
M <n?N
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Summation formulas and convergence theorems
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205
which can be made < ? for any ? > 0 uniformly in x and s provided that
M and N are sufficiently large. Hence the Cauchy criterion applies and
uniform convergence follows.
For fixed x ? R, analyticity of the sum function is a consequence of
Theorem A.3
Corollary B.1 (Dirichlet?s
N
test for uniform convergence) If bn ? 0
X
as n ? ?, and an (x) = O(1) uniformly in x ? R, then the series
n=1
?
X
bn an (x) is uniformly convergent on R.
n=1
Example B.2
If bn ? 0, then
?
X
bn sin 2?nx is uniformly convergent
n=1
in any interval not containing an integer. This follows from Exercise 7.8,
(7.14). In particular, the case bn = n1 establishes the uniform convergence of
?
1 X sin 2?nx
the Fourier series ?
for B1 (x) in any interval not containing
? n=1
n
an integer. Therefore, Lebesgue?s theorem allows to integrate it term by
term to obtain B2 (x) (and higher order periodic Bernoulli polynomials).
This example is a special case of the following.
Proposition B.1 The series for the polylogarithm function ls (x) defined
in the first instance for ? > 1 by
ls (x) =
?
X
e2?inx
ns
n=1
is uniformly convergent in any interval of x free from an integer (0 < x < 1)
for ? > 0.
Proof.
By Exercise 7.8, the partial sums satisfy
n
X
1
1
2?ikx ?i(n+1)x sin ?n =O
?
,
e
= e
sin ?x | sin ?x|
kxk
k=1
kxk indicating the distance to the nearest integer.
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Bibliography
[Ap1] T. M. Apostol, Remark on the Hurwitz zeta function, Proc. Amer. Math.
Soc. 2 (1951), 690?693.
[Ap2] T. M. Apostol, On the Lerch zeta-function, Pacific J. Math. 1 (1951),
161?167.
[Ap3] T. M. Apostol, Addendum to ?On the Lerch zeta-function?, Pacific J. Math.
2 (1952), 10.
[Ap4] T. M. Apostol, Introduction to analytic number theory, Springer, 1976.
[Ber6] B. C. Berndt, Identities involving the coefficients of a class of Dirichlet
series VI, Trans. Amer. Math. Soc. 160 (1971), 157?167.
[Ber3] B. C. Berndt, On the Hurwitz zeta-function, Rocky Mount. J. Math. 2
(1972), 151?157.
[Ber4] B. C. Berndt, Two new proofs of Lerch?s functional equation, Proc. Amer.
Math. Soc. 32 (1972), 403?408.
[Ber2] B. C. Berndt, The gamma function and the Hurwitz zeta-fu
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