On the Asymptotics to all Orders of the Riemann Zeta Function and of a Two-Parameter Generalization of the Riemann Zeta Function 1470450984, 9781470450984

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Number 1351

On the Asymptotics to all Orders of the Riemann Zeta Function and of a Two-Parameter Generalization of the Riemann Zeta Function Athanassios S. Fokas Jonatan Lenells

January 2022 • Volume 275 • Number 1351 (fifth of 6 numbers)

Number 1351

On the Asymptotics to all Orders of the Riemann Zeta Function and of a Two-Parameter Generalization of the Riemann Zeta Function Athanassios S. Fokas Jonatan Lenells

January 2022 • Volume 275 • Number 1351 (fifth of 6 numbers)

Library of Congress Cataloging-in-Publication Data Cataloging-in-Publication Data has been applied for by the AMS. See http://www.loc.gov/publish/cip/. DOI: https://doi.org/10.1090/memo/1351

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27 26 25 24 23 22 22

Dedicated to Trevor Stuart with deep gratitude

Contents Part 1.

Asymptotics to all Orders of the Riemann Zeta Function

Chapter 1. Introduction 1.1. The large t asymptotics of ζ(s) valid to all orders 1.2. The explicit form of certain sums 1.3. The explicit form of the difference of certain sums 1.4. Asymptotics of a two-parameter generalization of Riemann’s zeta function 1.5. Fourier coefficients of the product of two Hurwitz zeta functions 1.6. Several representations for the basic sum

1 3 3 5 6 7 7 9

Chapter 2. An Exact Representation for ζ(s)

11

Chapter 3. The Asymptotics of the Riemann Zeta Function for t ≤ η < ∞

15

Chapter 4. The Asymptotics of the Riemann Zeta Function for 0 < η < t

31

Chapter 5. Consequences of the Asymptotic Formulae

49

Part 2. Asymptotics to all Orders of a Two-Parameter Generalization of the Riemann Zeta Function

53

Chapter 6. An Exact Representation for Φ(u, v, β)

55

Chapter 7. The Asymptotics of Φ(u, v, β)

59

Chapter 8. More Explicit Asymptotics of Φ(u, v, β)

73

Chapter 9. Fourier coefficients of the product of two Hurwitz zeta functions 9.1. Corollaries 9.2. Asymptotics of the zeroth Fourier coefficient 9.3. Proof of Theorem 9.1 9.4. Proof of Theorem 9.2 9.5. Fourier coefficients of ζ(s, α) and ζ1 (s, α)

81 83 86 89 91 93

Part 3.

95

Representations for the Basic Sum

Chapter 10. Several Representations for the Basic Sum

97

Appendix A. The Asymptotics of Γ(1 − s) and χ(s)

107

Appendix B. Numerical Verifications B.1. Verification of Theorem 3.1

109 109 v

vi

CONTENTS

B.2. B.3. B.4. B.5.

Verification Verification Verification Verification

Bibliography

of of of of

Theorem 3.2 Theorem 4.1 Corollary 4.3 Theorem 4.4

109 110 110 110 113

Abstract We present several formulae for the large t asymptotics of the Riemann zeta function ζ(s), s = σ + it, 0 ≤ σ ≤ 1, t > 0, which are valid to all orders. A particular case of these results coincides with the classical results of Siegel. Using  these formulae, we derive explicit representations for the sum ba n−s for certain ranges of a and b. In addition, we present precise estimates relating this sum with d the sum c ns−1 for certain ranges of a, b, c, d. We also study a two-parameter generalization of the Riemann zeta function which we denote by Φ(u, v, β), u ∈ C, v ∈ C, β ∈ R. Generalizing the methodology used in the study of ζ(s), we derive asymptotic formulae for Φ(u, v, β).

Received by the editor December 1, 2015, and, in revised form, July 6, 2018, October 26, 2018, and November 5, 2018. Article electronically published on December 21, 2021. DOI: https://doi.org/10.1090/memo/1351 2020 Mathematics Subject Classification. Primary 11M06, 30E15; Secondary 33E20. Key words and phrases. Riemann zeta function, asymptotics. The authors are grateful to Professor D. Bump for several useful suggestions that led to a significant improvement of the manuscript. They are also grateful to Kostis Kalimeris for many helpful remarks. The first author is affiliated with the Department of Applied Mathematics and Theoretical Physics, University of Cambridge, Cambridge, CB3 0WA, United Kingdom and was supported by the EPSRC, UK via a senior fellowship and by the Guggenheim Foundation, USA. He is deeply grateful to his students Michail Dimakos (partially supported by the Onassis Foundation) and Dionyssis Mantzavinos (partially supported by the EPSRC, UK) for performing a large number of numerical experiments during the early stages of this project. He also expresses his sincere gratitude to Anthony Ashton, Bryce McLeod, Joe Keller, David Stuart, and Eugene Shargorodsky, for extensive discussions and important suggestions. The second author is affiliated with the Department of Mathematics, KTH Royal Institute of Technology, 100 44 Stockholm, Sweden and acknowledges support from the EPSRC, UK, the G¨ oran Gustafsson Foundation, Sweden, the Ruth and Nils-Erik Stenb¨ ack Foundation, Finland, the European Research Council, Grant Agreement No. 682537, and the Swedish Research Council, Grant No. 2015-05430. c 2022 American Mathematical Society

vii

Part 1

Asymptotics to all Orders of the Riemann Zeta Function

CHAPTER 1

Introduction It is well known, see for example theorem 4.11 in Titchmarsh [26], that the Riemann zeta function ζ(s), s = σ + it, σ, t ∈ R, satisfies the equation (1.1) ζ(s) =

   1 1 x1−s − , + O ns 1−s xσ

|t|
1,

σ > 0,

t → ∞.

Here we present a systematic investigation of asymptotic formulae for ζ(s). In this connection, in chapter 2 we derive the following exact formula (see theorem 2.1): ζ(1 − s) =

η [ 2π ]

n

s−1

n=1

   ∞eiφ1  ∞eiφ2 s−1

iπs iπs 1 ηs z dz − + +e 2 − + e 2 , iπ iπ s − (2π) s ez − 1 ηe 2 ηe 2

(1.2) 0 < η < ∞,

−

π π < φj < , 2 2

j = 1, 2,

η η where [ 2π ] denotes the integer part of 2π and the contours of integration in the first and second integrals in the rhs of equation (1.2) are the rays from η exp(−iπ/2) to ∞ exp(iφ1 ) and from η exp(iπ/2) to ∞ exp(iφ2 ), respectively.

1.1. The large t asymptotics of ζ(s) valid to all orders Equation (1.2) suggests a separate analysis for the cases t < η, t = η and t > η. The first two cases are analysed in theorems 3.1 and 3.2 of chapter 3. Theorem 3.1 presents the large t asymptotics of ζ(s) valid to all orders in the case of (1 + )t < η. For example, the formulae of theorem 3.1 yield the following equation for the leading asymptotic terms in the case when (1 + )t < η for some > 0: η [ 2π ]

ζ(s) =



n−s −

n=1

(1.3)

 +O

 , 2+σ t2

η



1  η 1−s 2iη −s t − iσ iη iη Re Li + ) + (e ) −i arg(1 − e 2 1 − s 2π (2π)1−s η (1 + )t < η < ∞,

0 ≤ σ ≤ 1,

t → ∞,

where the error term is uniform for η, σ in the above ranges and the polylogarithm Lim (z) is defined by Lim (z) =

∞  zk , km

k=1

3

m ≥ 1.

4

1. INTRODUCTION

Theorem 3.2 presents the large t asymptotics of ζ(s) valid to all orders in the case of η = t. The formula of theorem 3.2 yields the following equation for the leading asymptotic terms:  1−s t 1 1 − s 2π n=1

−it 2it−s eit iσ e [t + i(σ − 1)] − i Im Li1 (eit ) + Re Li2 (eit ) − − (2π)1−s 2t t  1 − Im Li3 (eit ) t

−s it 1 + i√ t e 1 + πt − i(3σ − 2) 1−s (2π) 2 3   i−1√  2 1  + √ π 6σ − 6σ + 1 + 45σ 3 − 45σ 2 + 4 135t 24 t     1+i √  4 3 2 − π 36σ − 24σ − 24σ + 12σ + 1 + O t−σ−2 , 3 576t 2 0 ≤ σ ≤ 1, t → ∞, t [ 2π ]

ζ(s) =

(1.4)



n−s −

where the error term is uniform for σ in the above range. The best estimate for the growth of ζ(s) as t → ∞, is not based on equation (1.1) but on the well known approximate functional equation, see for example equation (4.12.4) on page 79 of Titchmarsh [26]:   1  1 1 t −σ 2 −σ y σ−1 , xy = (1.5) , + χ(s) + O x + t ζ(s) = s 1−s n n 2π n≤x

n≤y

where 0 < σ < 1 and χ(s) is defined by χ(s) =

 πs (2π)s sin Γ(1 − s), π 2

with Γ(s) denoting the Gamma function. Building on Riemann’s unpublished notes, Siegel in his classical paper [24] presented the error termsof the rhs of equation (1.5) to all orders, only in the important case of x = y = t/2π. In theorems 4.1 and 4.4 we present analogous results for any x and y valid to all orders. Theorem 4.1 yields the following explicit √ result for the subleading term in the rhs of equation (1.5) when 1 < η = 2πy < t. For every > 0, there exists an A > 0 such that [t]

η

[ 2π ] iπ η   1 1 e−iπs Γ(1 − s) −([ ηt ]+1)iη iπ2 (s−1) s e 4 √ ζ(s) = e + χ(s) − e η ns n1−s 1 − e−iη i 2πt n=1 n=1    1 η , 1 < η < At 3 < ∞, O t5/6 −iπs − πt σ−1 +e Γ(1 − s)e 2 η × √  − At2 3  1 O e η + tη3/2 , t 3 < η < A t < ∞,

dist(η, 2πZ) > ,

0 ≤ σ ≤ 1,

t → ∞,

where the error term is uniform for all η, σ in√the above ranges. In corollary 4.3, similar formulas are presented for the case 2π t < η < 2π  t.

1.2. THE EXPLICIT FORM OF CERTAIN SUMS

5

√ The case of t < η < t is analyzed in theorem 4.4, which provides a formula valid to all orders in terms of the function Φ(τ, u) defined by  (1.6)

2

Φ(τ, u) = 01

eπiτ x +2πiux dx, eπix − e−πix

τ < 0,

u ∈ C,

where 0 1 denotes a straight line parallel to e3πi/4 which crosses the real axis between 0 and 1. The formula of theorem 4.4 yields the following result for the leading asymptotic terms: For every > 0, [t]

η

[ 2π ] η   1 1 (1.7) ζ(s) = + χ(s) s n n1−s n=1 n=1

η 2 2t η it iπ(s−1) [ ]πi−it− 2η 2 (2[ 2π ]π−η) + e−iπs Γ(1 − s) e 2 η s−1 e η 2π          σ πt η σ−1 η πi − iη Φ + O e− 2 5/6 × Φ+ , ∂2 Φ + 2 iη 2π t √

t < η < t, 0 ≤ σ ≤ 1, t → ∞,

where the error term is uniform for η, σ in the above ranges and Φ and ∂2 Φ are evaluated at the point   2πt 2t 2πt  η   t  1 − − − 2 , − 2 (1.8) . η η η 2π η 2 √ Remark 1.1. 1. In the particular case of η = 2πt, equation (1.7) agrees with the analogous formula of Siegel [24]. 2. Although Siegel suggested already in [24] the possibility of deriving a formula such as (1.7) based on the function (1.6), the authors have not been able to locate such a formula in the literature. In the case of σ = 1/2, an alternative asymptotic representation for ζ(s) involving a sum which is smoothly rather than sharply truncated is presented in [6]. An introduction to the Riemann-Siegel formula can be found in chapter 7 of [8].  2πt 3. Equation (1.7) is particularly useful in the case when η = b where b > 0 is a rational number. Indeed, in this case the function Φ and its derivatives evaluated at the point (1.8) can be computed explicitly (see equation (4.30) below). Therefore, in this case theorem 4.4 yields an explicit asymptotic expansion of ζ(s) to all orders. 4. Equation (1.3) is only useful as an asymptotic formula in the case when t/η → 0 as t → ∞ (otherwise the error term is as large as the retained terms). Therefore the asymptotic range where t < η and t/η = O(1) is not covered by this equation. However, this asymptotic sector is covered by corollary 4.3.

1.2. The explicit form of certain sums b Theorems 3.1 and 3.2 allow us to evaluate the sum a n−s , for certain a and b, to all orders, see theorems 5.1 and 5.2. For example, the leading order of such

6

1. INTRODUCTION

sums follows immediately from equations (1.3) and (1.4): For every > 0,

(1.9)

η2 [ 2π ]

n

−s

η

1 +1 n=[ 2π ]

  η2 1−s  η1 1−s 1 = − + O(η1−σ ), 1−s 2π 2π (1 + )t < η1 < η2 < ∞,

η 2π

(1.10)

[]

n

−s

t n=[ 2π ]+1

1  η 1−s 1 it πi = + e e4 1 − s 2π 2 (1 + )t < η < ∞,



0 ≤ σ ≤ 1, t 2π

 12 −s

0 ≤ σ ≤ 1,

t → ∞,

+ O(t−σ ), t → ∞,

uniformly with respect to η1 , η2 , η, and σ.

1.3. The explicit form of the difference of certain sums It is well known, see for example page 78 of Titchmarsh [26], that for large t, the following sums coincide to the leading order: 

n−s

and

χ(s) t 2πN

x 0,

and that the modified Hurwitz zeta function ζ1 (s, α) is defined by ζ1 (s, α) = ζ(s, α) − α−s = ζ(s, α + 1). For each choice of u, v ∈ C \ {1}, the product ζ1 (u, α)ζ1 (v, α) is a smooth function of α ∈ [0, 1]; hence it can be represented by its Fourier series  ζ1 (u, α)ζ1 (v, α) = (1.16) qn (u, v)e2πinα , n∈Z

where the Fourier coefficients qn (u, v) are defined by  1 qn (u, v) = ζ1 (u, α)ζ1 (v, α)e−2πinα dα,

n ∈ Z.

0

For u = v¯ = σ + it, the zeroth Fourier coefficient q0 (u, v) is given by the mean square average of the modified Hurwitz function:  1 q0 (σ + it, σ − it) = |ζ1 (σ + it, α)|2 dα. 0

Attempts to compute the large t asymptotics of q0 (σ +it, σ −it) have a long history, especially in the case when σ = 1/2. Increasingly refined asymptotic estimates were derived in an extensive series of papers (see [4, 14, 17, 21, 27, 28]) before the following formula was finally obtained independently in [1] and [29]: 2  1  1   ζ1 1 + it, α  dα = ln t + γ − 2 Re ζ( 2 + it) + O(t−1 ), (1.17) 1   2 2π 0 2 + it where γ denotes the Euler constant. An alternative derivation of (1.17) was presented in [15]. In the derivation of [15], the asymptotic formula (1.17) is obtained as an easy consequence of the following identity [15, Eq. (2.1) with N = 1]: (1.18)

q0 (u, v) =

1 + R0 (u, v) + R0 (v, u) − T0 (u, v) − T0 (v, u), u+v−1

where 0 < Re u, Re v < 2 and R0 (u, v) and T0 (u, v) are defined by Γ(1 − v) , Γ(u)  ∞ ∞  l1−u−v β u+v−2 (1 + β)−u−1 dβ.

(1.19)

R0 (u, v) = Γ(u + v − 1)ζ(u + v − 1)

(1.20)

T0 (u, v) =

u ζ(u) − 1 + 1−v 1−v

l=1

l

Actually, a version of the identity (1.18) is presented as the main result of [15], because from this identity many asymptotic results (such as (1.17)) can easily be obtained. A key point here is that, for u = σ + it and v = σ − it, the sum appearing in the definition (1.20) of T0 (u, v) is small as t → ∞. In chapter 9, we will analyze the large t asymptotics of the nth Fourier coefficient qn for any n ∈ Z by establishing a generalization of (1.18). We will first show

1.6. SEVERAL REPRESENTATIONS FOR THE BASIC SUM

that R0 and T0 are the analytic continuations of the integrals  ∞ (1.21a) α−v ζ1 (u, α)dα, Re(u + v) > 2, R0 (u, v) = 0  1

(1.21b)

T0 (u, v) =

α−v ζ1 (u, α)dα,

9

Re v < 1,

Re v < 1.

0

Then, we will prove (see Theorem 9.1) the following generalization of (1.18) which is valid for n ∈ Z and Re u, Re v < 2: (1.22)

qn (u, v) = bn (u + v) + Rn (u, v) + Rn (v, u) − Tn (u, v) − Tn (v, u),

where, for any n ∈ Z, bn (s), Rn (u, v), and Tn (u, v) are the analytic continuations of the following functions:  ∞ bn (s) = α−s e−2πinα dα, Re s > 1, 1  ∞ Rn (u, v) = α−v ζ1 (u, α)e−2iπnα dα, Re(u + v) > 2, Re v < 1, 

0 1

Tn (u, v) =

α−v ζ1 (u, α)e−2πinα dα,

Re v < 1.

0 1 Since b0 (s) = s−1 , it follows immediately from (1.21) that equation (1.22) reduces to (1.18) when n = 0. Just like the identity (1.18) can be used to find the large t asymptotics of the zeroth Fourier coefficient q0 , the identity (1.22) can be used to find the large t behavior of the nth Fourier coefficient qn . The function Φ(u, v, β) enters in this formulation, because it turns out (see equation (9.6b)) that for n = 0 the function Rn (u, v) is given by the following analog of (1.19):  πiv ie n ≥ 1, Γ(1 − v) sin(πu) , Φ(u, v, n) × 2ie Rn (u, v) = −πiv Γ(u) n ≤ −1. 2 sin(πu) ,

We will show in chapter 9 that the asymptotics of Tn can be obtained (as in the case n = 0) for any n = 0. Thus, the computation of the leading asymptotics of the Fourier coefficients qn , n = 0, reduces to the calculation of the large t asymptotics of Φ(u, v, n) for integer values of n. By substituting the asymptotic results for Φ obtained in chapter 7 into (1.22), we obtain in chapter 9 an asymptotic formula for the nth Fourier coefficient of |ζ1 (σ + it, α)|2 . These considerations also lead to Lindel¨of type asymptotic bounds on the function Φ(u, v, n) for large t (see Corollary 9.9): ⎧ ⎪ σ ∈ [0, 1/2), ⎨O(1), n ≥ 1, t → ∞. Φ(σ + it, σ − it, n) = O(ln t), σ = 1/2, ⎪ ⎩ 2σ−1 ), σ ∈ (1/2, 1], O(t 1.6. Several representations for the basic sum The remarkable fact about the large t-asymptotics of the Riemann zeta function is that, whereas the higher order terms in the asymptotic expansion can be computed explicitly, the sum appearing in the leading order is transcendental. In chapter 10 we present several integral representations of this basic sum with the

10

1. INTRODUCTION

hope that these representations may be useful for the estimation of this fundamental sum (some results in this direction can be found in [2]). Let Cηt denote the semicircle from iη to it with Re z ≥ 0: 

 i(η + t) t − η iθ  π π + e − δ, dist(t, 2πZ) > δ, and the contour in the second equality above denotes the principal value integral with respect to the points 

 t   η    +1≤n≤ . 2πn  n ∈ Z, 2π 2π Basic Assumption: In the statements of all results throughout this work, it will be assumed (also when not explicitly mentioned) that (1.24)

η > 0,

t > 0,

η∈ / 2πZ,

t∈ / 2πZ.

CHAPTER 2

An Exact Representation for ζ(s) In chapters 2 and 3, we assume that the branch cut for the logarithm runs along the negative real axis. By the standing assumption (1.24), we always have η, t ∈ / 2πZ. Theorem 2.1 (An exact representation for Riemann’s zeta function). Let ζ(s), s = σ + it, σ, t ∈ R, denote the Riemann zeta function. Then, η  [ 2π ] ns−1 ζ(s) = χ(s) n=1

(2.1)

   ∞eiφ1  ∞eiφ2 −z s−1  iπs 1 ηs e z dz − iπs 2 2 + +e − + e , iπ iπ s − (2π) s 1 − e−z ηe 2 ηe 2 π π 0 < η < ∞, − < φj < , j = 1, 2, 2 2

where

 πs (2π)s sin Γ(1 − s), π 2 with Γ(s), s ∈ C, denoting the Gamma function, and where the contours of the first and second integrals in the rhs of ( 2.1) are the rays from η exp(−iπ/2) to ∞ exp(iφ1 ) and from η exp(iπ/2) to ∞ exp(iφ2 ) respectively. An equivalent formula is η   [  ∞eiφ1  ∞eiφ2 −z s−1

2π ] iπs e z dz 1 ηs s−1 − iπs 2 2 n + +e − + e , ζ(1 − s) = iπ iπ s − (2π) s 1 − e−z ηe 2 ηe 2 n=1 (2.2)

χ(s) =

(2.3) 0 < η < ∞, −

π π < φj < , j = 1, 2. 2 2

Proof. We will first derive equation (2.1) in the particular case 0 < η < 2π. In this case, replacing η with α, (2.1) can be written as    ∞eiφ1  ∞eiφ2 −z s−1  iπs iπs e z χ(s) αs − ζ(s) = +e 2 dz , − + e 2 iπ iπ (2π)s s 1 − e−z αe− 2 αe 2 π −π < φj < , j = 1, 2. 2 2 We decompose the Hankel contour Hα in the expression (1.14) for ζ(s) into the union of three different contours (see figure 2.1), namely

(2.4)

0 < α < 2π;

Hα = L 1 ∪ L 2 ∪ L 3 , 11

12

2. AN EXACT REPRESENTATION FOR ζ(s)

iα

L1

L3 Re z

L2

−iα

Figure 2.1. The decomposition of Hα into L1 + L2 + L3 . iα 1 3 2 −iα Figure 2.2. The domains 1, 2, and 3.

where

π !  L1 = αeiθ  < θ < π ∪ {|z|eiπ | α < |z| < ∞)}, 2  π!  −iπ L2 = {|z|e | α < |z| < ∞} ∪ αeiθ  − π < θ < − , 2  ! π π  (2.5) . L3 = αeiθ  − < θ < 2 2 The integral along L1 can be written as follows:   ∞eiπ  ∞ z s−1 ez e−u us−1 du s−1 iπs dz = (2.6) z dz = e , iπ iπ −z − 1 1 − ez 1 − e−u L1 e αe 2 αe− 2 where the contours of the second and third integrals in (2.6) are the rays in the cut complex z-plane from α exp(iπ/2) to ∞ exp(iπ) and from α exp(−iπ/2) to ∞ respectively. Indeed, in the domain enclosed by L1 and by the ray from α exp(iπ/2) to ∞ exp(iπ), i.e., in the shaded domain 1 of figure 2.2, we have π z = reiθ , α ≤ r ≤ ∞, ≤ θ ≤ π; 2 thus, |ez | = er cos θ , α ≤ r ≤ ∞, 0 ≤ cos θ ≤ −1. Hence, Cauchy’s theorem implies the first equality of (2.6). Then, the substitution u = z exp(−iπ), yields the second equality in (2.4). Similarly,  (2.7) L2

z s−1 dz = e−z − 1



αe−

iπ 2

∞e−iπ

ez z s−1 dz = e−iπs 1 − ez



αe

∞

iπ 2

e−u us−1 du, 1 − e−u

2. AN EXACT REPRESENTATION FOR ζ(s)

13

where the curve L2 is defined in (2.5) and the contours of the second and third integrals in (2.7) are the rays in the cut complex z-plane from ∞ exp(−iπ) to α exp(−iπ/2) and from ∞ to α exp(iπ/2) respectively. Indeed, the first equality in (2.7) is the consequence of Cauchy’s theorem applied in the domain enclosed by L2 and by the ray from ∞ exp(−iπ) to α exp(−iπ/2) (i.e., in the shaded domain 2 of figure 2.2), whereas the second equality in (2.7) follows from the substitution u = z exp(iπ). The integral along L3 can be written as follows: ⎛ ⎞  ∞  αe iπ2   πs αs s−1 −z z ⎠ e (2.8) dz = − ⎝ , + z s−1 dz − 2i sin iπ −z −z −1 1−e 2 s L3 e ∞ αe− 2 where the curve L3 is defined in (2.5) and the contours in the first and second integrals in (2.8) are the rays in the cut complex z-plane from α exp(−iπ/2) to ∞ and from ∞ to α exp(iπ/2). Indeed,    z s−1 e−z s−1 dz = − (2.9) z dz − z s−1 dz. −z − 1 −z e 1 − e L3 L3 L3 The second term in the rhs of (2.9) yields the second term in the rhs of (2.8). Furthermore, in the domain enclosed by L3 and by the two rays (α exp(−iπ/2), ∞) and (∞, α exp(iπ/2)), i.e., in the shaded domain 3 of figure 2.2, we have π π z = reiθ , α ≤ r ≤ ∞, − < θ < ; 2 2 thus |e−z | = e−r cos θ , α ≤ r ≤ ∞, cos θ ≥ 0. Hence, Cauchy’s theorem implies that the first term of the rhs of (2.9) equals the first term in the rhs of (2.8). Adding equations (2.6)–(2.8) we find equation (2.4) but with φ1 = φ2 = 0. However, Cauchy’s theorem implies that the rays (α exp(−iπ/2), ∞) and (∞, α exp(iπ/2)) can be replaced with the rays (α exp(−iπ/2), ∞ exp(iφ1 )) and

(∞ exp(iφ2 ), α exp(iπ/2)),

respectively, and then we find (2.4). In order to derive equation (2.1) with 2π < η < ∞, we introduce the curves −α Cαη , C−η and Cˆηα , which are the following semi-circles in the complex z-plane, see figure 2.3: 

 i(η + α) (η − α) iθ  π π + e − < θ < Cαη = (2.10a) , 2 2 2 2 

 −i(η + α) (η − α) iθ  π π −α + e − < θ < = (2.10b) , C−η 2 2 2 2 

 i(η + α) (η − α) iθ   π π + e  −π < θ < ∪ 0, 0 ≤ σ ≤ 1, N ≥ 2,

t → ∞,

where the error term is uniform for all η, , σ, N in the above ranges and (2N + 1)!! is defined by (2N + 1)!! = 1 · 3 · 5 · · · · · (2N − 1)(2N + 1). For N = 3 equation ( 3.1) simplifies to η [ 2π ]

(3.2)

ζ(s) =

1  η 1−s 1 − s 2π n=1

−s 2iη t − iσ + −i arg(1 − eiη ) + Re Li2 (eiη ) 1−s (2π) η  ' 1& + 2 it2 + (2σ + 1)t − iσ(σ + 1) Im Li3 (eiη ) η    1 + 8  1 + O 3+σ t3 + , η

(1 + )t < η < ∞, > 0, 0 ≤ σ ≤ 1, t → ∞, 

n−s −

15

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

16

where the error term is uniform for all η, , σ in the above ranges and the polylogarithm Lim (z) is defined by Lim (z) =

(3.3)

∞  zk , km

m ≥ 1.

k=1

Similarly, it is straightforward for any N ≥ 4 to derive an asymptotic formula for ζ(s) analogous to ( 3.2) with an error term of order    1 + 2(N +1)  1 N O N +σ t + . η

Proof. Suppose first that 0 < t ≤ η < ∞, > 0, 0 ≤ σ ≤ 1, N ≥ 2. All error terms of the form O(·) will be uniform with respect to η, , σ, N unless otherwise specified. The proof is based on equation (2.3), i.e., η [ 2π ]

(3.4) ζ(1 − s) =



ns−1 −

n=1

1  η s + GL (t, σ; η) + GU (t, σ; η), s 2π

where GL and GU are defined by iπs  ∞eiφ1 e−z z s−1 dz e 2 (3.5) GL (t, σ; η) = , (2π)s −iη 1 − e−z

0 < η < ∞,

−

π π < φ1 < , 2 2

0 < η < ∞,

−

π π < φ2 < , 2 2

0 < η < ∞.

and e− 2 (3.6) GU (t, σ; η) = (2π)s iπs



∞eiφ2 iη

e−z z s−1 dz , 1 − e−z

The assumption η ∈ / 2πZ implies that GL and GU are well-defined. The asymptotics of GL Using the expansion (3.7)

∞  1 = e−nz , 1 − e−z n=0

Re z > 0,

in the definition (3.5) of GL , we find (3.8)

iφ1 iπs ∞  e 2  ∞e e−nz z s−1 dz, GL (t, σ; η) = (2π)s n=1 −iη

−

π π < φ1 < , 2 2

0 < η < ∞.

The interchange of the integration and the summation in (3.8) is allowed because the sum on the rhs of (3.7) converges absolutely and uniformly in Re z ≥ δ for any δ > 0. Integration by parts shows that the integral in (3.8) can be written as  j   ∞eiφ1 N −1 σ−1   1 d z  e−nz z s−1 dz = e−nz+it ln z it dz it  n− z n− z −iη j=0 z=−iη  ∞eiφ1 (3.9) + e−nz+it ln z DN dz, −iη

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

17

where DN = DN (z, n, σ, t) is short-hand notation for  N 1 d DN := (3.10) z σ−1 . dz n − it z We claim that

iφ1   iπs ∞  e 2  ∞e −nz+it ln z 2 σ−1−N e D dz = O (2N − 1)!!(N + 1) η . N (2π)s n=1 −iη

(3.11)

In order to derive equation (3.11) we first note that we can write DN in the form DN =

(3.12)

N  N  b=0 c=0

where

(N ) Abc ,

(N )

Abc z σ−1

tb (nz)N −b σ c (nz − it)2N

b = 0, . . . , N , c = 0, . . . , N , are integers which satisfy

(N )

|Abc | ≤ (2N − 1)!! = 1 · 3 · 5 · · · · · (2N − 1),

b = 0, . . . , N,

c = 0, . . . , N.

Indeed, if DN can be written as in (3.12), then the computation   b N −b c 1 σ d (N ) σ−1 t (nz) Abc z dz n − it (nz − it)2N z (N )

= Abc z σ−1

tb (nz)N −b σ c [it(b − N ) − itσ − nz(1 + b + N ) + nzσ)] (nz − it)2(N +1)

shows that DN +1 also can be written as in (3.12). In order to estimate the lhs of (3.11), we choose φ1 = 0. Then, since t ≤ |z| and |n − it z | ≥ n for all z on the contour, we can estimate    N  N    t b (nz)N σ c   DN = O (2N − 1)!!|z|σ−1  nz 2N  z 2N (n − it z) b=0 c=0   1 = O (2N − 1)!!|z|σ−1−N (N + 1)2 N , z = u − iη, u ∈ [0, ∞). n Thus, the lhs of (3.11) is   ∞  ∞  − πt −nz it ln z 2 σ−1−N 1 2 |e e |(2N − 1)!!(N + 1) |z| |dz| O e nN n=1 −iη    ∞ ∞  1 2 −n Re z σ−1−N = O (2N − 1)!!(N + 1) e |z| |dz| nN −iη n=1    ∞ ∞  1 −nu = O (2N − 1)!!(N + 1)2 η σ−1−N e du nN 0 n=1   ∞  1 = O (2N − 1)!!(N + 1)2 η σ−1−N nN +1 n=1   = O (2N − 1)!!(N + 1)2 η σ−1−N . This proves (3.11).

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

18

Equations (3.8), (3.9), and (3.11) imply that GL satisfies (3.13)

j   iπs ∞ N −1 d z σ−1  e 2   −nz+it ln z 1 e GL (t, σ; η) =  (2π)s n=1 j=0 dz n − it n − it z z z=−iη   + O (2N − 1)!!(N + 1)2 η σ−1−N , t ≤ η < ∞, t → ∞.

The asymptotics of GU We now let > 0 and suppose that 0 < (1 + )t < η < ∞. In analogy with equations (3.8) and (3.9) we now find

(3.14)

iφ2 iπs ∞  e− 2  ∞e GU (t, σ; η) = e−nz z s−1 dz, (2π)s n=1 iη

−

π π < φ2 < , 2 2

0 < η < ∞,

and 

∞eiφ2

−nz s−1

e iη

z

dz =

N −1 

e

j=0



(3.15)

 −nz+it ln z

∞eiφ2

+

1 n−

it z

d dz

j

 z σ−1   n − it z z=iη

e−nz+it ln z DN dz,

iη

where DN is defined by (3.10). We claim that

(3.16)

iφ2   iπs ∞   1 + 2N e− 2  ∞e −nz+it ln z σ−N e DN dz = O (2N − 1)!!N η . (2π)s n=1 iη

Indeed, since DN can be written in the form given by (3.12), Jordan’s lemma implies that we may choose φ2 = π/2. Then, using the estimate    N  N    t b (nz)N σ c   DN = O (2N − 1)!!|z|σ−1  nz 2N  z 2N (n − it z) b=0 c=0   nN = O (2N − 1)!!|z|σ−1−N (N + 1)2 , z = iλ, (n − ηt )2N

λ ∈ [η, ∞),

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

19

we find that the lhs of (3.16) is  ∞   πt O e2 n=1

i∞



|e−nz+it ln z DN dz|

iη

 ∞   πt =O e2 n=1

i∞

e−

πt 2

nN |dz| (n − ηt )2N  N



(2N − 1)!!|z|σ−1−N (N + 1)2

iη

  2 = O (2N − 1)!!(N + 1)

∞

∞ 

n 1 2N (n − η 1+ ) n=1   σ−N  1 + 2N 2 η = O (2N − 1)!!(N + 1) N −σ

   1 + 2N = O (2N − 1)!!N η σ−N .

λ

σ−1−N

dλ

In the third equality above we have used the fact that

(3.17)

∞  n=1

nN (n −

1 2N 1+ )

= ≤

∞  1 nN (1 − n=1

1 (1 −

1 2N 1+ )

1 1 2N (1+)n ) ∞ 

  1 + 2N 1 = O , nN

n=1

N ≥ 2.

This proves (3.16). We next claim that (3.16) can be modified as follows:

(3.18)

iφ2 iπs ∞  e− 2  ∞e e−nz+it ln z DN dz (2π)s n=1 iη    1 + 2(N +1) σ−N −1 η . = O (2N + 1)!!N

Indeed, integration by parts shows that the lhs of (3.16) equals iπs ∞ e− 2  −nz+it ln z 1 e (2π)s n=1 n−

   D N it  z

z=iη

+

iφ2 iπs ∞  e− 2  ∞e e−nz+it ln z DN +1 dz. (2π)s n=1 iη

By (3.16), the second term is    1 + 2(N +1) σ−N −1 η , O (2N + 1)!!(N + 1)

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

20

while, in view of (3.12) and (3.17), the first term satisfies iπs ∞ N N e− 2  −niη+it ln iη 1   (N ) tb (niη)N −b σ c e Abc (iη)σ−1 t s (2π) n=1 (niη − it)2N n − η b=0 c=0   N  ∞ N   tb nN −b (N ) = O η σ−1−N |Abc | η b (n − ηt )2N +1 n=1 b=0 c=0   ∞  nN σ−1−N 2 =O η (2N − 1)!!(N + 1) 1 2N +1 (n − 1+ ) n=1    1 + 2N +1 = O η σ−1−N (2N − 1)!!(N + 1)2 .

This proves (3.18). Equations (3.14), (3.15), and (3.18) imply that GU satisfies (3.19)

j   iπs ∞ N −1 1 d z σ−1  e− 2   −nz+it ln z GU (t, σ; η) = e  (2π)s n=1 j=0 dz n − it n − it z z z=iη    1 + 2(N +1) η σ−N −1 , + O (2N + 1)!!N (1 + )t < η < ∞.

Proof of (3.1) Substituting the expressions (3.13) and (3.19) for GL and GU into (3.4), we find η [ 2π ]

ζ(1 − s) =



n=1

ns−1 −

1  η s s 2π

iπs ∞ N −1 e 2   −nz+it ln z + e (2π)s n=1 j=0



1 n−

it z

d dz

j

 z σ−1   n − it z z=−iη

j   iπs ∞ N −1 d z σ−1  1 e− 2   −nz+it ln z + e  (2π)s n=1 j=0 dz n − it n − it z z z=iη    1 + 2(N +1) + O (2N + 1)!!N η σ−N −1 ,

(1 + )t < η < ∞, t → ∞. Replacing σ by 1 − σ and taking the complex conjugate of the resulting equation, we find (3.1).

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

21

Proof of (3.2) Letting N = 3 in (3.1), we find η [ 2π ]

ζ(s) =



n=1

n−s −

1  η 1−s 1 − s 2π

iπ(1−s) ∞  1 1 i(q(σ − 1) + σ) e− 2 −s −iηn 1 + 2 + (iη) e 1−s (2π) n 1 + q ηn (1 + q)3 n=1    1 −q 2 (σ − 1)2 + q −2σ 2 + σ + 2 − σ(σ + 1) + 2 3 η n (1 + q)5

iπ(1−s) ∞  1 1 1 i(q(σ − 1) − σ) e 2 −s + 2 + (−iη) eiηn (2π)1−s n 1 − q ηn (1 − q)3 n=1    1 q 2 (σ − 1)2 + q −2σ 2 + σ + 2 + σ(σ + 1) − 2 3 η n (1 − q)5   1 + 8  t + O η −σ−3 . , q=

nη The above sums can be expanded as series in q with coefficients expressible in terms of the polylogarithms Lim (z) defined by (3.3). For example, k ∞ ∞   eiηn eiηn  t = n(1 − q) n=1 n nη n=1 k=0   ∞ ∞ k  t eiηn = η nk+1 n=1 k=0 ∞  k    t = Lik+1 eiη , η ∞ 

k=0

where the interchange of the two sums can be justified by separating the terms where k = 0 and noticing that the remaining double sum is absolutely convergent. More precisely, N k k M  M  N N   eiηn  eiηn  t eiηn  t + lim = lim lim lim N →∞ M →∞ N →∞ M →∞ n nη n n nη n=1 n=1 n=1 k=0 k=1 k ∞ ∞   eiηn  t = Li1 (eiη ) + n nη n=1 k=1   ∞ ∞   t k eiηn = Li1 (eiη ) + η nk+1 k=1 n=1 ∞  k    t = Lik+1 eiη . η k=0

22

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

Similarly, for any j ≥ 0 and l, m ≥ 1,

 k+j ∞  ∞ ∞   t eiηn qj eiηn  k + m − 1 = l (1 − q)m l k n n nη n=1 n=1 k=0  k+j ∞   k+m−1 t = Lik+l+j (eiη ). k η k=0

Letting η → −η in this equation, we find  k+j ∞ ∞    k+m−1 e−iηn (−q)j t = Lik+l+j (e−iη ). − l (1 + q)m k n η n=1 k=0

This gives the following expression for the Riemann zeta function: η [ 2π ]

1  η 1−s 1 − s 2π n=1

  k k iπ(1−s) ∞  ∞  e− 2 iσ  k + 2 t t −s −iη + (iη) Li (e ) + Lik+2 (e−iη ) − − k+1 k (2π)1−s η η η k=0 k=0  k+1 ∞   k+2 i(σ − 1) t Lik+3 (e−iη ) − − k η η k=0  k ∞  σ(σ + 1)  k + 4 t Lik+3 (e−iη ) − − k η2 η k=0  k+1 ∞  2 2σ − σ − 2  k + 4 t Lik+4 (e−iη ) + − k η2 η k=0   k+2 ∞  (σ − 1)2  k + 4 t −iη Li (e ) − − k+5 k η2 η k=0

   k  iπ(1−s) ∞ ∞  k e 2 t t iσ  k + 2 −s iη + (−iη) Lik+1 (e ) − Lik+2 (eiη ) k (2π)1−s η η η k=0 k=0  k+1 ∞  t i(σ − 1)  k + 2 Lik+3 (eiη ) + k η η k=0  k ∞  t σ(σ + 1)  k + 4 − Lik+3 (eiη ) k η2 η k=0  k+1 ∞  t 2σ 2 − σ − 1  k + 4 Lik+4 (eiη ) + 2 k η η k=0   k+2 ∞  2  k+4 t (σ − 1) iη Lik+5 (e ) − k η2 η k=0   1 + 8  + O η −σ−3 .

ζ(s) =



n−s −

By including only the first few terms in the sums over k, we find an expansion for ζ(s) in powers of ηt . For example, including only the terms of order larger than

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

23

3

t O( η3+σ ) we have

η [ 2π ]

ζ(s) =

1  η 1−s 1 − s 2π n=1  2

t iη −s t −iη −iη − Li3 (e−iη ) Li1 (e ) − Li2 (e ) + 1−s (2π) η η iσ 3iσt + Li2 (e−iη ) − 2 Li3 (e−iη ) η η  i(σ − 1)t σ(σ + 1) −iη −iη + Li (e ) − Li (e ) 3 3 η2 η2 

2 t iη −s t iη iη + Li3 (eiη ) Li1 (e ) + Li2 (e ) + (2π)1−s η η iσ 3iσt − Li2 (eiη ) − 2 Li3 (eiη ) η η  i(σ − 1)t σ(σ + 1) iη iη + Li (e ) − Li (e ) 3 3 η2 η2     1 + 8 1 + O 3+σ t3 + . η



n−s −

After simplification we find

η [ 2π ]

ζ(s) =

1  η 1−s 1 − s 2π n=1  2

t 2iη −s t iη iη Re Li + (e ) + (e ) + i Im Li3 (eiη ) i Im Li 1 2 (2π)1−s η η iσ 3σt − Re Li2 (eiη ) + 2 Im Li3 (eiη ) η η  (σ − 1)t iσ(σ + 1) iη iη − Im Li3 (e ) − Im Li3 (e ) η2 η2     1 + 8 1 + O 3+σ t3 + , η



n−s −

  The identity Li1 eiη = − ln(1 − eiη ) now yields (3.2).



3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

24

Theorem 3.2 (The asymptotic expansion to all orders for the case t = η). Let ζ(s), s = σ + it, σ, t ∈ R, denote the Riemann zeta function. Then, t [ 2π ]

ζ(s) =



n=1

n−s −

1 1−s



t 2π

1−s

j   iπ(1−s) ∞ N −1   1 d z −σ  e− 2 −nz−it ln z + e  (2π)1−s n=1 j=0 dz n + it n + it z z z=it j   iπ(1−s) ∞ N −1   d z −σ  e 2 1 −nz−it ln z + e  (2π)1−s n=2 j=0 dz n + it n + it z z z=−it  1−s    2N ck (1 − σ)Γ( k+1 t it 2N −σ−N 2 ) + e + O (2N + 1)!!N 2 t , k+1 2π t 2 k=0 0 ≤ σ ≤ 1,

(3.20)

N ≥ 2,

t → ∞,

where the error term is uniform for all σ, N in the above ranges and the coefficients ck (σ) are given by equation ( 3.30) below. The first few of the ck ’s are given by 1−i , 2 i c1 (σ) = − iσ, 3  1+i 2 c2 (σ) = − 6σ − 6σ + 1 , 12 (3.21)  1  −45σ 3 + 90σ 2 − 45σ + 4 , c3 (σ) = 135  i−1 36σ 4 − 120σ 3 + 120σ 2 − 36σ + 1 , c4 (σ) = 432  i  c5 (σ) = 189σ 5 − 945σ 4 + 1575σ 3 − 987σ 2 + 168σ + 8 , 5670  1+i  c6 (σ) = 1080σ 6 − 7560σ 5 + 18900σ 4 − 20160σ 3 + 8190σ 2 − 450σ − 139 . 194400 c0 (σ) =

For N = 3 equation ( 3.20) simplifies to  1−s t 1 ζ(s) = n − 1 − s 2π n=1

−it −s it ' e & 2 2it e iσ Re Li2 (eit ) −t − it(σ − 1) + (σ − 1)2 − i Im Li1 (eit ) + − 1−s 2 (2π) 2t t  it + σ + σ 2 2 + 3σ 3i it it it + i Im Li3 (e ) + Re Li4 (e ) + 2 Im Li5 (e ) t2 t2 t t [ 2π ]



−s

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

(3.22)

t−s eit 1 + i √ 1 + πt − i(3σ − 2) (2π)1−s 2 3    √ i−1 1  + √ π 6σ 2 − 6σ + 1 + 45σ 3 − 45σ 2 + 4 135t 24 t  1+i √  π 36σ 4 − 24σ 3 − 24σ 2 + 12σ + 1 − 3 576t 2  i  189σ 5 − 315σ 3 + 42σ 2 + 84σ − 8 + 2 2835t 1−i √  π 1080σ 6 + 1080σ 5 − 2700σ 4 − 1440σ 3 + 5 103680t 2     2 0 ≤ σ ≤ 1, + 1710σ + 270σ − 139 + O t−σ−3 ,

25

t → ∞,

where the error term is uniform for all σ in the above range and the polylogarithm Lim (z), m ≥ 1, is defined by ( 3.3). Similarly, it is straightforward for any N ≥ 4 to derive an asymptotic formula for ζ(s) analogous to ( 3.22) with an error term of  order O t−σ−N . Proof. Setting η = t in (3.4), we find t [ 2π ]

(3.23) ζ(1 − s) =



n

s−1

n=1

1 − s



t 2π

s + GL (t, σ; t) + GU (t, σ; t),

0 < t < ∞.

Equation (3.13) is valid also when η = t and gives the asymptotics of GL (t, σ; t). On the other hand, using the expansion (3.7), we write (1)

(2)

GU (t, σ; t) = GU (t, σ; t) + GU (t, σ; t), where e− 2 = (2π)s iπs

(3.24)

(1) GU (t, σ; t)



∞eiφ2

e−z z s−1 dz,

π π < φ2 < , 2 2

0 < t < ∞,

π π < φ2 < , 2 2

0 < t < ∞.

−

it

and (3.25) (2)

GU (t, σ; t) =

iφ2 iπs ∞  e− 2  ∞e e−nz z s−1 dz, (2π)s n=2 it

(2)

−

The asymptotics of GU can be found using integration by parts, whereas the (1) asymptotics of GU will be computed by considering the critical point at z = it. (2) The asymptotics of GU Repeating the steps that led to (3.19) but with the sum over n only going from 2 to ∞ and with t = η, we find the following analog of (3.19): j   iπs ∞ N −1 d z σ−1  e− 2   −nz+it ln z 1 (2) e GU (t, σ; t) =  (2π)s n=2 j=0 dz n − it n − it z z z=it   + O (2N + 1)!!N 22N tσ−N −1 . (3.26)

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

26

(1)

The asymptotics of GU (1) Letting in the definition (3.24) of GU , φ2 = 0 and ρ ∈ [0, ∞),

z = it + tρ, we find e− 2 = (it)s−1 te−it (2π)s iπs

(1) GU



∞

e−t(ρ−i ln(1−iρ)) (1 − iρ)σ−1 dρ.

0

Letting v = ρ − i ln(1 − iρ),

(3.27) we find

 (1)

GU =

t 2π

s

e−it

 γ

e−tv

(1 − iρ(v))σ dv, ρ(v)

where ρ(v) is defined by inverting (3.27) and γ denotes the image of the contour [0, ∞) under (3.27). In order to ascertain that the value of ρ(v) is well-defined, we consider in detail the map φ defined by φ : C \ [−i, −i∞) → C, ρ → v = ρ − i ln(1 − iρ), where [−i, −i∞) is a branch cut and the principal branch is chosen for the logarithm, i.e., (3.28)

v = ρ − i ln |1 − iρ| + arg(1 − iρ),

arg(1 − iρ) ∈ (−π, π).

The function φ satisfies φ(ρ) = 0 iff ρ = 0; ρ φ (ρ) = = 0 iff ρ = 0; i+ρ iρ2 + O(ρ3 ), v = φ(ρ) = − ρ → 0. 2 Moreover, we claim that φ maps the first quadrant of the complex ρ-plane bijectively onto the region delimited by the contour γ = φ([0, ∞)) and the positive imaginary axis in the complex v-plane, see figure 3.1. Indeed, it is clear that φ maps [0, i∞) bijectively onto [0, i∞). Writing ρ = ρ1 + iρ2 , we have ∂ Re φ |ρ|2 + ρ2 ∂ Re φ ρ1 = 2 , = 2 , ∂ρ1 ρ1 + (ρ2 + 1)2 ∂ρ2 ρ1 + (ρ2 + 1)2 This shows that each level curve of the function Re φ in the first quadrant intersects the positive ρ-axis in a unique point and that the vector   −ρ1 , |ρ|2 + ρ2 (3.29) is tangent to the level curve passing through ρ. Since ∂ Im φ −ρ1 = 2 , ∂ρ1 ρ1 + (ρ2 + 1)2

∂ Im φ |ρ|2 + ρ2 = 2 , ∂ρ2 ρ1 + (ρ2 + 1)2

we infer that the derivative of Im φ in the direction of (3.29) equals |ρ|2 , showing that Im φ is strictly increasing along each level curve. This proves the claim. Thus, if v belongs to the region delimited by the contour γ = φ([0, ∞)) and the positive imaginary axis, we may define ρ(v) as the unique inverse image of v under

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

Im ρ

27

Im v

φ

Re ρ −i

Re v −i

γ

Figure 3.1. The function φ maps the indicated shaded areas in the complex ρ-plane bijectively onto the corresponding shaded areas in the v-plane. φ which belongs to the first quadrant. Using analyticity to deform the contour γ to the positive real axis, we obtain  s  ∞ t (1 − iρ(v))σ (1) −it dv. e e−tv GU = 2π ρ(v) 0 −1 We claim that there exist coefficients {ck (σ)}N such that 0 N −1  √ (1 − iρ(v))σ 1  =√ (3.30) ck (σ)v k/2 + rN ( v) , v ≥ 0, ρ(v) v k=0 √ where rN ( v) satisfies  N +1  √  ∞ Γ( 2 ) −tv rN ( v) √ dv = O e (3.31) . N +1 v t 2 0 Indeed, since φ has a double zero at ρ = 0, there exists a neighborhood V of ρ = 0 in the complex ρ-plane and a Riemann surface Σ which is a√two-sheeted cover √ of V with local parameter λ := v, such that the map ρ → v is bijective and  holomorphic V → Σ. The function φ is analytic away from √ [−i, −i∞) and φ (ρ) = 0 for all ρ = 0, thus the Riemann surface Σ and the map v → ρ can be analytically extended as long as φ−1 (v) ∩ [−i, −i∞) = ∅. The function φ maps the segments [−i + 0, −i∞ + 0) and √ [−i − 0, −i∞ − 0) onto the lines Re v = −π and Re v = π respectively, hence v → ρ is well-defined and analytic for all |v| < π. Thus, the map eσ ln(1−iρ) (1 − iρ)σ =λ f : λ → λ ρ ρ √ is holomorphic from the open disk {0 < |λ| < π} ⊂ Σ to C. Let

f (k) (0) , k = 0, . . . , N − 1, k! and define rN (λ) so that (3.30) holds. Then rN (λ) satisfies ck (σ) =

rN (λ) =

∞  f (k) (0) k λ k!

k=N

28

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

√ for 0 ≤ |λ| < π. In order to prove (3.31), it is sufficient to show that there exists a constant C independent of N such that √ N |rN ( v)| ≤ Cv 2 , (3.32) v ≥ 0, N ≥ 2, 0 ≤ σ ≤ 1. (k)

Cauchy’s estimates imply that the numbers ck (σ) = f k!(0) are uniformly bounded for all k ≥ 0 and √ σ ∈ [0, 1]. Thus, an application of Cauchy’s estimates in a disk of radius |λ| = r < π yields  N k ∞   |λ| |λ| 1 |rN (λ)| ≤ Mr ≤ Mr , |λ| < r, r r 1 − |λ| k=N r √ where Mr = max|λ|=r |f (λ)|. Choosing any r ∈ (1, π) we see that such a C exists for 0 ≤ v ≤ 1. On the other hand, equations (3.28) and (3.30) imply that such a C exists also for v ≥ 1. This proves (3.32) and hence also (3.30) and (3.31). Equation (3.30) together with the identity  ∞ Γ(α + 1) e−tv v α dv = , α > −1, tα+1 0 (1)

imply the following asymptotic expansion of GU :  s  N +1  N −1  Γ( 2 ) ck Γ( k+1 t (1) 2 ) e−it + O GU (t, σ; t) = (3.33) . N +1 k+1 2π 2 t 2 −σ t k=0 Proof of (3.20) Equations (3.13), (3.26), and (3.33) give the asymptotics of GL and GU . Substitution into (3.23) yields t [ 2π ]

ζ(1 − s) =



n

s−1

n=1

1 − s



t 2π

s

iπs ∞ N −1 e 2   −nz+it ln z + e (2π)s n=1 j=0 iπs ∞ N −1 e− 2   −nz+it ln z + e (2π)s n=2 j=0



+

t 2π

s

−it

e

 

1 n− 1 n−

2N  ck (σ)Γ( k+1 ) 2

k=0

t

k+1 2

it z

d dz

it z

d dz

j j

 z σ−1   n − it z z=−it  z σ−1   n − it z z=it

  2N σ−N −1 + O (2N + 1)!!N 2 t ,

where we have replaced N by 2N + 1 in (3.33) and used that Γ(N + 1) = N ! ≤ (2N + 1)!! to eliminate one of the error terms. Replacing σ by 1 − σ and taking the complex conjugate of the resulting equation, we find (3.20). Proof of (3.22) Letting N = 3 in (3.20) and using the expressions in (3.21) for {ck }60 , we find that the term on the rhs of (3.20) involving the ck ’s yields the term involving the second curly bracket on the rhs of (3.22). On the other hand, using that 1 n+

it z

d z −σ z 1−σ (nσz + i(σ − 1)t) =− it dz n + z (nz + it)3

3. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR t ≤ η < ∞

and 

1 n+

it z

d dz

2

29

    z 1−σ n2 σ(σ + 1)z 2 + in 2σ 2 − σ − 2 tz − (σ − 1)2 t2 z −σ = , (nz + it)5 n + it z

long but straightforward computations show that the two terms involving the double sums on the rhs of (3.20) yield the term involving the first curly bracket on the rhs of (3.22). This proves (3.22).  Remark 3.3. Using the basic identity (1.2), it is possible to analyze the case of 0 < η < t in a way that is similar to the case of η > t. Actually, this approach is implemented in [9] where the asymptotics of the Hurwitz zeta function to all orders is computed. However, in the next section we find it convenient to analyze the case of 0 < η < t by implementing the approach introduced by Siegel. We expect that Siegel’s approach can be used also to obtain asymptotic formulas for the Hurwitz and Lerch zeta functions. In this regard, we note that an analog of the approximate functional equation (1.5) for the Lerch zeta function is derived in [10] and that a Riemann-Siegel type integral representation for the Lerch zeta function is presented in [3]. The large t asymptotics of the second moment of the Lerch function is derived in [11]. Another analog of (1.5) for the Lerch function as well as an alternative proof of the large t asymptotics of its second moment can be found in [18].

CHAPTER 4

The Asymptotics of the Riemann Zeta Function for 0 < η < t In this chapter, we consider the asymptotics of ζ(s) as √ t → ∞ with √ 0 < η < t. Theorem 4.1 and its corollary treat the cases < η < t and 2π t < η < 2π  t under the assumption that dist(η, 2πZ) >

for some

> 0. The case when η is of √ the same order as t and the case when dist(η, 2πZ) → 0 are covered by theorem 4.4 and its corollary. Throughout this chapter, we assume that the branch cut for the logarithm runs along the positive real axis. Recall the standing assumption (1.24) that η, t ∈ / 2πZ. Theorem 4.1 (The asymptotics to all orders for the case < η < For every > 0, there exists a constant A > 0 such that

√ t).

(4.1) [t]

η

[ 2π ] η   1 1 ζ(s) = + χ(s) ns n1−s n=1 n=1

 k+ 12   [ N 2−1 ] ϕ(2k) (0) k 2η 2 1 e−iπs Γ(1 − s) −([ ηt ]+1)iη iπ (s−1) s−1 iπ  e i e2 η e4 Γ k+ − 2πi (2k)! t 2 k=0

−iπs

− πt 2

Γ(1 − s)e η  ⎧  N   ⎪ η 6 ⎪ √ , ⎨O 2N t t  ×  2 N2+1  ⎪ − At 2 ⎪ η ⎩O N e + Ntη ,

+e

σ−1

1

< η < t 3 < ∞, 1

t3 < η
,

1≤N
0 be given and suppose that 4 < η < t, 0 ≤ σ ≤ 1, m = [t/η], dist(η, 2πZ) > 4 , and N ≥ 1. All error terms of the form O(·) will be uniform with respect to η, σ, N (but not with respect to ). In order to analyze the integral I2 , we write  it (w−iη)2 e 2η2 ϕ(w − iη)dw, I2 = e−(m+1)iη (iη)s−1 C2

where 2 z it t t (s−1) ln(1+ iη )− ηt z− 2η 2 z +( η −[ η ])z

ϕ(z) =

e

Defining φ(z) by (4.4)

.

ez − e−iη 



2 z it (s−1) ln 1+ iη − ηt z− 2η 2z

φ(z) = e

,

we have φ(z)e( η −[ η ])z . ez − e−iη t

ϕ(z) =

t

We split the contour C2 as follows: C2 = C2 ∪ C2r , where C2 denotes the segment of C2 of length 2 which consists of the points within a distance from iη. We write I2 = I2 + I2r , where I2 = e−(m+1)iη (iη)s−1



it

e 2η2 C2

(w−iη)2

ϕ(w − iη)dw

34

4. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR 0 < η < t

and



I2r = e−(m+1)iη (iη)s−1

it

e 2η2

(w−iη)2

C2r

ϕ(w − iη)dw.

We claim that I2r can be estimated as follows:   σ 2 πt η − t I2r = O e− 2 √ e 12η2 . (4.5) t iπ

Indeed, the change of variables w = iη + λe 4 gives    −  η  2 πt iπ − t λ2 I2r = O e− 2 η σ−1 + e 2η2 |ϕ(λe 4 )|dλ . − η2

We have (4.6)



 ( t −[ t ])z  e η η     ez − e−iη  = O(1),

iπ

η . 2

< |λ|
, 0 ≤ σ ≤ 1, t → ∞, η where the error terms are uniform for all η, σ, N within the above ranges, and the function ψ(z) is defined by dist

iηz

ψ(z) =

iη 2

e−s ln(1+ 2πt )+ 8π2 t z ez − e

2πit η

2

η −[ 2π ]z

.

Proof. We replace σ by 1 − σ in (4.1) and take the complex conjugate of both sides. We then multiply the resulting equation by χ(s) and use the identities ζ(¯ s) = ζ(s),

Γ(¯ s) = Γ(s),

χ(s)χ(1 − s) = 1,

(4.25)

χ(¯ s) = χ(s),

χ(s)ζ(1 − s) = ζ(s).

This yields the following equation [t]

ζ(s) = χ(s)

η 

η [ 2π ]

1 n1−s

+

n=1 iπ(1−s)

 1 ns n=1

Γ(s) ([ ηt ]+1)iη iπs −s − iπ e e 2 η e 4 2πi 2 z it t  2 k+ 12   [ N 2−1 ]  d2k  e−s ln(1− iη )+ 2η2 z −[ η ]z  1 1 k 2η  × Γ k+ (−i) dz 2k z=0 ez − eiη (2k)! t 2

+ χ(s)

e

k=0

+ χ(s)eiπ(1−s) Γ(s)e− 2 η −σ  ⎧   2N  N6 η 1 ⎪ ⎪ √ ,

< η < t 3 < ∞, 1 ≤ N < At ⎨O t η3 , t  × N +1   √ 2 2 1 ⎪ − At ⎪ ⎩O N e η2 + Ntη , t 3 < η < t < ∞, 1 ≤ N < At η2 , πt

dist(η, 2πZ) > , Replacing η by

2πt η ,

we find (4.24).

0 ≤ σ ≤ 1,

t → ∞, 

40

4. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR 0 < η < t

√ Theorem 4.4 (The asymptotics to all orders for the case t < η < t). For every > 0, there exists a constant A > 0 such that [t]

(4.26)

η

[ 2π ] η   1 1 ζ(s) = + χ(s) ns n1−s n=1 n=1

η 2 2t η it iπ(s−1) [ ]πi−it− 2η 2 (2[ 2π ]π−η) + e−iπs Γ(1 − s) e 2 η s−1 e η 2π SN (s, η)

√

t < η < t,

   N πt 3N 6 η σ √ + O e− 2 , t t 0 ≤ σ ≤ 1,

1 ≤ N < At,

t → ∞,

where the error term is uniform for all η, σ, N in the above ranges and the function SN (s, η) is defined by k n     n η  (4.27) SN (s, η) = an 2 πi − iη k 2π n=0 k=0   2πt 2t 2πt  η   t  1 n−k − − × ∂2 Φ − 2 , − 2 η η η 2π η 2 N −1 

with ∂2 denoting differentiation with respect to the second argument. The coefficients an are determined by the recurrence formula (4.28)

iη(n + 1)an+1 = (σ − n − 1)an −

it an−2 , η2

n = 0, 1, 2, . . . ,

together with the initial conditions a−2 = a−1 = 0 and a0 = 1, and the function Φ(τ, u) is defined by  (4.29)

2

Φ(τ, u) = 01

eπiτ x +2πiux dx, eπix − e−πix

τ < 0,

u ∈ C,

with the contour 0 1 denoting a straight line parallel to e3πi/4 which crosses the real axis between 0 and 1.2 In the particular case when p, q > 0 are integers, we have

(4.30)

2 In

 q−1    2p p 1 Φ − ,u = (−1)n e−πin q −2πinu q 1 − (−1)q e−πiqp−2πiqu n=0  p−1 1 2 (−1)q e−πiqp−2πiqu 3πi  πiq (u+n+ ) 2  + e 4 e p . p/q n=0

the particular case of η = paper [18, Eq. (1.4)].

√

2πt, an analog of (4.26) is stated without proof in the recent

4. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR 0 < η < t

41

For N = 3 equation ( 4.26) simplifies to (4.31) [t]

η

[ 2π ] η   1 1 ζ(s) = + χ(s) s n n1−s n=1 n=1

η 2 2t η it iπ(s−1) [ ]πi−it− 2η 2 (2[ 2π ]π−η) + e−iπs Γ(1 − s) e 2 η s−1 e η 2π        σ−1 η πi − iη Φ × Φ+ ∂2 Φ + 2 iη 2π  2         (σ − 2)(σ − 1) 2 η η − πi − iη ∂ πi − iη Φ + 2 2 Φ + 2 Φ ∂ 2 2 2η 2 2π 2π   σ √ πt η + O e− 2 ,

t < η < t, 0 ≤ σ ≤ 1, t → ∞, t

where the error term is uniform for all η, σ in the above ranges and Φ and its partial derivatives are evaluated at the point ( 1.8).  2πt Remark 4.5. 1. If η = b where b = p/q > 0 is a rational number, then equations (4.26)-(4.30) provide an explicit asymptotic expansion of ζ(s) to all orders. The particular case b = 1 is the famous case analyzed by Siegel in [24] using ideas from Riemann’s unpublished notes. We note that, building on the work of [16], an elementary (but formal) derivation of the Riemann-Siegel formula is presented in [5]. 2. The results of theorem 4.4 are convenient for √ studying the higher-order t. However, if the order of η asymptotics of ζ(s) in the case when η = constant × √ is strictly smaller or larger than the order of t, the results of theorem 4.4 are less convenient, because in this case the asymptotics of ζ(s) involves the asymptotics of the sums in (4.30) as p and/or q tend to infinity. In this case, the alternative representation of the asymptotics of theorem 4.1, which avoids the appearance of the above sums, is usually more convenient. 3. The error term in (4.26) is uniform with respect to η, σ, N . More precisely, this means that equation (4.26) is equivalent to the following statement: For every

> 0, there exist constants A > 0, C > 0, K > 0 such that the inequality  η [ ηt ] [ 2π   ] 1 1  − χ(s) ζ(s) −  ns n1−s n=1 n=1   η 2 2t η it iπ(s−1) [ ]πi−it− (2[ ]π−η)  2π 2η 2 − e−iπs Γ(1 − s)e 2 η s−1 e η 2π SN (s, η)   N πt 3N 6 η σ √ , ≤ C|e−iπs Γ(1 − s)|e− 2 t t holds for all t > K and all σ, N, η such that 0 ≤ σ ≤ 1, 1 ≤ N < At, and √

t < η < t. 4. The main difference between the proofs of theorems 4.1 and 4.4 can be explained as follows. Consider the representation (4.3) of ζ(s). If η is of strictly smaller order than t1/2 , the main contribution to the asymptotics of ζ(s) comes

42

4. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR 0 < η < t

from the part of the contour C2 that lies within a distance of the critical point iη; the remaining part of C2 gives a contribution which is suppressed by a factor −const× ηt2 of the form e cf. Eq. (4.5). Thus, the proof of theorem 4.1 relies on a direct study of the integral near the critical point √ iη using the method of steepest descent. On the other hand, in the case when η ∼ t, the asymptotic expansion of ζ(s) depends on the integral along all of the contour C2 . The contour C2 is a line segment of total length η centered on the critical point iη. In the proof of theorem 4.4, we handle this integral by relating it to the function Φ(τ, u) defined in (4.29). 5. The coefficients an satisfy the estimate  [n]  t3 an = O , (not uniformly in n); ηn indeed, assuming this estimate up to n, we find  an+1 = O

n

t[ 3 ] η n+1



 +O

n−2

t t[ 3 ] η 3 η n−2



 =O

n+1  t[ 3 ] , η n+1

(not uniformly in n).

Proof of Theorem 4.4. Let t > 0, 0 ≤ σ ≤ 1, and m = [t/η]. As in the proof of theorem 4.1, ζ(s) is given by (4.3) where the contributions from I1 , I3 , and I4 are exponentially small, and the error terms of the form O(·) are uniform with respect to the variables η, σ, N as t → ∞. It remains to analyze the integral I2 . In the neighborhood of w = iη we have   2 w − iη 1 w − iη w = (s − 1) − (s − 1) ln + ··· iη iη 2 iη =

t it (w − iη) + 2 (w − iη)2 + · · · . η 2η

Hence we write t

w

e(s−1) ln iη = e η

2 it (w−iη)+ 2η 2 (w−iη)

φ(w − iη),

where φ(z) is defined by (4.4). Define {an }∞ 0 by φ(z) =

∞ 

an z n ,

|z| < η,

n=0

The identity dφ = dz



 s−1 t it − − 2 z φ(z) iη + z η η

implies (iη + z)

∞  n=1

 nan z

n−1

= s − 1 − (iη + z)



t itz + 2 η η

  ∞

an z n .

n=0

Hence the coefficients an are determined in succession by the recurrence formula (4.28) supplemented with the conditions a−2 = a−1 = 0 and a0 = 1. The first few

4. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR 0 < η < t

43

coefficients are given by σ−1 (σ − 2)(σ − 1) , a2 = − , iη 2η 2 −2t + i(σ − 3)(σ − 2)(σ − 1) a3 = , 6η 3 (σ − 4)(σ − 3)(σ − 2)(σ − 1) + 2i(4σ − 7)t a4 = . 24η 4

a0 = 1,

a1 =

Representing φ(z) in the form φ(z) =

N −1 

an z n + rN (z),

n=0

we find

 φ(w) zN dw, |z| < η, N 2πi Γ w (w − z) where Γ is a contour contained in the disk of radius η centered at the origin which encircles the points 0 and z once. 3 Let |z| < 47 η (so that 21 20 |z| < 5 η) and let Γ be a circle with center w = 0 and radius ρN , where 21 3 |z| ≤ ρN ≤ η. 20 5 Equation (4.7) implies the following analog of (4.8): rN (z) =

Re ln φ(z) ≤ |σ − 1| ln Hence rN (z) = O



3

5tρN 2πρN 6η 3 |z|N ρ−N N e ρN − |z|

8 5t|z|3 + , 5 6η 3



|z| ≤

3 η. 5

  5tρ3 N N −N 6η3 = O |z| ρN e ,

|z|
0 such that (4.39) I2S =



t

C2

(iη)s−1

eη

(w−iη)+ 2i

t η2

−1 (w−iη)2 −mw N 

ew − 1

an (w − iη)n dw + O(e−

πt 2 −At

),

n=0

where C2 denotes the infinite straight line of which C2 is a part. Indeed, in view of (4.38), if we replace C2 by C2 , the integral multiplying an in the expression for I2S changes by   O η σ−1 e−

(4.40)

πt 2

∞

− 2ηt 2 λ2 n

e

λ dλ .

η 2

We write the integrand as − 4ηt 2 λ2 n

− 4ηt 2 λ2

λ ×e

e



.

2n t η,

and so it decreases throughout The first factor is steadily decreasing for λ > the interval of integration provided that n < N < At, with A sufficiently small. The term in (4.40) is then (4.41)  O η

− 4ηt 2 σ−1 − πt 2

e

e

η2 4

 η n  2

∞

− 4ηt 2 λ2

e

dλ

η 2

Also, by (4.32), choosing z such that |z| ≤

  2  πt − t η η n η √ . = O η σ−1 e− 2 e 4η2 4 2 t

20 2 13 21 ( 5t ) η,

we find

−n

(4.42)

an = (rn (z) − rn+1 (z))z   n   n+1 3 5et 3 5et n n+1 −n =O |z| − |z| |z| 2nη 3 2(n + 1)η 3  n 5et 3 =O , 1 ≤ n ≤ N − 1. 2nη 3

Multiplying (4.41) by an and summing from 0 to N − 1, we find that the total error is    n  N −1  n   t 5et 3 η η σ−1 − πt − e 2 16 √ a0 + O η 2 2nη 3 t n=1    n  N −1   t 5et 3 η σ−1 − πt − e 2 16 √ 1 + . =O η 16n t n=1

46

4. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR 0 < η < t n

Now the factor (t/n) 3 increases steadily up to n = t/e, and so if n < At, where A < 1/e, it is 1 1 O(e 3 tA ln A ). Hence if N < At, with A sufficiently small, the total error is O(e−

πt 2 −At

).

This proves (4.39). Finally, we analyze the sum 2 t it  N −1 (w−iη)+ 2η 2 (w−iη) −mw  eη (iη)s−1 (w − iη)n dw. (4.43) an w −1  e C2 n=0 The integral in (4.43) may be expressed as  η η η 2 t it t (w + 2[ 2π ]πi − iη)n (w+2[ 2π ]πi−iη)+ 2η 2 (w+2[ 2π ]πi−iη) −[ η ]w dw, − eη w e −1 L where L is a line in the direction arg w = π/4, passing between 0 and 2πi. This is n! times the coefficient of ξ n in the following expression: (4.44)  η η η 2 t it t dw (w+2[ 2π ]πi−iη)+ 2η 2 (w+2[ 2π ]πi−iη) −[ η ]w+ξ(w+2[ 2π ]πi−iη) eη − w −1 e L  η η 2 t it it 2πt η t dw 2[ η ]πi−it+ 2η w2 +( 2t 2 (2[ 2π ]πi−iη) +ξ(2[ 2π ]πi−iη) η − η 2 [ 2π ]−[ η ]+ξ)w = −e η 2π e 2η2 w −1 e L   η η 2 t it 1 2πt 2t 2πt  η   t  η 2[ 2π ]πi−it+ 2η 2 (2[ 2π ]πi−iη) =e − +ξ− 2πiΦ − 2 , − 2 η η η 2π η 2 η

× eξ(2[ 2π ]πi−iη) , where the function Φ(τ, u) is defined by (4.29), i.e.,   2 2 iτ 1 eπiτ x +2πiux e− 4π w +(u+ 2 )w 1 dw, dx = Φ(τ, u) = πix − e−πix 2πi −L ew − 1 01 e

τ < 0,

u ∈ C.

We rewrite the expression on the rhs of (4.44) as   ∞  η 2 t it 2πt 2t 2πt  η   t  1 ξ l 2[ η ]πi−it+ 2η 2 (2[ 2π ]πi−iη) e η 2π 2πi ∂2l Φ − 2 , − 2 − − η η η 2π η 2 l! l=0

×

∞  ξ k (2[

η 2π ]πi

− iη)k

k!

k=0

.

It follows that the expression in (4.43) is given by t

(4.45)

2πi(iη)s−1 e η

η η 2 it 2[ 2π ]πi−it+ 2η 2 (2[ 2π ]πi−iη)

SN (s, η)

where SN (s, η) is defined in (4.27). Equations (4.37), (4.39), and (4.45) show that I2 satisfies t

2[

η

]πi−it+

it

(2[

η

]πi−iη)2

2π 2η 2 I2 = 2πi(iη)s−1 e η 2π SN (s, η)    N6 σ  πt πt 3N η √ + O(e− 2 −At ). + O e− 2 t t Equation (4.26) follows by substituting this expression into (4.3).

4. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR 0 < η < t

47

In order to prove (4.30) we suppose that u ∈ C and τ < 0. We claim that Φ satisfies the two recursion relations 3πi

πi 1 2 e 4 Φ(τ, u) = Φ(τ, u + 1) −  e− τ (u+ 2 ) |τ |

(4.46) and

Φ(τ, u) = 1 − eπiτ −2πiu Φ(τ, u − τ ).

(4.47)

Repetitive use of these two equations yields the following identities: 3πi N −1 e 4  − πi (u+n+ 1 )2 2 e τ Φ(τ, u) = Φ(τ, u + N ) −  |τ | n=0

(4.48) and (4.49)

Φ(τ, u) =

N −1 

2

(−1)n eπin

τ −2πinu

+ (−1)N eπiN

2

τ −2πiN u

Φ(τ, u − N τ )

n=0

valid for all N ≥ 0. If τ = −p/q, we apply (4.48) with N = p and (4.49) with N = q to find 3πi p−1 e 4  − πi (u+n+ 1 )2 2  Φ(τ, u) = Φ(τ, u + p) − e τ |τ | n=0

(4.50) and (4.51)

Φ(τ, u) =

q−1 

2

(−1)n eπin

τ −2πinu

+ (−1)q eπiq

2

τ −2πiqu

Φ(τ, u + p).

n=0

Eliminating Φ(τ, u + p) from these two equations, we find (4.30). It remains to prove (4.46) and (4.47). In order to prove (4.46), we note that 

2

Φ(τ, u + 1) − Φ(τ, u) =

eπiτ x 01



2

eπiτ x

= 01

1 2 − πi τ (u+ 2 )

e2πi(u+1)x − e2πiux dx eπix − e−πix +2πi(u+ 12 )x

dx

 eπiτ (x+

=e

u+ 1 2 τ

01

= e− τ

πi

(u+ 12 )2



2

eπiτ x dx. 01

The identities



2

eπiτ x dx = 01

imply equation (4.46).



3πi

2 dy e 4 = e−πiy  |τ | |τ | 01

)2

dx

48

4. THE ASYMPTOTICS OF THE RIEMANN ZETA FUNCTION FOR 0 < η < t

The identity (4.47) is established as follows:  2 eπiτ x +2πiux dx Φ(τ, u) = πix − e−πix 01 e  2 eπiτ x +2πiux =1+ dx πix − e−πix −10 e  2 eπiτ (x−1) +2πiu(x−1) =1+ dx πi(x−1) − e−πi(x−1) 01 e  2 eπiτ x +2πi(u−τ )x πiτ −2πiu =1−e dx eπix − e−πix 01 = 1 − eπiτ −2πiu Φ(τ, u − τ ), where the second equality follows from the fact that the residue of the integrand at 1 .  x = 0 is 2πi √ Theorem 4.4 is valid for

t < η < t. As a corollary, we can find an analogous √ t result valid for 2π < η <  . Corollary 4.6 (The asymptotics to all orders for the case 2π < η < For every > 0, there exists a constant A > 0 such that (4.52) [t]

√ t  ).

η

[ 2π ] η   1 1 ζ(s) = + χ(s) ns n1−s n=1 n=1

  s  iη 2 t t t 2 iπs η 2πt −iη[ ]+it+ ([ ]− ) iπ(1−s) η 2t η η + χ(s)e Γ(s) e 2 e SN 1 − s¯, 2πt η    N6 2πt 1−σ  ( η ) 3N πt √ + O e− 2 , t t √ t , 0 ≤ σ ≤ 1, 1 ≤ N < At, t → ∞, 2π < η <

where the error term is uniform for all η, σ, N in the above ranges and SN is given by ( 4.27).

Proof. We replace σ by 1 − σ in (4.26) and take the complex conjugate of both sides. We then multiply the resulting equation by χ(s) and use the identities (4.25). This yields the following equation [t]

η [ 2π ]

 1 ns n=1 n=1

η 2 it iπs − 2t [ η ]πi+it+ 2η 2 (2[ 2π ]π−η) + χ(s)eiπ(1−s) Γ(s) (e 2 η −s e η 2π SN (1 − s¯, η)

ζ(s) = χ(s)

η 

1

n1−s

+

√

t < η < t, Replacing η by

2πt η ,

   N 3N 6 η 1−σ − πt √ +O e 2 , t t 0 ≤ σ ≤ 1,

we find (4.52).

1 ≤ N < At,

t → ∞. 

CHAPTER 5

Consequences of the Asymptotic Formulae Using theorems 3.1 and 3.2 we can compute several interesting sums. Theorem 5.1. Define the polylogarithm Lim (z) by ( 3.3). The following relation holds: (5.1) η2   [ 2π ] 1  η2 1−s  η1 1−s n−s = − 1 − s 2π 2π η1 n=[ 2π ]+1

iη1 j  iπ(1−s) ∞ N −1   e− 2 d z −σ  1 −nz−it ln z + e   (2π)1−s n=1 j=0 dz n + it n + it z z z=iη2   j −iη1 iπ(1−s) ∞ N −1   d z −σ  e 2 1 −nz−it ln z + e   (2π)1−s n=1 j=0 dz n + it n + it z z z=−iη2    1 + 2(N +1) + O (2N + 1)!!N η1−σ−N ,

(1 + )t < η1 < η2 < ∞, > 0, 0 ≤ σ ≤ 1, N ≥ 2, t → ∞,

where the error term is uniform for all η1 , η2 , , σ, N in the above ranges. For N = 3 equation ( 5.1) simplifies to (5.2) η

2] [ 2π

 η

1 ]+1 n=[ 2π

n

−s

  1  η2 1−s  η1 1−s = − 1 − s 2π 2π

2iη1−s t − iσ + Re Li2 (eiη1 ) −i arg(1 − eiη1 ) + (2π)1−s η1  ' 1& + 2 it2 − 3iσt − (σ − 1)t − iσ(σ + 1) Im Li3 (eiη1 ) η1 −s 2iη2 t − iσ − Re Li2 (eiη2 ) −i arg(1 − eiη2 ) + (2π)1−s η2  ' 1& 2 iη2 + 2 it − 3iσt − (σ − 1)t − iσ(σ + 1) Im Li3 (e ) η2    1 + 8  1 3 + O 3+σ t + ,

η1 (1 + )t < η1 < η2 < ∞, > 0, 0 ≤ σ ≤ 1, t → ∞,

where the error term is uniform for all η1 , η2 , , σ in the above ranges. 49

50

5. CONSEQUENCES OF THE ASYMPTOTIC FORMULAE

Proof. Replacing in equation (3.1) η with η1 and subtracting the resulting equation from the equation obtained from (3.1) by replacing η with η2 , we find (5.1). Equation (5.2) follows in a similar way from (3.2).  Theorem 5.2. Define the polylogarithm Lim (z) by ( 3.3). The following relation holds: (5.3) η  1−s   [ 2π ] t 1  η 1−s n−s = − 1 − s 2π 2π t n=[ 2π ]+1

it j  iπ(1−s) ∞ N −1   d z −σ  e− 2 1 −nz−it ln z + e   (2π)1−s n=1 j=0 dz n + it n + it z z z=iη   j iπ(1−s) ∞ N −1   d z −σ  e 2 1 −nz−it ln z − e  it  (2π)1−s n=1 j=0 n + z dz n + it z z=−iη  j  iπ(1−s) ∞ N −1   d z −σ  e 2 1 −nz−it ln z + e  (2π)1−s n=2 j=0 dz n + it n + it z z z=−it  1−s  2N ck (1 − σ)Γ( k+1 t 2 ) + eit k+1 2π t 2 k=0  2(N +1)  (2N + 1)!!N ( 1+ (2N + 1)!!N 22N  ) +O + , tσ+N η σ+N (1 + )t < η < ∞, > 0, 0 ≤ σ ≤ 1, N ≥ 2, t → ∞,

where the error term is uniform for all η, , σ, N in the above ranges and the coefficients ck (σ) are defined in ( 3.30). For N = 3 equation ( 5.3) simplifies to (5.4) η [ 2π ]



t n=[ 2π ]+1

n

−s

 1−s   ! t 1  η 1−s = − + ··· 1 − s 2π 2π     1 + 8  1 1 + O 3+σ t3 + + 3+σ , η

t (1 + )t < η < ∞, > 0,

0 ≤ σ ≤ 1,

t → ∞,

where the error term is uniform for all η, , σ in the above ranges and {· · · } denotes an explicit expression involving σ, t, and the polylogarithms Lim (eit ), m = 1, . . . , 5. Proof. Equation (5.3) follows by subtracting equation (3.20) from equation (3.1). Similarly, equation (5.4) follows by subtracting equation (3.22) from equation (3.2).  The following equation is given on page 78 of Titchmarsh [26]:   1 1 (5.5) ∼ χ(s) . s 1−s n n t t x 0 such that (5.6) [ ηt ]

η

1 

n=[ ηt ]+1

1 = χ(s) ns

2] [ 2π



1 n1−s

η

1 ]+1 n=[ 2π

η 2 2t η it iπ(s−1) [ ]πi−it− 2η 2 (2[ 2π ]π−η) + e−iπs Γ(1 − s) e 2 η s−1 e η 2π SN (s, η)   η2 N 3N 6 η σ  − πt √ +O e 2 ,   t t η=η1 √

t < η1 < η2 < t, 0 ≤ σ ≤ 1, 1 ≤ N < At, t → ∞,

2

where the error term is uniform for all η1 , η2 , σ, N in the above ranges and the function SN (s, η) is defined by ( 4.27). For N = 3 equation ( 5.6) simplifies to [ ηt ]

(5.7)

1 

n=[ ηt ]+1 2

η

1 = χ(s) ns

2] [ 2π

 η

1 ]+1 n=[ 2π

1 n1−s

η 2 t it iπ(s−1) 2[ η ]πi−it− 2η 2 (2[ 2π ]π−η) Γ(1 − s) e 2 η s−1 e η 2π +e        σ−1 η πi − iη Φ × Φ+ ∂2 Φ + 2 iη 2π      (σ − 2)(σ − 1) 2 η − πi − iη ∂2 Φ ∂2 Φ + 2 2 2η 2 2π 2      η2 ησ η  − πt πi − iη Φ + O e 2 + 2 ,   2π t η=η1 √

t < η1 < η2 < t, 0 ≤ σ ≤ 1, t → ∞, −iπs

where the error term is uniform for all η1 , η2 , σ in the above ranges, Φ is defined by ( 4.29), and Φ and its partial derivatives are evaluated at the point ( 1.8). Proof. Replacing in equation (4.26) η with η1 and subtracting the resulting equation from the equation obtained from (4.26) by replacing η with η2 , we find (5.6). Equation (5.7) follows in a similar way from (4.31).  Remark 5.4. The relation (5.5) is a particular case of (5.6) (let x = t/η2 and N = t/η1 ).

Part 2

Asymptotics to all Orders of a Two-Parameter Generalization of the Riemann Zeta Function

CHAPTER 6

An Exact Representation for Φ(u, v, β) The two-parameter generalization Φ(u, v, β) of ζ(s) was defined in (1.12). In Theorem 2.1, we derived an exact representation for ζ(s). In this chapter, we prove Theorem 6.1 which provides an analogous representation for Φ(u, v, β). Recall that we make the basic assumption (1.24); in particular η ∈ / 2πZ. The branch cut for the logarithm is assumed to run along the negative real axis. Theorem 6.1 (An exact representation for Φ). Let Φ(u, v, β) be defined by ( 1.12). Then −iπu

Φ(u, v, β) = (e

−iπv

− 1)e

  −iπv iπu +e (1 − e )

η [ 2π ] iπ 2

(2πe )

u+v−1



mu−1 (m + β)v−1

m=1

 dz +(e − 1) z u−1 (z + 2iπβ)v−1 z e −1 −iη iη  ' z u−1 & (z − 2iπβ)v−1 + e−iπv e−z (z + 2iπβ)v−1 dz + −z − 1 L3 e  ' z u−1 & (z − 2iπβ)v−1 + e−iπv e−z (z + 2iπβ)v−1 dz, + (e−iπu − 1) −z −1 ˆα e C η ∞eiφ1

−iπu



∞eiφ2

(6.1) u, v ∈ C,

β > 0,

0 < η < 2πβ,

−

π π < φj < , j = 1, 2, 2 2

where 0 < α < 2π min(1, β) and the contours L3 and Cˆηα with the orientations shown in figures 2.1 and 2.3 are defined in ( 2.5) and ( 2.10c), respectively. Proof. We decompose the contour Hα in the definition (1.12) of Φ into the union of the three contours {Lj }31 defined in (2.5) with the orientation shown in figure 2.1: Hα = L 1 ∪ L 2 ∪ L 3 . Proceeding in analogy with the proof of Theorem 2.1, we write the integral along L1 as follows:   ∞eiπ dz dz u−1 v−1 = z (z − 2iπβ) z u−1 (z − 2iπβ)v−1 −z −z − 1 e e −1 L1 iα  ∞ dζ ζ u−1 (−ζ − 2iπβ)v−1 ζ = eiπu e −1 −iα  ∞ dζ = eiπu e−iπ(v−1) (6.2) , ζ u−1 (ζ + 2iπβ)v−1 ζ e −1 −iα 55

56

6. AN EXACT REPRESENTATION FOR Φ(u, v, β)

iη Cˆηα

Cαη iα Re z

−iα −α C−η

−iη

−α Figure 6.1. The contours Cαη , C−η and Cˆηα .

where the first equality is a consequence of Cauchy’s theorem, the second equality is a consequence of the substitution z = eiπ ζ, and the third equality uses the fact that −ζ − 2iπβ is in the third quadrant. Similarly, the integral along L2 can be written in the form (6.3)  z u−1 (z − 2iπβ)v−1 L2

dz = e−iπu e−iπ(v−1) e−z − 1



iα

ζ u−1 (ζ + 2iπβ)v−1 ∞

dζ . eζ − 1

The starting point for deriving this identity is the application of Cauchy’s theorem in the domain enclosed by L2 and by the ray from ∞e−iπ to −iα, i.e., in the shaded domain 2 of figure 2.2. The integral along L3 can be written as follows:   dz dζ = −e−iπ(v−1) z u−1 (z − 2iπβ)v−1 −z ζ u−1 (ζ + 2iπβ)v−1 ζ e −1 e −1 L3 L3    v−1 (ζ − 2iπβ)v−1 −iπ(v−1) (ζ + 2iπβ) + e ζ u−1 + dζ e−ζ − 1 eζ − 1 L3   ∞  iα  dζ −iπ(v−1) + ζ u−1 (ζ + 2iπβ)v−1 ζ =−e e −1 −iα ∞  ' ζ u−1 & (ζ − 2iπβ)v−1 + e−iπv e−ζ (ζ + 2iπβ)v−1 dζ, + (6.4) −ζ −1 L3 e where the first equality is an identity and the second equality follows from an application of Cauchy’s theorem in the domain enclosed by L3 and the two rays (−iα, ∞) and (∞, iα), i.e., in the shaded domain 3 of figure 2.2. Adding equations (6.2)-(6.4) we obtain  ∞  πu  iπu  ∞ dz −iπv − iπu 2 2 sin +e Φ(u, v, β) = − 2ie z u−1 (z + 2iπβ)v−1 z e 2 e −1 −iα iα  u−1 & ' z (6.5) (z − 2iπβ)v−1 + e−iπv e−z (z + 2iπβ)v−1 dz. + −z −1 L3 e −α ˆ α Defining the contours Cαη , C−η , Cη as in (2.10) with the orientations shown in

6. AN EXACT REPRESENTATION FOR Φ(u, v, β)

57

figure 6.1, the second integral in the rhs of (6.5) can be rewritten as the sum of an integral along the contour Cαη plus an integral along the ray (iη, ∞eiφ1 ). Similarly, the first integral in the rhs of (6.5) can be rewritten as the sum of an integral along −α the curve −C−η plus an integral along the ray (−iη, ∞eiφ2 ). Hence the first two terms in the rhs of (6.5) yield the second line in the rhs of (6.1), as well as the additional term I defined by (6.6)     πu  iπu dz −iπv − iπu 2 2 . I = −2ie sin +e z u−1 (z + 2iπβ)v−1 z −e η −α 2 e −1 C−η Cα −α Letting z = ζe−iπ in the integral involving C−η and using the fact that −ζ + 2iπβ is in the first quadrant we find (6.7)   iπu dz u−1 v−1 dz − iπu iπv 2 2 =e . z (z + 2iπβ) e z u−1 (z − 2iπβ)v−1 −z −e z −1 −α e e −1 ˆα C C−η η

Using the above equation in the rhs of (6.6) and then adding and subtracting in the resulting equation the term   πu iπu dz e− 2 e−iπv , z u−1 (z + 2iπβ)v−1 z −2i sin 2 e −1 ˆα C η we find that I is given by   πu iπu dz − 2 −iπv I = − 2i sin z u−1 (z + 2iπβ)v−1 z e e η 2 e −1 α ˆ Cα ∪Cη    (z − 2iπβ)v−1 e−iπv (z + 2iπβ)v−1 − z u−1 + . −z e −1 ez − 1 ˆα C η That is, −iπu

I = (e

 (6.8)

+ ˆα C η

 −iπv − 1) e

z u−1 (z + 2iπβ)v−1

η ˆα Cα ∪C η

dz ez − 1

 ' z u−1 & v−1 −iπv −z v−1 (z − 2iπβ) + e e (z + 2iπβ) . e−z − 1

Cauchy’s theorem implies that the first integral in the curly bracket in (6.8) equals η [ 2π ] iπ 2

(2πe )u+v−1



mu−1 (m + β)v−1 .

m=1

Thus (6.5) with the aid of equation (6.8) becomes equation (6.1).



CHAPTER 7

The Asymptotics of Φ(u, v, β) √ In Theorem 4.1, we established an asymptotic formula for ζ(s) for < η < t. In this chapter, we establish an analogous asymptotic formula for Φ. Let u = σ1 + it and v = σ2 − it. The function Φ(u, v, β) was defined in (1.12) by  dz (7.1) Φ(u, v, β) = , u, v ∈ C, β ∈ R \ {0}, z u−1 (z − 2iπβ)v−1 −z e −1 Hα where Hα , 0 < α < 2π min(1, |β|), denotes the Hankel contour (1.13) surrounding the negative real axis in the counterclockwise direction, and the complex powers are defined by wa = ea(ln |w|+i arg w) with arg w ∈ (−π, π], i.e., the branch cut runs along the negative real axis. In this and the following chapter, we prefer to have the branch cut along the positive real axis. Therefore we change variables z = −w in (7.1) to get (7.2) Φ(u, v, β) = −e−iπ(u+v)

 wu−1 (w + 2iπβ)v−1 ˆα H

dw , −1

ew

u, v ∈ C, β ∈ R \ {0},

ˆ α denotes the Hankel contour surrounding the positive real axis in the where H counterclockwise direction, i.e.,   ˆ α = {r + i0 | α < r < ∞} ∪ αeiθ | 0 < θ < 2π ∪ {r − i0 | α < r < ∞} , (7.3) H and the complex powers are defined by wa = ea(ln |w|+i arg w) with arg w ∈ [0, 2π), i.e., the branch cut now runs along the positive real axis. Using the identity m  1 e−mw = , e−nw + w w e − 1 n=1 e −1

m = 1, 2, . . . ,

we may write Φ(u, v, β) = −e−iπ(u+v)

 m  n=1



(7.4)

wu−1 (w + 2iπβ)v−1 e−nw dw

ˆα H

+ ˆα H

 wu−1 (w + 2iπβ)v−1 e−mw dw . ew − 1

We are interested in the asymptotic behavior of Φ(u, v, β) as t → ∞. Thus we note that the second integral on the rhs of (7.4) has critical points at the solutions of   d ln wu−1 (w + 2iπβ)v−1 e−mw = 0, dw 59

60

7. THE ASYMPTOTICS OF Φ(u, v, β)

that is, at

 (−2iπβm + u + v − 2)2 + 8iπβm(u − 1) . w= 2m Anticipating that m will grow like tα , α > 0, we find that for large t the critical points are approximately given by (   2t iβπ − 1 ± 1 + . βπm By performing a steepest descent analysis we can find the asymptotics of Φ to all orders. The idea is that given some η > 0, we choose m so that the critical point for large t lies at iη. Solving the equation (   2t η = βπ − 1 + 1 + βπm for m we find that m should be given by   1 1 − . t η η + 2πβ But since m has to be an integer, we instead use the approximate definition   1 1 − m = [x] where x = t . η η + 2πβ √ Theorem 7.1 (The asymptotics of Φ(u, v, β) to all orders for < η < t). For every > 0, there exists a constant A > 0 such that m   − eiπ(σ1 +σ2 ) Φ(u, v, β) = wu−1 (w + 2iπβ)v−1 e−nw dw −2iπβm + u + v − 2 ±

n=1

+ i(2π)σ1 +σ2 −1 e 2

πi

ˆα H

η [ 2π ]

(σ1 +σ2 )



nu−1 (n + β)v−1

n=1

− e−(m+1)iη (iη)u−1 (iη + 2πiβ)v−1 e 4

iπ

−k− 12    ϕ(2k) (0)  t  1 1 1 k i × − Γ k+ (2k)! 2 η2 (η + 2πβ)2 2 [ N 2−1 ]

k=0

+ η σ1 −1 (η + 2πβ)σ2 −1 ⎧    22N  N6 η ⎪ ⎪ √ , ⎨O t t   ×  3t − N2+1 ⎪ − At 2 ⎪ η + 4N η2 , ⎩O N e (7.5) β > 0,

< η < 2πβ − < ∞,

1

< η < t 3 < ∞, 1

t3 < η
,

1≤N
8 , and N ≥ 1 (since > 0 is arbitrary, we can replace 8 with at the end). All error terms of the form O(·) will be uniform with respect to η, σ1 , σ2 , β, N in the given ranges, but not with respect to . We let A > 0 denote a generic constant which can change within a computation. Let   1 1 − x=t , m = [x]. η η + 2πβ √ Since η < t and η < 2πβ, we have √ t 1 t x= ≥ . η η 2πβ +1 2 ˆ α into the In particular, m → ∞ as t → ∞. We deform the Hankel contour H straight lines Cj , j = 1, . . . , 4 joining ∞, cη + iη(1 + c), −cη + iη(1 − c), −cη − iη, ∞, where 0 < c ≤ 1/2 is an absolute constant, see figure 4.1. Then (7.6) η [ 2π  4 ]  wu−1 (w + 2iπβ)v−1 e−mw u−1 v−1 dw = − 2πi (2πin) (2πi(n + β)) + Ij , ew − 1 ˆα H η j=1 n=−[ 2π ] n =0

where

 Ij = Cj

wu−1 (w + 2iπβ)v−1 e−mw dw, ew − 1

j = 1, . . . , 4.

Note that η [ 2π ]

−2πi



(2πin)u−1 (2πi(n + β))v−1 = −(2π)σ1 +σ2 −1 e

η [ 2π ] iπ 2 (u+v−1)

η n=−[ 2π ]



nu−1 (n + β)v−1

n=1

n =0

− (2π)

σ1 +σ2 −1

η [ 2π ]

e

πi 2 v

e

3πi 2 (u−1)



nu−1 (−n + β)v−1

n=1

= i(2π)σ1 +σ2 −1 e 2

πi

η [ 2π ]

(σ1 +σ2 )



& ' nu−1 (n + β)v−1 − eπiu (−n + β)v−1

n=1

= i(2π)

σ1 +σ2 −1

η [ 2π ]

e

πi 2 (σ1 +σ2 )



nu−1 (n + β)v−1 + O(e−At ).

n=1

In view of (7.4) and (7.6) this gives the first two terms on the rhs of (7.5). It remains to analyze the contributions from the Ij ’s. We first prove that I1 , I3 , I4 are exponentially small as t → ∞. Let w = ρeiφ , ˜ 0 < φ < 2π, and w + 2iπβ = ρ˜eiφ , 0 < φ˜ < 2π. Then |wu−1 | = ρσ1 −1 e−tφ ,

˜

|(w + 2iπβ)v−1 | = ρ˜σ2 −1 etφ .

Also, xη =

tQ , 1+Q

where

Q :=

2πβ > 1. η

62

7. THE ASYMPTOTICS OF Φ(u, v, β)

Analysis of the integral I4 For w ∈ C4 we have |ew − 1| > A, and

ρ ≥ η,

ρ˜ ≥ 2πβ − η,

φ ≥ π + arctan

η , cη

2πβ − η Q−1 φ˜ ≤ π − arctan = π − arctan . cη c

Hence ˜

|wu−1 (w + 2iπβ)v−1 | = ρσ1 −1 ρ˜σ2 −1 e−t(φ−φ) ≤ η σ1 −1 (2πβ − η)σ2 −1 e−t(arctan c +arctan 1

= O(e−t(arctan c +arctan 1

(7.7)

Q−1 c )

Q−1 c )

).

Thus, 

wu−1 (w + 2iπβ)v−1 e−mw dw ew − 1 C4    ∞ −t(arctan 1c +arctan Q−1 ) −mw 1 c =O e e dw1

I4 =

−cη

−t(arctan

= O(e

= O(e−t(arctan

Q−1 1 c +arctan c )+mcη Q−1 1 c +arctan c )+xcη

where F (Q) = arctan Since

c2 4

) = O(e−tF (Q) ),

Q−1 cQ 1 + arctan − . c c 1+Q

< Q, we have F  (Q) =

Hence

)

(Q +

4cQ − c3 > 0, (c2 + (Q − 1)2 )

1)2

Q > 1.

  1 c inf F (Q) ≥ F (1) = arctan − > A > 0. Q>1 c 2

It follows that I4 = O(e−At ). Analysis of the integral I3 For w = w1 + iw2 ∈ C3 we have φ = π − arctan

w2 , cη

w2 + 2πβ φ˜ = π − arctan . cη

The function φ− φ˜ assumes its maximum at w2 = −πβ and decreases symmetrically as w2 moves away from this point. Its minimum on C3 is therefore assumed at the upper endpoint where w2 = (1 − c)η. For w2 = (1 − c)η we have (7.8) w2 w2 + 2πβ − arctan = arctan φ − φ˜ = arctan cη cη



1−c+Q c

 − arctan

1−c . c

7. THE ASYMPTOTICS OF Φ(u, v, β)

63

Hence

 1−c+Q 1−c wu−1 (w + 2iπβ)v−1 e−mw = O (cη)σ1 −1 (cη)σ2 −1 e−t(arctan( c )−arctan c ) emcη    1−c+Q 1−c = O e−t(arctan( c )−arctan c ) excη = O e−tF (Q) , 

where F (Q) = arctan

1−c+Q c



 − arctan

1−c c

 −

cQ . 1+Q

Since c < 1 + Q, we have F  (Q) = −

2c2 (c − Q − 1) > 0, (Q + 1)2 (c2 + (1 + Q − c)2 )

Hence

 inf F (Q) ≥ F (1) = arctan

Q>1

2−c c



 − arctan

1−c c

Q > 1.  −

c > A > 0. 2

It follows that I3 = O(e−At ). Analysis of the integral I1 For w ∈ C1 we have w = w1 + i(1 + c)η and w1 ≥ cη. Also, ρ ≥ η,

ρ˜ ≥ η + 2πβ,

1+c (1 + c)η = arctan , w1 P 1+c+Q (1 + c)η + 2πβ (7.9) , = arctan φ˜ = arctan w1 P where P := w1 /η ≥ c. There exists A > 0 such that φ = arctan

1 ≤ (1 − A)ec ≤ (1 − A)ew1 , and so |ew − 1| ≥ ew1 − 1 ≥ Aew1 . Thus   wu−1 (w + 2iπβ)v−1 e−mw ˜ = O η σ1 −1 (η + 2πβ)σ2 −1 e−t(φ−φ)−(m+1)w1 w e −1   ˜ = O η σ1 −1 (η + 2πβ)σ2 −1 e−t(φ−φ)−xw1   = O η σ1 −1 (η + 2πβ)σ2 −1 e−tF (P,Q) , where F (P, Q) = arctan Now

1+c 1+c+Q P − arctan +P − . P P 1+Q

∂F 1 P > 0, −  = (c+Q+1)2 ∂Q (Q + 1)2 P + 1 2 P

P ≥ c,

Q ≥ 1,

so F assumes its minimum over the domain {P ≥ c, Q ≥ 1} on the boundary where Q = 1. For definiteness, let us henceforth set 1 c= . 8 Then P ] ∂[F (P, 1) − 16 c > 0 for P ≥ c, and F (c, 1) − > 0, ∂P 16

64

7. THE ASYMPTOTICS OF Φ(u, v, β)

so that F (P, 1) > P/16 for P ≥ c. Hence  wu−1 (w + 2iπβ)v−1 e−mw dw I1 = ew − 1 C1    ∞ = O η σ1 −1 (η + 2πβ)σ2 −1 e−tP/16 dw1 cη    ∞ σ1 σ2 −1 −tP/16 = O η (η + 2πβ) e dP c   = O η σ1 (η + 2πβ)σ2 −1 t−1 e−ct/16 . It follows that I1 = O(e−At ). Analysis of the integral I2 It remains to analyze the integral I2 . We write  it 1 ( 1 − )(w−iη)2 −(m+1)iη u−1 v−1 I2 = e (iη) (iη + 2πiβ) e 2 η2 (η+2πβ)2 ϕ(w − iη)dw, C2

where 2 z z 1 1 (u−1) ln(1+ iη )+(v−1) ln(1+ iη+2πiβ )−xz− it 2 ( η 2 − (η+2πβ)2 )z +(x−m)z

ϕ(z) =

e

Defining φ(z) by (7.10)

.

ez − e−iη 



2 z z 1 1 (u−1) ln 1+ iη +(v−1) ln(1+ iη+2πiβ )−xz− it 2 ( η 2 − (η+2πβ)2 )z

φ(z) = e

,

we have ϕ(z) =

φ(z)e(x−m)z . ez − e−iη

We split the contour C2 as follows: C2 = C2 ∪ C2r , where C2 denotes the segment of C2 of length 2 which consists of the points within a distance from iη. We write I2 = I2 + I2r , where I2 = e−(m+1)iη (iη)u−1 (iη + 2πiβ)v−1 and I2r = e−(m+1)iη (iη)u−1 (iη + 2πiβ)v−1



it

e2

2 1 ( η12 − (η+2πβ) 2 )(w−iη)

C2



it

e2

2 1 ( η12 − (η+2πβ) 2 )(w−iη)

C2r

ϕ(w − iη)dw

ϕ(w − iη)dw.

We claim that I2r can be estimated as follows:  σ1  t2 η (η + 2πβ)σ2 −1 − 32η r 2 √ I2 = O e (7.11) . t iπ

Indeed, the change of variables w = iη + λe 4 gives    −  √2cη  2 1 iπ − 2t ( η12 − (η+2πβ) r σ1 −1 σ2 −1 2 )λ 4 I2 = O η (η + 2πβ) + |ϕ(λe )|dλ . e √ − 2cη



7. THE ASYMPTOTICS OF Φ(u, v, β)

Now there exists an A > 0 such that  (x−m)z    e iπ   z = λe 4 , (7.12)  ez − e−iη  < A,

< |λ|


1 16 .

I2r

  = O η σ1 −1 (η + 2πβ)σ2 −1

Hence √ 2cη

2

tλ − 16η 2

e

 dλ .



Splitting the integrand as 2 t − 16η 2λ

e

2 t − 32η 2λ

=e

2 t − 32η 2λ

×e

and noting that √



2cη

2 t − 32η 2λ

e

 dλ ≤

∞

2 t − 32η 2λ

e



dλ =

0

we find I2r

√ η 8π √ , t

  2 t σ1 −1 σ2 −1 − 32η2  η √ =O η (η + 2πβ) e . t

This proves (7.11). We now consider I2 . In view of the assumption dist(η, 2πZ) > 8 , we have  (x−m)w   e   (7.16) |z| < 2 .  ez − e−iη  = O(1), In particular, ϕ(z) is analytic for |z| < 2 . Thus we can write ϕ(z) =

∞  n=0

bn z n =

N −1  n=0

where (7.17)

|z| < 2 ,

bn z n + sN (z),

sN (z) =

zN 2πi

 Γ

ϕ(w)dw wN (w − z)

and Γ is a counterclockwise contour which encircles 0 and z but none of the poles of ϕ. We claim that  N −1  it 1 ( 1 − )(w−iη)2 bn (w − iη)n dw I2 = e−(m+1)iη (iη)u−1 (iη + 2πiβ)v−1 e 2 η2 (η+2πβ)2 C2

n=0

(7.18)

⎧    N η ⎪ σ1 −1 σ2 −1 22N 6 √ ⎪ (η + 2πβ) , ⎨O η t t   +  − N2+1 ⎪ ⎪ , ⎩O η σ1 −1 (η + 2πβ)σ2 −1 4N3tη2

1≤N < 1

t3 < η
t1+ . We recall that u = σ1 + it and v = σ2 − it. ) Theorem 8.1 (The asymptotics to all orders of the integral Hˆ α wu−1 × (w + 2iπβ)v−1 e−nw dw). For every > 0, there exists a constant A > 0 such that  iπ wu−1 (w + 2iπβ)v−1 e−nw dw = −(iη)u−1 (iη + 2πiβ)v−1 e−inη e 4 ˆα H

−k− 12    φ(2k) (0)  t  1 1 1 k i × − Γ k+ (2k)! 2 η2 (η + 2πβ)2 2 k=0    N 7N 6 1 √ , + O η σ1 (η + 2πβ)σ2 −1 t → ∞, t t [ N 2−1 ]

(8.1) t1+ < β < ∞,

n = 1, 2, . . . , [t],

σ1 ∈ [0, 1],

σ2 ∈ [0, 1],

1 ≤ N < At,

ˆ α denotes the Hankel contour defined in ( 7.3), η and φ(z) are given by where H (   2t η = βπ − 1 + 1 + (8.2) , βπn (8.3)

2 z z 1 1 (u−1) ln(1+ iη )+(v−1) ln(1+ iη+2πiβ )−nz− it 2 ( η 2 − (η+2πβ)2 )z

φ(z) = e

,

and the error terms are uniform with respect to n, σ1 , σ2 , β, N in the given ranges. Proof. Let > 0 be given and suppose that β > t1+ , n = 1, . . . , [t], σ1 ∈ [0, 1], σ2 ∈ [0, 1], and N ≥ 1. All error terms of the form O(·) will be uniform with respect to n, σ1 , σ2 , β, N (but not with respect to ). Let η be given by (8.2). The function x( 1 + 1/x − 1) increases from 0 to 1/2 as x goes from 0 to ∞. Hence, using that (  2t βπn 2t η= 1+ −1 n 2t βπn we infer that, given any δ > 0, we have t t (8.4) (1 − δ) < η < n n for all sufficiently large t. 73

74

8. MORE EXPLICIT ASYMPTOTICS OF Φ(u, v, β)

ˆ α into the As in the proof of Theorem 7.1, we deform the Hankel contour H straight lines Cj , j = 1, . . . , 4 joining ∞, iη + (1 + i)cη, iη − (1 + i)cη, −cη − iη, ∞, where 0 < c ≤ 1/2 is an absolute constant. This implies 

wu−1 (w + 2iπβ)v−1 e−nw dw =

ˆα H

where

4 

Ij ,

j=1



wu−1 (w + 2iπβ)v−1 e−nw dw.

Ij = Cj

We first prove that I1 , I3 , I4 are exponentially small as t → ∞. We let Q = > Ant → ∞ as t → ∞. 2πβ/η. Then, by (8.4), Q > 2πβn t Using (7.7) and (8.4), we find that I4 is exponentially small:     ∞ Q−1 1 I4 = wu−1 (w + 2iπβ)v−1 e−nw dw = O e−t(arctan c +arctan c ) e−nw1 dw1 −cη

C4 −t(arctan

Q−1 1 c +arctan c )+ncη

−t(arctan

1 c −c)

= O(e = O(e

−At

) = O(e

−t(arctan

) = O(e

Q−1 1 c +arctan c −c)

)

). ˜

We next consider I3 . Letting w = ρeiφ and w + 2iπβ = ρ˜eiφ , equations (7.8) and (8.4) yield wu−1 (w + 2iπβ)v−1 e−nw = O((cη)σ1 −1 (cη)σ2 −1 e−t(arctan( = O(e−t(arctan( where

 F (Q) = arctan

1−c+Q )−arctan 1−c c c −c)

1−c+Q c



 − arctan

1−c+Q )−arctan 1−c c c )

encη )

) = O(e−tF (Q) ),

1−c c

 − c.

Since Q → ∞ and F (Q) > A > 0 for all sufficiently large Q, it follows that I3 = O(e−At ). For w ∈ C1 we have w = w1 + i(1 + c)η with w1 ≥ cη. Equations (7.9) and (8.4) imply   ˜ wu−1 (w + 2iπβ)v−1 e−nw = O η σ1 −1 (η + 2πβ)σ2 −1 e−t(φ−φ)−nw1    ˜ nw1  ˜ = O e−t(φ−φ+ t ) = O e−t(φ−φ+(1−δ)P )   = O e−tF (P,Q) where P = w1 /η ≥ c and F (P, Q) = arctan

1+c+Q 1+c − arctan + (1 − δ)P. P P

Now the rhs of the inequality F (P, Q) − AP > arctan

1+c π − + (1 − δ)P − AP P 2

8. MORE EXPLICIT ASYMPTOTICS OF Φ(u, v, β)

75

π is an increasing function of P ≥ c and arctan 1+c c − 2 + (1 − δ)c − Ac > 0 for δ and A small enough. Hence F (P, Q) > AP . It follows that  I1 = wu−1 (w + 2iπβ)v−1 e−nw dw C1   ∞    ∞     −tAP −tAP e dw1 = O η e dP = O ηt−1 e−cAt = O e−cAt . =O cη

c

1 It remains to analyze the integral I2 . Using that n = t( η1 − η+2πβ ), we write  it 1 1 1 ( 1 − )(w−iη)2 e 2 η2 (η+2πβ)2 φ(w − iη)dw, I2 = (iη)u−1 (iη + 2πiβ)v−1 e−itη( η − η+2πβ ) C2

where φ(z) is defined in (8.3). Define {aj }∞ 0 by φ(z) =

∞ 

|z| < η.

aj z j ,

j=0

Then φ(z) =

N −1 

aj z j + rN (z),

|z| < η,

j=0

where

 φ(w) zN dw, N 2πi Γ w (w − z) and Γ is a counterclockwise contour contained in the disk of radius η centered at the origin which √ encircles the points 0 and z once. 21 Let α ∈ ( 2c, 1) be a constant. Let |z| < 20α 21 η (so that 20 |z| < αη) and let Γ be a circle with center w = 0 and radius ρN , where 21 |z| ≤ ρN ≤ αη. 20 By (7.13),     2tρ3 2tρ3 N N 2πρN |z|N N −N 3(1−α)η3 3 3(1−α)η e rN (z) = O = O |z| ρN e , |z| ≤ αη. ρN N (ρN − |z|) rN (z) =

2tρ3

2et N/3 for ρ = ( (1−α)N )1/3 η; The function ρ−N e 3(1−α)η3 has the minimum ( (1−α)N η3 ) 2t ρN can have this value if  1 (1 − α)N 3 21 |z| ≤ η ≤ αη. 20 2t Hence, (8.5)    N3  1 2et 2α3 20 (1 − α)N 3 N η. rN (z) = O |z| , N ≤ t, |z| ≤ (1 − α)N η 3 1−α 21 2t

For |z|
0, there exists a constant A > 0 such that m   wu−1 (w + 2iπβ)v−1 e−nw dw = J + O(t1/6 β σ2 −1 ), t → ∞, n=1

ˆα H

m = 1, . . . , [t],

(8.15)

t1+ < β < ∞,

σ1 ∈ [0, 1],

σ2 ∈ [0, 1],

where J is defined by ( (8.16) J = e

iπ 2 (σ1 +σ2 )

e

iπ 4

 it m  √ √ 2t π (πβ)σ1 +σ2 −1 ( R − 1)σ1 ( R + 1)σ2 F, 2t πnβ n=1

with 2t , R := 1 + πnβ

(8.17)

√ (1 + R)−2it iπnβ(1−√R) F := e , R1/4

and the error term is uniform with respect to m, β, σ1 , σ2 in the given ranges. Proof. Summing equation (8.1) with N = 2 from n = 1 to n = m, we find   m m   1  σ1 u−1 v−1 −nw σ2 −1 w (w + 2iπβ) e dw = J + O 5/6 η (η + 2πβ) , t ˆ n=1 Hα n=1 where J =−

m 

(iη)

n=1

u−1

(iη + 2πiβ)

(  2t η = βπ − 1 + 1 + . βπn 

v−1 −inη

e

  − 12 √ t 1 1 e − π, 2 2 2 η (η + 2πβ) iπ 4

8. MORE EXPLICIT ASYMPTOTICS OF Φ(u, v, β)

79

Straightforward algebra shows that J can be written as in (8.16). Equation (8.15) follows because, by (8.4),   m m  σ      t 1 σ2 −1 η σ1 (η + 2πβ)σ2 −1 = O β = O tσ1 β σ2 −1 t1−σ1 = O tβ σ2 −1 . n n=1 n=1  Substituting the result of corollary 8.2 into the asymptotic expansion of theorem 7.1, we arrive at the following asymptotic formula for Φ(u, v, β). For brevity of presentation, we state the result only to leading order. Corollary 8.3 (The asymptotics of Φ(u, v, β)). For every > 0, there exists a constant A > 0 such that ( π − iπ (σ1 +σ2 ) iπ 2 4 Φ(u, v, β) = −e e (πβ)σ1 +σ2 −1 2t 1 [t( η1 − η+2πβ )]  it  √ √ 2t × ( R − 1)σ1 ( R + 1)σ2 F πnβ n=1 − i(2π)

σ1 +σ2 −1 − πi 2 (σ1 +σ2 )

e

η [ 2π ]



nu−1 (n + β)v−1 + O(t1/6 β σ2 −1 ),

t → ∞,

n=1

(8.18) 1 ≤ η < t 3 − < ∞, 1

t1+ < β < ∞,

dist(η, 2πZ) > ,

σ1 ∈ [0, 1],

σ2 ∈ [0, 1],

where R and F are defined in ( 8.17) and the error term is uniform with respect to η, β, σ1 , σ2 in the given ranges. Proof. Letting N = 2 in Theorem 7.1 and using Corollary 8.2 we find equation (8.18) but with the error term   1 η O(t1/6 β σ2 −1 ) + η σ1 −1 (η + 2πβ)σ2 −1 O t− 3 √ t − 12 iπ     √ 1 1 (iη)u−1 (iη + 2πiβ)v−1 e 4 t 1 1 + e− [t( η − η+2πβ )]+1 iη − π. −iη 2 2 1−e 2 η (η + 2πβ) Since this error term is    σ1 σ2 −1 − 5     1/6 σ2 −1  σ1 −1 σ2 −1 η 6 √ = O t1/6 β σ2 −1 +O η β +O η t β O t β t the result follows. 

CHAPTER 9

Fourier coefficients of the product of two Hurwitz zeta functions The erratic behavior of the Riemann zeta function in the critical strip makes it difficult to obtain good asymptotic information on ζ(s) for 0 ≤ Re s ≤ 1. Much of the literature on the asymptotics of ζ(s) therefore instead focuses on the asymptotics of various related quantities in which the erratic behavior has been smoothed out by averaging. The prime example is the study of the large T behavior of the moments (see [26, chapter VII] and e.g. [7, 25])  T |ζ(σ + it)|2k dt. 1

A second example involves the small δ behavior of the integral  ∞ |ζ(σ + it)|2k e−δt dt, 0

where δ > 0, see [26, Section 7.12]. Another class of examples, which is relevant for the present chapter, considers the large t behavior of various quantities involving the Hurwitz zeta function ζ(s, α) averaged with respect to the extra parameter α. Recall that the Hurwitz zeta function ζ(s, α) is defined as the analytic continuation of the sum ∞  1 , Re s > 1, Re α > 0. ζ(s, α) = (n + α)s n=0 For each α with Re α > 0, ζ(s, α) is a meromorphic function of s ∈ C with a simple pole at s = 1 and no other poles. For each s ∈ C \ {1}, ζ(s, α) is analytic in the half-plane Re α > 0. These properties of the Hurwitz function follow easily from the representation  eαz z s−1 Γ(1 − s) (9.1) dz, s ∈ C \ {1}, Re α > 0, ζ(s, α) = z 2πi H1 1 − e where H1 is the Hankel contour surrounding the negative real axis defined in (1.13). For many purposes it is more convenient to work with the modified Hurwitz function ζ1 (s, α) defined by (9.2)

ζ1 (s, α) = ζ(s, α) − α−s = ζ(s, α + 1),

which is regular at α = 0 (in fact, analytic for Re α > −1). For each s ∈ C \ {1}, ζ1 (s, α) is a smooth function of α ∈ [0, 1]. Thus, for each choice of u, v ∈ C \ {1}, the product ζ1 (u, α)ζ1 (v, α) can be represented by its Fourier series  ζ1 (u, α)ζ1 (v, α) = (9.3) qn (u, v)e2πinα , 0 < α < 1, n∈Z 81

82

9. FOURIER COEFFICIENTS OF THE PRODUCT

where the Fourier coefficients qn (u, v) are defined by  1 ζ1 (u, α)ζ1 (v, α)e−2πinα dα, qn (u, v) =

n ∈ Z.

0

By standard Fourier analysis, the series in (9.3) converges in L2 ([0, 1]) and converges pointwise for each α ∈ (0, 1). In this chapter, we will show the following two theorems on the Fourier coefficients qn (u, v). The proofs are presented in sections 9.3 and 9.4, respectively. Theorem 9.1 (Expression for qn (u, v)). For each n ∈ Z, the nth Fourier coefficient qn (u, v) can be expressed as follows for u, v ∈ C \ {1} with Re u, Re v < 2: (9.4)

qn (u, v) = bn (u + v) + Rn (u, v) + Rn (v, u) − Tn (u, v) − Tn (v, u),

where, for any n ∈ Z, bn (s), Rn (u, v), and Tn (u, v) are the meromorphic continuations of the functions  ∞ (9.5a) bn (s) = α−s e−2πinα dα, Re s > 1, 1  ∞ (9.5b) Rn (u, v) = α−v ζ1 (u, α)e−2iπnα dα, Re(u + v) > 2, Re v < 1,  (9.5c)

0 1

Tn (u, v) =

α−v ζ1 (u, α)e−2πinα dα,

Re v < 1.

0

These meromorphic continuations are given by  Γ(1 − s, 2πin)(2πin)s−1, n = 0, bn (s) = (9.6a) 1 n = 0, s−1 , ⎧ πiv ie Φ(u,v,n) ⎪ ⎪ 2 sin(πu) , Γ(1 − v) ⎨ ie−πiv Φ(u,v,n) (9.6b) Rn (u, v) = × , 2 sin(πu) ⎪ Γ(u) ⎪ ⎩Γ(u + v − 1)ζ(u + v − 1), 

(9.6c)

Tn (u, v) =

n ≥ 1, n ≤ −1, n = 0,

u ζ(u) − 1 + α1−v ζ1 (u + 1, α)e−2iπnα dα 1−v 1−v 0  2iπn 1 1−v α ζ1 (u, α)e−2iπnα dα, Re v < 2, + 1−v 0 1

n ∈ Z,

where Φ(u, v, n) is the function defined in ( 1.12) and Γ(s, z) denotes the incomplete Gamma function:  ∞  ∞ Γ(s, z) = r s−1 e−r dr, Γ(s, 0) = Γ(s) = r s−1 e−r dr. z

0

Theorem 9.2 (Large t asymptotics of qn (u, v)). Let u = σ1 +it and v = σ2 −it. For each integer n ≥ 1, the nth Fourier coefficient qn (u, v) satisfies the following asymptotic formula as t → ∞: ζ(u) ζ(v) Γ(1 − v) ieπiv Φ(u, v, n) − − Γ(u) 2 sin(πu) 1−v 1−u ζ(u + 1)u ζ(v + 1)v − − + O(t−1 ), σ1 , σ2 ∈ [0, 1], (1 − v)(2 − v) (1 − u)(2 − u)

(9.7) qn (u, v) = bn (u + v) +

where the error term is uniform for σ1 , σ2 in the given range.

9.1. COROLLARIES

83

For n = 0, the following formulas hold as t → ∞: • For σ1 , σ2 ∈ [0, 1] with σ1 + σ2 = 1,

  Γ(1 − u) Γ(1 − v) 1 q0 (u, v) = + Γ(σ1 + σ2 − 1)ζ(σ1 + σ2 − 1) + σ1 + σ2 − 1 Γ(v) Γ(u) ζ(v) ζ(u + 1)u ζ(v + 1)v ζ(u) − − − + O(t−1 ), (9.8) − 1 − v 1 − u (1 − v)(2 − v) (1 − u)(2 − u) where the error term is uniform for all σ1 , σ2 ∈ [0, 1] such that σ1 +σ2 = 1. • For σ1 , σ2 ∈ [0, 1] with σ1 + σ2 = 1, ζ(u) ζ(v) t +γ− − 2π 1−v 1−u ζ(u + 1)u ζ(v + 1)v − − + O(t−1 ), (1 − v)(2 − v) (1 − u)(2 − u)

(9.9)

q0 (u, v) = ln

where the error term is uniform for all σ1 , σ2 ∈ [0, 1] such that σ1 +σ2 = 1, and γ denotes the Euler constant. Remark 9.3. The results of theorem 9.1 and 9.2 are well-known in the case when n = 0. More precisely, the formulas (9.9) and (9.8) can be found in Theorem 1 and Theorem 2 of [29], respectively. They are included here for the sake of comparison; the new content lies in the case of nonzero n. Note however that the ζ(u+1)u ζ(v+1)v and (1−u)(2−u) in (9.8) and (9.9) are missing in [29]. These terms terms (1−v)(2−v) must be included in order to make the asymptotic formulas uniform for σ1 and/or σ2 near 0, because it is known that ζ(1 + it) is unbounded as t → ∞; in fact, ζ(1 + it) = Ω(ln ln t), see Titchmarsh [26, Theorem 8.5]. Remark 9.4. The integral in (9.5b) converges whenever Re(u + v) > 2 and Re v < 1, because for each s ∈ C \ {1}, ζ1 (s, α) satisfies the large α asymptotics α−s α1−s − + O(α−s−1 ), s−1 2 uniformly as α → ∞ in the half-plane Re α > 0, see e.g. [19, Eq. (25.11.43)]. (9.10)

ζ1 (s, α) =

9.1. Corollaries By substituting the asymptotic formula for Φ(u, v, n) found in corollary 7.2 into (9.7), we obtain the following corollary of theorem 9.2. Corollary 9.5. Let u = σ1 + it and v = σ2 − it. For each integer n ≥ 1, Γ(1 − v) ieπiv Γ(u) 2 sin(πu)

m   wu−1 (w + 2iπn)v−1 e−jw dw × − e−πi(σ1 +σ2 )

qn (u, v) = bn (σ1 + σ2 ) + (9.11)

j=1 πi 4

ˆα H

− πi 2 (σ1 +σ2 )

(−1)m e e

π σ1 +σ2 − 2 (1 + 2n)v 1

−3/2



 + O(t ) 2 2n(1 + n)t ζ(u) ζ(v) ζ(u + 1)u ζ(v + 1)v − − − − + O(t−1 ), 1−v 1 − u (1 − v)(2 − v) (1 − u)(2 − u) σ1 , σ2 ∈ [0, 1], t → ∞,

+

84

9. FOURIER COEFFICIENTS OF THE PRODUCT

t where m = [ π(1+(2n) −1 ) ] and the error term is uniform for σ1 , σ2 ∈ [0, 1].

In the special case of σ1 = σ2 , we obtain asymptotic estimates for the Fourier coefficients  1 |ζ1 (u, α)|2 e−2πinα dα Qn (u) = 0

of |ζ1 (s, α)|2 as corollaries. Corollary 9.6 (Large t asymptotics of Q0 (u)). For σ ∈ [0, 1] with σ = 1/2, 1 + 2Γ(2σ − 1)ζ(2σ − 1) sin(πσ)t1−2σ 2σ − 1 Im ζ(σ + 1 + it) ζ(σ + it) −2 + O(t−1 ), (9.12) − 2 Re t → ∞, 1 − σ + it t where the error term is uniform with respect to σ in the given range. For σ = 1/2,   ζ( 1 + it) 1 t Q0 + it = ln + γ − 2 Re 12 (9.13) t → ∞. + O(t−1 ), 2 2π + it 2 Q0 (σ + it) =

Proof. Let u = σ + it and v = σ − it. Then (see (A.1)) Γ(1 − u) Γ(1 − v) + = 2 sin(πσ)t1−2σ + O(t−2σ−1 ) Γ(v) Γ(u) uniformly for σ ∈ [0, 1]. Hence (9.12) and (9.13) follow from equations (9.8) and (9.9), respectively.  Corollary 9.7. As t → ∞, (9.14)

⎧ 1−2σ ⎪ ), ⎨O(t Q0 (σ + it) = O(ln t), ⎪ ⎩ O(1),

σ ∈ [0, 1/2), σ = 1/2, σ ∈ (1/2, 1].

If the error terms O(t1−2σ ) and O(1) on the rhs are replaced with O(t1−2σ ln t) and O(ln t), respectively, then ( 9.14) holds uniformly for σ ∈ [0, 1]. Proof. Equation (9.14) follows immediately from corollary 9.6. From this corollary it also follows that the error terms in (9.14) are uniform with respect to σ ∈ [0, 12 − δ] ∪ { 12 } ∪ [ 12 + δ, 1] for any fixed δ > 0. We will complete the proof by showing that (9.15)

Q0 (σ + it) = O(t1−2σ ln t),

t → ∞,

uniformly for σ ∈ [0, 1/2) and (9.16)

Q0 (σ + it) = O(ln t),

t → ∞,

uniformly for σ ∈ (1/2, 1]. By (9.12), we have 1 + 2Γ(2σ − 1)ζ(2σ − 1) sin(πσ)t1−2σ + O(1), t → ∞, 2σ − 1 uniformly for σ ∈ [0, 1/2) ∪ (1/2, 1]. Since   σ 1 + O((1 − 2σ)t1−2σ ln t), σ ∈ [0, 1/2), 1−2σ 1−2σ   t =1−2 t (ln t)dσ = 1 + O((2σ − 1) ln t), σ ∈ (1/2, 1], 1/2 Q0 (σ + it) =

9.1. COROLLARIES

85

uniformly for σ in the given ranges, and since 1 + 2Γ(2σ − 1)ζ(2σ − 1) sin(πσ), 2σ − 1

2Γ(2σ − 1)ζ(2σ − 1) sin(πσ)|2σ − 1|,

are bounded functions of σ ∈ [0, 1], we find (9.15) and (9.16).



Corollary 9.8 (Large t asymptotics of Qn (u)). Let u = σ + it. For each integer n ≥ 1, Qn (u) = Q−n (u) = bn (2σ) − t1−2σ ie−πiσ

m   j=1

(9.17)

+ t 2 −2σ 1

m

πi 4

wu−1 (w + 2iπn)u¯−1 e−jw dw

ˆα H

2σ− 12

Im ζ(σ + 1 + it) i(−1) e π (1 + 2n)u¯ ζ(σ + it)  −2 − 2 Re 1 − σ + it t 2 2n(1 + n)

+ O(t−1 ) + O(t− 2 −2σ ), 1

t → ∞,

σ ∈ [0, 1],

t where m = [ π(1+(2n) −1 ) ] and the error term is uniform for σ in the given range.

Proof. Let u = σ + it and v = σ − it. By (A.1), the coefficient of Φ(u, v, n) in (9.7) satisfies   πi Γ(1 − v) ieπiv = t1−2σ e 2 (2σ+1) 1 + O(t−2 ) , Γ(u) 2 sin(πu)

t → ∞,

uniformly for σ ∈ [0, 1]. Hence there exist constants T ≥ 1 and c > 0 such that    Γ(1 − v) ieπiv  1−2σ   ,  Γ(u) 2 sin(πu)  ≥ ct for all t ≥ T and all σ ∈ [0, 1]. Since the terms not involving Φ on the rhs of (9.7) are O(1), and clearly |Qn (u)| ≤ Q0 (u) for each integer n, we conclude from (9.7) and corollary 9.7 that  O(t1−2σ ln t), σ ∈ [0, 1/2), 1−2σ t Φ(u, v, n) = O(Q0 (u)) + O(1) = (9.18) O(ln t), σ ∈ [1/2, 1]. uniformly for σ ∈ [0, 1]. Equation (9.17) now follows from (9.11).



From (9.18) and corollary 9.7, we also obtain an estimate for Φ. Corollary 9.9. For each integer n ≥ 1, ⎧ ⎪ ⎨O(1), (9.19) Φ(σ + it, σ − it, n) = O(ln t), ⎪ ⎩ O(t2σ−1 ),

σ ∈ [0, 1/2), σ = 1/2, σ ∈ (1/2, 1].

If the error terms O(1) and O(t2σ−1 ) on the rhs are replaced with O(ln t) and O(t2σ−1 ln t), respectively, then ( 9.19) holds uniformly for σ ∈ [0, 1].

86

9. FOURIER COEFFICIENTS OF THE PRODUCT

9.2. Asymptotics of the zeroth Fourier coefficient In order to put theorems 9.1 and 9.2 in context, let us first consider the case n = 0 of the zeroth Fourier coefficient. The study of the large t asymptotics of the zeroth Fourier coefficient  1 Q0 (s) = |ζ1 (s, α)|2 dα, s = σ + it, 0

i.e., of the mean square of ζ1 (s, α) as a function of α ∈ [0, 1], has a long history. Koksma and Lekkerkerker showed in the 1952 paper [17] that Q0 ( 12 + it) = O(ln t) and that    1 1 1 1−2σ +O t + ln t uniformly for < σ ≤ 1. Q0 (σ + it) = 2σ − 1 2σ − 1 2 In the case when σ = 1/2, Balasubramanian established [4] the asymptotic formula   1 Q0 + it = ln t + O(ln ln t) 2 and in the 1980s the error term in this formula was improved to O(1) by Rane [21]. Sitaramachandrarao was awarded a prize (announced in Hardy-Ramanujan Journal 10 (1987), p. 28) for proving the more precise result   1 t Q0 + it = ln + γ + O(t−3/16 (ln t)3/8 ). 2 2π Zhang independently obtained [27] a similar result but with the slightly larger error term O(t−3/16 (ln t)11/8 ), which was later improved to O(t−7/36 (ln t)25/18 ) in [28]. Finally, in [1] and [29], Andersson and Zhang independently arrived at the asymptotic expansion stated in (9.13). Around the same time as Andersson and Zhang obtained the expansion (9.13), Katsurada and Matsumoto started using Atkinson’s dissection argument applied to the product ζ1 (u, α)ζ1 (v, α) with two independent complex variables u and v in order to analyze Q0 , see [14]. In [15], they refined their approach and found an alternative proof of (9.13) together with a number of other related results. The derivation of [15] is based on the following expression for the zeroth Fourier coefficient (see [15, Eq. (2.1)]): (9.20)

q0 (u, v) =

1 + R0 (u, v) + R0 (v, u) − T0 (u, v) − T0 (v, u), u+v−1 − N + 1 < Re u, Re v < N + 1,

where (9.21)

R0 (u, v) = Γ(u + v − 1)ζ(u + v − 1)

Γ(1 − v) Γ(u)

and, for any integer N ≥ 0, T0 (u, v) = (9.22)

N −1 

(u)k (ζ(u + k) − 1) (1 − v)k+1 k=0  ∞ (u)N  1−u−v ∞ u+v−2 + l β (1 + β)−u−N dβ, (1 − v)N l l=1

9.2. ASYMPTOTICS OF THE ZEROTH FOURIER COEFFICIENT

87

with the Pochhammer symbol (a)k being defined by (a)k =

Γ(a + k) = a(a + 1)(a + 2) · · · (a + k − 1). Γ(a)

It is not hard to see that T0 is independent of the choice of N (see Lemma 9.10). It appears that the identity (9.20) provides the most powerful approach available for determining the asymptotics of Q0 (s). For example, the expansion of corollary 9.6 for Q0 (σ + it) with σ = 1/2 is easily obtained from (9.20) by taking u = v¯ = σ + it and N = 1, and noting that the sum over l in (9.22) is O(t−1 ); the expansion for σ = 1/2 is obtained in a similar way by taking N = 1 and σ = (1 + δ)/2 and letting δ → 0, see [15] or the proof of theorem 9.2. The expression (9.4) for qn given in theorem 9.1 is the natural generalization to 1 clearly any integer n ∈ Z of the expression (9.20) for q0 . Indeed, since b0 (s) = s−1 ) ∞ −s is the analytic continuation of the integral 1 α dα, this follows from (9.5) and the following lemma. Lemma 9.10. The functions R0 and T0 defined in ( 9.21) and ( 9.22) are the analytic continuations of the integrals  ∞ α−v ζ1 (u, α)dα, Re(u + v) > 2, Re v < 1, R0 (u, v) = 

0 1

T0 (u, v) =

α−v ζ1 (u, α)dα,

Re v < 1.

0

Proof. Employing the integral representation  ∞ −αr u−1 e r 1 (9.23) dr, Re u > 1, ζ1 (u, α) = Γ(u) 0 er − 1

Re α > −1,

and using Fubini’s theorem to change the order of integration, we find, for Re(u + v) > 2 and Re v < 1,  ∞  ∞ u−1  ∞ r 1 −v α ζ1 (u, α)dα = α−v e−αr dαdr. r −1 Γ(u) e 0 0 0 The change of variables β = αr shows that the rhs can be rewritten as  ∞ u+v−2  ∞ r 1 β −v e−β dβdr. Γ(u) 0 er − 1 0 Since the β-integral equals Γ(1 − v), we arrive at   ∞ Γ(1 − v) ∞ r u+v−2 −v dr α ζ1 (u, α)dα = Γ(u) er − 1 0 0 Γ(1 − v) = Γ(u + v − 1)ζ(u + v − 1), Re(u + v) > 2, Γ(u)

Re v < 1,

which proves the statement for R0 . Integrating by parts repeatedly using the facts that ζ1 (s, 1) = ζ(s) − 1 and (9.24)

∂α ζ1 (s, α) = −sζ1 (s + 1, α),

88

9. FOURIER COEFFICIENTS OF THE PRODUCT

we obtain, for any integer N ≥ 1 and Re v < 1,  1  1 u ζ(u) − 1 + α−v ζ1 (u, α)dα = α1−v ζ1 (u + 1, α)dα 1 − v 1 − v 0 0  1 N −1  (u)k (ζ(u + k) − 1) (u)N = ··· = (9.25) + αN −v ζ1 (u + N, α)dα. (1 − v)k+1 (1 − v)N 0 k=0

The statement for T0 will follow once we prove that  (9.26)

1

α

N −v

ζ1 (u + N, α)dα =

∞ 

0

l=1

 l

1−u−v

∞

β u+v−2 (1 + β)−u−N dβ

l

for Re v < N + 1 and Re u > −N + 1. But for Re u > −N + 1 we can use the representation (9.27)

ζ1 (s, α) =

∞  l=1

1 , (l + α)s

Re s > 1,

Re α > −1,

to write the left-hand side of (9.26) as ∞  

1

αN −v (l + α)−u−N dα.

0

l=1

The change of variables α = l/β completes the proof of (9.26) and hence also of the lemma.  Remark 9.11. The formula for Tn given in (9.6c) is obtained by integrating by parts once in the integral (9.5c). If we instead integrate by parts N times, we find   1 N −1  (−1)k ∂ k  α−v ζ1 (u, α)e−2πinα dα = (ζ1 (u, α)e−2πinα ) Tn (u, v) = k (1 − v) ∂α k+1 0 α=1 k=0 N  1 N ∂ (−1) + (9.28) αN −v N (ζ1 (u, α)e−2πinα )dα (1 − v)N 0 ∂α k   N −1   k (u)j (2πin)k−j (ζ(u + j) − 1) = j (1 − v)k+1 j=0 k=0

+

 N    N (u)j (2πin)N −j j=0

j

(1 − v)N

1

αN −v ζ1 (u + j, α)e−2πinα dα.

0

Using that (cf. (9.26)), for 2 ≤ j ≤ N with Re v < N + 1 and Re u > −j + 1,  1 αN −v ζ1 (u + j, α)e−2πinα dα 0

=

∞  l=1

l1−u−v+N −j

 l

∞

β u+v−2−N +j (1 + β)−u−j e−2πinl/β dβ,

9.3. PROOF OF THEOREM 9.1

89

this leads to the following expression for Tn when Re u > −1 and Re v < N + 1, which is the natural extension to any integer n of the expression (9.22) for T0 : N −1  N   k     k (u)j (2πin)k−j N (u)j (2πin)N −j (ζ(u + j) − 1) + Tn (u, v) = j j (1 − v)k+1 (1 − v)N j=2 k=0 j=0  ∞ ∞  (9.29) × l1−u−v+N −j β u+v−2−N +j (1 + β)−u−j e−2πinl/β dβ l=1



N

+

(2πin) (1 − v)N

l 1

0 N −1

+

u(2πin) (1 − v)N

αN −v ζ1 (u, α)e−2πinα dα 

1

αN −v ζ1 (u + 1, α)e−2πinα dα.

0

9.3. Proof of Theorem 9.1 Lemma 9.12 (Fourier series of ζ1 (u, α)ζ1 (v, α)). Suppose u, v ∈ C \ {1} and δ ∈ (0, π/2). Then the Fourier series representation of ζ1 (u, ·)ζ1 (v, ·) ∈ C ∞ ([0, 1]) is given by  ζ1 (u, α)ζ1 (v, α) = (9.30) qn (u, v)e2πinα , n∈Z

where (9.31)



e−(sgn n)iδ ∞

qn (u, v) =

 −u−v  α + α−v ζ1 (u, α) + α−u ζ1 (v, α) e−2πinα dα,

1

n ∈ Z \ {0}, and



1

q0 (u, v) =

ζ1 (u, α)ζ1 (v, α)dα. 0 2

The series in ( 9.30) converges in L ([0, 1]) and converges pointwise for each α ∈ (0, 1). If Re(u + v) > 2, the Fourier coefficients can be expressed as  ∞  −u−v  α qn (u, v) = (9.32) + α−v ζ1 (u, α) + α−u ζ1 (v, α) e−2πinα dα, 1

n ∈ Z,

u, v ∈ C \ {1},

Re(u + v) > 2.

Proof. Equation (9.31) follows from (9.2) and easy algebra. Indeed, for u, v ∈ C \ {1} and n ≥ 1, we compute  1 ζ1 (u, α)ζ1 (v, α)e−2πinα dα qn (u, v) = 0



e−iδ ∞

= 0



−

e−iδ ∞

 ζ1 (u, α)ζ1 (v, α)e−2πinα dα

1

e−iδ ∞

=



ζ(u, α + 1)ζ(v, α + 1)e−2πinα dα

0



e−iδ ∞

− 1

(ζ(u, α) − α−u )(ζ(v, α) − α−v )e−2πinα dα.

90

9. FOURIER COEFFICIENTS OF THE PRODUCT

Changing variables β = α + 1 in the integral from 0 to e−iδ ∞, we infer that  e−iδ ∞  e−iδ ∞ −2πinβ qn (u, v) = ζ(u, β)ζ(v, β)e dβ − ζ(u, α)ζ(v, α)e−2πinα dα 1



1 e−iδ ∞

+  =

 −v  α ζ(u, α) + α−u ζ(v, α) − α−u−v e−2πinα dα

1 e−iδ ∞

 −v  α ζ1 (u, α) + α−u ζ1 (v, α) + α−u−v e−2πinα dα,

1

which proves (9.31) for n ≥ 1; the same computation applies also when n ≤ −1 if δ is replaced by −δ. The expression in (9.32) follows because, by (9.10), ζ(s, α) = α1−s −s s−1 + O(α ) as α → ∞, so if Re(u + v) > 3 and u, v = 1, the above proof can be repeated for each n ∈ Z with δ = 0; the resulting formula can then be extended to Re(u + v) > 2 by analytic continuation.  Proof of Theorem 9.1. Equations (9.4) and (9.5) are a direct consequence of (9.32). It remains to establish the meromorphic continuations in (9.6). For n = 0, this was carried out already in section 9.2 (see Lemma 9.10 and equation (9.25) with N = 1). Suppose n is a nonzero integer. Making the change of variables r = 2πinα in the integral in (9.5a) and deforming the contour back to the real axis, we obtain, for Re s > 1,  ∞ r −s (2πin)s−1 e−r dr = Γ(1 − s, 2πin)(2πin)s−1. bn (s) = 2πin

This proves (9.6a). To determine the analytic continuation of Rn , we substitute the integral representation (9.23) for ζ1 into (9.5b) and change the order of integration. This gives, for Re(u + v) > 2 and Re v < 1,  ∞  ∞ u−1  ∞ r 1 α−v ζ1 (u, α)e−2πinα dα = α−v e−α(r+2πin) dαdr. Γ(u) 0 er − 1 0 0 The change of variables β = α(r + 2πin) shows that the rhs can be rewritten as  ∞  ∞ u−1 r 1 v−1 (r + 2πin) β −v e−β dβdr, Γ(u) 0 er − 1 0 where we have used the decay of e−β to deform the contour back to the positive real axis. Since the β-integral equals Γ(1 − v), we arrive at the formula   ∞ Γ(1 − v) ∞ r u−1 (r + 2πin)v−1 dr, (9.33) α−v ζ1 (u, α)e−2πinα dα = r −1 Γ(u) e 0 0 Re(u + v) > 2, Re v < 1. The integral on the rhs can be expressed in terms of Φ:   ∞ u−1 1 , r v−1 e−iπ(u+v) −eiπ(u−v) (r + 2πin) dr = Φ(u, v, n) × (9.34) 1 r −1 e , 0 e−iπ(u−v) −eiπ(u+v)

n > 0, n < 0,

for Re u > 1 and v ∈ C. Indeed, for Re u > 1 the contribution from the circular part of the Hankel contour Hα in (1.12) vanishes in the limit α → 0; hence (1.12)

9.4. PROOF OF THEOREM 9.2

can be written as iπu

Φ(u, v, n) = (e

−iπu

−e

 ) 0

∞

91

r u−1 (−r − 2πin)v−1 dr. er − 1

For n > 0 (resp. n < 0), −r − 2πin lies in the third (resp. second) quadrant, and so (−r − 2πin)v−1 = e−(sgn n)iπ(v−1) (r + 2πin)v−1 , which proves (9.34). The expression (9.6b) for Rn follows from (9.33) and (9.34). The expression (9.6c) for Tn is the special case N = 1 of (9.28).  9.4. Proof of Theorem 9.2 Lemma 9.13. Let u = σ1 + it and v = σ2 − it. Then, for every N ≥ 1, Rn (v, u) = O(t−N ) uniformly for n ≥ 1 and σ1 , σ2 ∈ [0, 1] as t → ∞. Proof. This is an immediate consequence of the expression (9.6b) for Rn , the estimate (7.30) of Φ(v, u, β), and the fact that (see (A.2)) Γ(1 − u) ieπiu = O(t1−σ1 −σ2 e−2πt ), Γ(v) 2 sin(πv)

t → ∞,

uniformly for σ1 , σ2 ∈ [0, 1].



Lemma 9.14. Let u = σ1 + it, v = σ2 − it, and n ∈ Z. Then, as t → ∞, (9.35a) (9.35b)

ζ(u) − 1 (ζ(u + 1) − 1)u + + O(t−1 ), 1−v (1 − v)(2 − v) ζ(v) − 1 (ζ(v + 1) − 1)v + + O(t−1 ), Tn (v, u) = 1−u (1 − u)(2 − u)

Tn (u, v) =

uniformly for σ1 , σ2 ∈ [0, 1]. Proof. Consider the expression (9.29) for Tn for some integer N ≥ 3. An integration by parts shows that the integral with respect to dβ in (9.29) can be written as follows for Re v < N + 1:  ∞ (1 + l)1−u−j u+v−2−N +j l (9.36) β u+v−2−N +j (1 + β)−u−j e−2πinl/β dβ = u+j−1 l  ∞ (1 + β)1−u−j ∂  u+v−2−N +j −2πinl/β  dβ = E1 + E2 + E3 , β − e 1 − u − j ∂β l where we have used the short-hand notation (1 + l)1−u−j u+v−2−N +j E1 := l , u+j−1  ∞ (1 + β)1−u−j u+v−3−N +j −2πinl/β β E2 := (u + v − 2 − N + j) e dβ, u+j−1 l  ∞ (1 + β)1−u−j u+v−4−N +j −2πinl/β β E3 := 2πinl e dβ. u+j−1 l Substituting (9.36) into (9.29), we see that Tn (u, v) can be expressed for Re u > −1 and Re v < N + 1 as N −1  5 k     k (u)j (2πin)k−j (9.37) (ζ(u + j) − 1) + Fm , Tn (u, v) = j (1 − v)k+1 m=1 j=0 k=0

92

9. FOURIER COEFFICIENTS OF THE PRODUCT

where we have used the short-hand notation N   ∞  N (u)j (2πin)N −j  1−u−v+N −j Fm := l Em , j (1 − v)N j=2 l=1  1 (2πin)N F4 := αN −v ζ1 (u, α)e−2πinα dα, (1 − v)N 0  u(2πin)N −1 1 N −v F5 := α ζ1 (u + 1, α)e−2πinα dα. (1 − v)N 0

m = 1, 2, 3,

The first two terms on the rhs of (9.35a) come from the terms with k = j = 0 and k = j = 1 in the double sum in (9.37). All other terms in this double sum are O(t−1 ) by standard estimates of ζ(s). Thus the asymptotic formula (9.35a) for Tn (u, v) will follow if we can show that Fm = O(t−1 ) uniformly for σ1 , σ2 ∈ [0, 1] for each m = 1, . . . , 5. The terms F1 , F2 , and F3 are easily estimated: |F1 | ≤

N ∞ C   −1 l (1 + l)1−σ1 −j = O(t−1 ), t j=2 l=1

|F2 | ≤ C

N  ∞ 

l1−σ1 −σ2 +N −j

C t

|F3 | ≤ C

∞ N  

l2−σ1 −σ2 +N −j

≤

C t



∞

l

j=2 l=1 ∞ N  

(1 + β)1−σ1 −j σ1 +σ2 −3−N +j β dβ |u + j − 1|

l−σ1 −j = O(t−1 ),

j=2 l=1 N  ∞ 

∞

l

j=2 l=1

≤



(1 + β)1−σ1 −j σ1 +σ2 −4−N +j β dβ |u + j − 1|

l−σ1 −j = O(t−1 ),

j=2 l=1

uniformly for σ1 , σ2 ∈ [0, 1]. Integrating by parts twice and using (9.24), we find, for σ1 , σ2 ∈ [0, 1],  1 (9.38) αN −v e−2πinα ζ1 (u, α)dα = ∂α f (v, 1)ζ1 (u, 1) + uf (v, 1)ζ1 (u + 1, 1) 0



1

f (v, α)ζ1 (u + 2, α)dα,

+ u(u + 1) 0

where f (v, α) is defined by



α



f (v, α) = 0

α1

α2 N −v e−2πinα2 dα2 dα1 .

0

The functions f (v, α) and ∂α f (v, α) are uniformly bounded for α ∈ [0, 1] and Re v ∈ [0, 1]. Also, by a trivial estimate of the sum in (9.27), ζ1 (u + 2, α) is uniformly bounded for α ∈ [0, 1] and Re u ∈ [0, 1]. Since ζ1 (u, 1) = ζ(u) − 1 and ζ1 (u + 1, 1) = ζ(u + 1) − 1, we conclude that the rhs of (9.38) is O(t2 ) uniformly for σ1 , σ2 ∈ [0, 1]. It follows that F4 = O(t2−N ) = O(t−1 ) uniformly for σ1 , σ2 ∈ [0, 1] as t → ∞. A similar argument shows that F5 also is O(t2−N ). This completes the

9.5. FOURIER COEFFICIENTS OF ζ(s, α) AND ζ1 (s, α)

93

proof of the estimate (9.35a) for Tn (u, v); the estimate (9.35b) for Tn (v, u) follows by interchanging u and v in the above arguments.  Proof of Theorem 9.2. Employing lemma 9.13 and lemma 9.14 in the expression (9.4) for qn (u, v) with n ≥ 1, we find formula (9.7). The formula (9.8) for the asymptotics of q0 (u, v) when σ1 + σ2 = 1 follows immediately by substituting the expression (9.6b) for R0 into equation (9.4) and using Lemma 9.14. To derive the formula (9.9) for the asymptotics of q0 (u, v) when σ1 + σ2 = 1, we first set u = σ + δ + it and v = 1 − σ + δ − it in equation (9.4) and take the limit δ → 0; this yields q0 (u, v) = − ln 2π + γ +

ψ(u) + ψ(v) + T0 (u, v) + T0 (v, u) 2

where u = σ + it, v = 1 − σ − it, and ψ(s) = Γ (s)/Γ(s). The formula (9.9) then −1 follows from Lemma 9.14 and the fact that ψ(σ ± it) = ln t ± πi ) uniformly 2 + O(t for σ ∈ [0, 1] as t → ∞.  9.5. Fourier coefficients of ζ(s, α) and ζ1 (s, α) In this section, we consider, for completeness, the Fourier coefficients of ζ(s, α) and ζ1 (s, α). In particular, we will see that the coefficients bn (s) defined in (9.5a) are the Fourier coefficients of ζ1 (s, α), cf. [2, 22]. We first compute the Fourier coefficients an (s) of ζ(s, α) defined by  1 an (s) = ζ(s, α)e−2πinα dα, n ∈ Z. 0

The coefficients an (s) are well-defined whenever Re s < 1, because the relation ζ(s, α) = ζ(s, α + 1) + α−s implies that ζ(s, α) ∼ α−s as α ↓ 0; hence ζ(s, ·) ∈ L1 ([0, 1]) iff Re s < 1. For 0 < Re s < 1, the next lemma was proved in [22] using different techniques. Lemma 9.15 (Fourier series of ζ(s, α)). Suppose Re s < 1. Then the Fourier series representation of ζ(s, ·) ∈ L1 ([0, 1]) ∩ C ∞ ((0, 1]) is given by   Γ(1 − s)(2πin)s−1 , n ∈ Z \ {0}, 2πinα an (s)e , an (s) = (9.39) ζ(s, α) = 0, n = 0. n∈Z Proof. For Re s < 0, a computation using the representation (9.1) for ζ(s, α) gives  1   eαz z s−1 Γ(1 − s) 1 −2πinα −2πinα an (s) = ζ(s, α)e dα = e dzdα z 2πi H1 1 − e 0 0   1 Γ(1 − s) z s−1 = eα(z−2πin) dα dz 2πi 1 − ez H1 0   ez−2πin − 1 z s−1 z s−1 Γ(1 − s) Γ(1 − s) dz = dz = − z 2πi 2πi H1 z − 2πin 1 − e H1 z − 2πin  Γ(1 − s)(2πin)s−1 , n ∈ Z \ {0}, = (9.40) Re s < 0, 0, n = 0,

94

9. FOURIER COEFFICIENTS OF THE PRODUCT

where, in the last step, we have deformed the Hankel contour H1 to infinity and used the residue theorem. By writing  1  1 an (s) = ζ(s, α + 1)e−2πinα dα + α−s e−2πinα dα, Re s < 0, n ∈ Z, 0

0

we see that each Fourier coefficient an (s) is analytic for Re s < 1. Since the expression in (9.39) for an (s) also is analytic for Re s < 1 for each n, the lemma follows from (9.40) and analytic continuation.  Remark 9.16. Since ζ(s, ·) ∈ L1 ([0, 1])∩C ∞ ((0, 1]), the Fourier series in (9.39) converges pointwise for each α ∈ (0, 1). Easy estimates show that the convergence is absolute and uniform for s in compact subsets of Re s < 0 and α ∈ [0, 1]. This reflects the fact that ζ(s, α) can be extended continuously to a function ζ(s, ·) ∈ C([0, 1]) with ζ(s, 0) = ζ(s, 1) if Re s < 0 (see (9.2)). Lemma 9.17 (Fourier series of ζ1 (s, α)). Suppose s ∈ C \ {1}. Then the Fourier series representation of ζ1 (s, ·) ∈ C ∞ ([0, 1]) is given by  (9.41) bn (s)e2πinα , ζ1 (s, α) = n∈Z

where bn (s), n ∈ Z, are the coefficients in ( 9.6a). Proof. We need to show that   1 Γ(1 − s, 2πin)(2πin)s−1, n = 0, −2πinα ζ1 (s, α)e dα = (9.42) 1 n = 0, 0 s−1 ,

s ∈ C \ {1}.

Actually, since both sides of (9.42) are analytic for s ∈ C \ {1} for each n ∈ Z, it is enough to show (9.42) for Re s < 1. By the definition (9.2) of ζ1 , we have  1  1 −2πinα ζ1 (s, α)e dα = an (s) − α−s e−2πinα dα, Re s < 1. 0

0

The change of variables r = 2πinα shows that  1 − α−s e−2πinα dα = (Γ(1 − s, 2πin) − Γ(1 − s))(2πin)s−1 ,

Re s < 1.

0

Recalling the expression for an (s) in (9.39), the lemma follows.



Remark 9.18. Since ζ1 (s, ·) ∈ C ∞ ([0, 1]), the Fourier series in (9.41) converges pointwise for each α ∈ (0, 1). For fixed s, the incomplete Gamma function admits the asymptotic expansion (see e.g. [19, Eq. (8.11.2)])   N  (s − 1) · · · (s − k) s−1 −z −N −1 + O(z ) , z → ∞, Γ(s, z) = z e 1+ zk k=1

where the error term is uniform for | arg z| ≤ bn (s) = (2πin)

−1

(1 + O(n

3π 2 −1

− δ. Hence )),

n → ∞,

showing that the Fourier series for ζ1 is not absolutely convergent for any s ∈ C\{1}.

Part 3

Representations for the Basic Sum

CHAPTER 10

Several Representations for the Basic Sum The purpose of this chapter is to present integral representations of the basic b s−1 for certain values of a and b. an Let the contour Ctη , t < η, denote the semicircle from it to iη with Re z ≥ 0. Splitting the contour of the second integral in the rhs of equation (1.2) into the contour Cηt plus the ray from it to ∞ exp(iφ2 ), we find

sum

η [ 2π ]

ζ(1 − s) =



n=1

ηs e− 2 − − s(2π)s (2π)s iπs

n

s−1

+ GL (t, σ; η) + GU (t, σ; t),

(10.1)

 Ctη

z s−1 dz ez − 1 0 ≤ σ ≤ 1,

0 < t < η,

where GL and GU are defined in (3.5) and (3.6) respectively. The term GU (t, σ; t) was computed to all orders as t → ∞ in theorem 3.2, see equations (3.26) and (3.33). The term GL (t, σ; η) with t < η was computed to all orders as t → ∞ in (3.13). Thus, by comparing equation (10.1) with the representation obtained in [η/2π] theorem 3.2, it follows that we can express the fundamental sum n=[t/2π]+1 ns−1 in terms of the integral appearing in the rhs of (10.1), where the error is computed to all orders. For brevity of presentation, we state this result only to leading order. Lemma 10.1 (An integral representation for the basic sum). Let Ctη , t < η, denote the semicircle from z = it to z = iη with Re z ≥ 0. For every > 0, η [ 2π ]

(10.2)



e− 2 = (2π)s iπs

n

s−1

t n=[ 2π ]+1



z s−1 ηs iη s−1 dz + + ln(1 − eiη ) z s(2π)s (2π)s Ctη e − 1   1 its−1 e−it t it − ln(1 − e ) + O + , (2π)s t2−σ η 2−σ (1 + )t < η < ∞, 0 ≤ σ ≤ 1, t → ∞,

where the error term is uniform for all η, σ in the above ranges. Proof. Letting σ → 1 − σ in (1.4) and taking the complex conjugate of the resulting equation, we find  s 1 t its−1 e−it n − + (−eit + 2i Im Li1 (eit )) ζ(1 − s) = s s 2π (2π) n=1  √   1 + i π 6σ 2 − 6σ + 1 i ts−1 e−it 1 − i √ √ πt + − iσ − + +O(tσ−2 ). (2π)s 2 3 24 t t [ 2π ]



s−1

97

98

10. SEVERAL REPRESENTATIONS FOR THE BASIC SUM

The lemma follows by subtracting this equation from (10.1), employing the following equations which follow from (3.13), (3.33), and (3.26) respectively:   t iη s−1 iη ln(1 − e ) + O 2−σ , (1 + )t < η, (not uniformly in ), GL (t, σ; η) = − (2π)s η  √   s−1 −it √ 1−i 1 + i π 6σ 2 − 6σ + 1 t e i (1) √ πt + − iσ − +O(tσ−2 ), GU (t, σ; t) = (2π)s 2 3 24 t its−1 e−it (2) ln(1 − e−it ) + O(tσ−2 ), GU (t, σ; t) = (2π)s and using the identity ts−1 e−it its−1 e−it its−1 e−it ln(1 − e−it ) + 2 Im Li1 (eit )) = ln(1 − eit ). s s (2π) (2π) (2π)s



It is possible to derive an alternative integral representation for the related sum [t/2π] s−1 . [η/2π]+1 n Lemma 10.2 (An alternative integral representation for the basic sum). For every δ > 0, (10.3) t [ 2π ]



ns−1 =

η n=[ 2π ]+1

0 < η < t,

t ρs−1 2 dρ + O(tσ−1 ),  s iρ (2π) η e − 1

dist(η, 2πZ) > δ,

dist(t, 2πZ) > δ,

0 ≤ σ ≤ 1,

t → ∞,

where the error term is uniform for all η, σ in the above ranges and the contour in the integral denotes the principal value integral with respect to the points 

 η   t    2πn  n ∈ Z, +1≤n≤ . 2π 2π Proof. Let Cˆtη denote the semicircle from it to iη with Re z ≤ 0, defined in equation (2.10c) with η and α replaced by t and η respectively, see figure 10.1. Let Rηt (s) and Lηt (s) denote the following: iπs  z s−1 e− 2 t (10.4) dz, s∈C Rη (s) = (2π)s Cηt ez − 1 and e− 2 (2π)s iπs

(10.5)

Lηt (s) =

 ˆη C t

z s−1 dz, ez − 1

s ∈ C.

Cauchy’s theorem applied in the interior of the disk whose boundary is Cηt ∪ Cˆtη , yields t [ 2π ]

(10.6)

Lηt (s)

+

Rηt (s)



=

ns−1 ,

s ∈ C.

η n=[ 2π ]+1

On the other hand, the Plemelj formulas imply (10.7)

Rηt (s) − Lηt (s) =

t ρs−1 2 dρ.  (2π)s η eiρ − 1

10. SEVERAL REPRESENTATIONS FOR THE BASIC SUM

99

it Cˆtη

Cηt iη Re z

Figure 10.1. The contours Cηt and Cˆtη . Using (10.7) to replace Rηt in (10.6), we find t [ 2π ]



(10.8)

ns−1 =

η n=[ 2π ]+1

t ρs−1 2 dρ + 2Lηt .  s iρ (2π) η e − 1

Replacing the contour Cˆtη in (10.5) by the union of the following three segments: {iteiθ | 0 ≤ θ ≤ π/4}, we obtain Lηt

πi

{iρe 4 | η ≤ ρ ≤ t},

{iηeiθ | 0 ≤ θ ≤ π/4},

 π iθ(s−1) s−1  t πi (s−1) s−1 4 e πi t e4 ρ 1 iθ e 4 dρ = dθ − ite πi iθ s ite (2π) e −1 0 η eiρe 4 − 1   π4 iθ(s−1) s−1 e η − iηeiθ dθ iθ iηe e −1 0 1 [J1 + J2 + J3 ]. =: (2π)s

The assumption dist(t, 2πZ) > δ implies that  π4 −θt σ−1 πt e t tσ−1 |J1 | ≤ tdθ = (1 − e− 4 ) = O(tσ−1 ). A A 0 Similar computations show that J2 is exponentially small and, in view of the assumption dist(η, 2πZ) > δ, that J3 is O(η σ t−1 ). Thus, (10.9)

Lηt = O(tσ−1 ), 

and the lemma follows from (10.8).

Remark 10.3. Lemma 10.2 implies that the leading behavior of ζ(s) is characterized by the following integral: t

I(η, t, σ) = 

η

ρs−1 dρ. eiρ − 1

100

10. SEVERAL REPRESENTATIONS FOR THE BASIC SUM

Remark 10.4. The estimate (10.9) is a consequence of the fact that the integral appearing in the definition of Lηt (s) does not possess any stationary points. Indeed, using ∞  1 emz , Re z < 0, = − ez − 1 m=0 it follows that − iπs 2



e

ˆη C t

∞  t  z s−1 dz = eimρ+it ln ρ ρσ−1 dρ. ez − 1 η m=0

Candidates for stationary points occur at ρ∗ = −t/m and the inequality η ≤ ρ∗ ≤ t implies the non-existence of any stationary points. The next lemma gives an alternative representation for the fundamental integral Rηt (s) defined in (10.4). Lemma 10.5. Let Cηt denote the semicircle from iη to it with Re z ≥ 0 defined by equation ( 1.23). Then, for every δ > 0, (10.10) Rηt (s)

 t  e −1 us−1 du +O , iue−i − 1 t1−σ η e dist(η, 2πZ) > δ, dist(t, 2πZ) > δ,

e−is = (2π)s

0 < η < t,



t

0 ≤ σ ≤ 1,

0 < < 1,

t → ∞,

where the error term is uniform for all η, σ, in the above ranges. Proof. Let ∈ (0, 1). We deform the contour Cηt on the lhs of (10.10) so that it consists of the following three pieces: {iηe−iθ | 0 ≤ θ ≤ },

{iue−i | η ≤ u ≤ t},

{ite−iθ | 0 ≤ θ ≤ }.

This yields   t   −iθs   −iθs iπs z s−1 dz us−1 du e dθ e dθ −is s s − iη = e + it . e− 2 −i −iθ −iθ z iue iηe ite −1 −1 −1 Cηt e − 1 η e 0 e 0 e The assumption that dist(η, 2πZ) > δ implies that there exists an A such that −iθ |eiηe − 1| ≥ A for all θ ∈ [0, ], ∈ (0, 1), and η > 0. Thus,    e−iθs  eθt   θ ∈ [0, ],  eiηe−iθ − 1  ≤ A , and so

   

0



  et − 1 e−iθs ≤ . dθ −iθ iηe At e −1 

Similarly, because of the assumption dist(t, 2πZ) > δ,      et − 1 e−iθs  ≤ . dθ   ite−iθ − 1 At 0 e This proves the lemma.



10. SEVERAL REPRESENTATIONS FOR THE BASIC SUM

it

101

z∗

t−η

iη

Figure 10.2. The contour of integration of the integral in the rhs of equation (10.14). Lemma 10.5 implies that (10.11)

Rηt (s) =

1 lim (2π)s →0



t→0

η

t

us−1 du , eiue−i − 1

which in view of (10.6) and (10.9) implies the following alternative representation for the basic sum: t [ 2π  t ] us−1 du 1 = lim ns−1 + O(tσ−1 ). −i →0 η eiue (2π)s t→0 −1 η n=[ 2π ]+1

In fact, in the remaining part of this chapter we will present arguments which suggest1 that on the critical line the assumption t → 0 in (10.11) can be relaxed to 2 t → 0, so that the following equation is valid:  t us−1 du 1 (10.12) , lim Rηt (s) = s iue (2π) 2→0 η e −i − 1  t→0

where it is assumed that dist(η, 2πZ) > δ, η = O(1), dist(t, 2πZ) > δ, and σ = 1/2. Indeed, consider the integral  e−z z s−1 dz s t − iπs 2 (2π) Rη (s) = e . 1 − e−z Cηt Instead of the contour Cηt , we consider the finite ray from the point iη to the point z ∗ , see figure 10.2, where π iπ π z ∗ = iη + (t − η)ei( 2 −) = [η + (t − η)e−i ]e 2 , 0≤ < . 2 We claim that  it −z s−1 3 iπs e z dz e− 2 (10.13) = O(tσ et ). −z 1 − e ∗ z 1 The

rigorous justification of some of these arguments remains open.

102

10. SEVERAL REPRESENTATIONS FOR THE BASIC SUM

Indeed, the assumption dist(t, 2πZ) > δ implies that there exists an A > 0 independent of η, such that |1 − e−z | > A on the contour from z ∗ to it. Thus, using the parametrization z = ite−iθ , 0 ≤ θ ≤ , in the lhs of (10.13), we can prove (10.13) as follows:         −t sin θ θt  − iπs it e−z z s−1 dz   s  e−ite−iθ e−iθs dθ  e e dθ σ e 2  = it  ≤ t −iθ     −z −ite 1−e A 1−e z∗ 0  0   3 3 = O tσ etθ dθ = O(tσ et ). 0

where we have used that 0 ≤ θ ≤ 1.

|θ − sin θ| ≤ θ 3 ,

It follows from (10.13) that   z∗ −z s−1 iπs e−z z s−1 dz e z − iπs 2 e− 2 (10.14) = e lim dz. iπ →0 1 − e−z 1 − e−z 2 Cηt ηe 2  t→0

We next employ the following parametrization which maps z ∗ to ρ = 0:   π   iπ η iπ η z = ηe 2 + t 1 − − ρ ei( 2 −) = η + t 1 − − ρ e−i e 2 , t t η 0 0. η η Then, the rhs of equation (10.14) is given by  1− ηt −i  eit(ρ−1)e ρ −iψ s−1 s −i s−1 −iμ(s−1) e (10.17a) t lim e M e dρ, 1 − −i →0 M 1 − eit(ρ−1)e 0 2  t→0

where (10.17b)

ψ(t, ) = − μ.

The functions M and μ are given by    2  η η2

2 η η 2 4 M (t, ) = 1 − − 2 +O

− 2 (10.18a) , 2 t t t t

→0

and

η + O( 2 ),

→ 0. t Indeed, consider the triangle D1 formed by the intersection of the following finite

(10.18b)

μ(t, ) = −

10. SEVERAL REPRESENTATIONS FOR THE BASIC SUM

0

103

1 μ

t η

t ηM

−1

Figure 10.3. The triangle D1 . rays, see figure 10.3: (0, 1),

    t 1, 1 + − 1 e−i , η



 t 0, M e−iμ . η

The cosine rule for the triangle D1 implies the following identities:    2 2  t t t =1+ M −1 −2 − 1 cos(π − ) η η η 2    t t −1 +2 − 1 cos

=1+ η η   2   t t = 1+ −1 − 1 (1 − cos ). −2 η η Thus,

*  + + t 2 − 1 (1 − cos ) η t t+ M = ,1 − , 2 t η η η2

or

M=



1−2

η η2 − 2 t t

 (1 − cos ),

which yields (10.18a). The sine rule for the triangle D1 implies the identity sin μ sin(π − ) = . t −1 ηM

t η

Hence, 1 − ηt sin , M which, using the expression (10.18a) for M , yields (10.18b). Equation (10.18b) implies that ψ > 0, thus, we can define (W, w) via the equation ρ i(π−ψ) ρ −iψ (10.19) 1 − = 1+ = W eiw , W (ρ, t, ) > 0, w(ρ, t, ) > 0. e e M M The functions W and w are given by  2   2 2

η

η W (ρ, t, ) = 1 − ρ + O (10.20a) ,

→0 +O t t2 sin μ =

104

10. SEVERAL REPRESENTATIONS FOR THE BASIC SUM

ρ M

W

ψ

w 0

1 Figure 10.4. The triangle D2 .

and (10.20b)

w(ρ, t, ) =

  2   2 2 

η

η

η ρ , 1+O +O t 1−ρ t t2

→ 0.

Indeed, consider the triangle D2 formed by the intersection of the following finite rays, see figure 10.4:    ρ i(π−ψ) (0, 1), 1, 1 + , 0, W eiw . e M The cosine rule for the triangle D2 implies the following identities: ρ2 ρ W 2 = 1 + 2 − 2 cos ψ M M  ρ 2 2ρ = 1− (1 − cos ψ). + M M Hence,  ρ W = 1− M

-

2ρ 1 +  Mρ 2 (1 − cos ψ). 1− M

Thus, using the definition (10.17b) of ψ, together with equation (10.18b), we find ρ

ρ 1 M W =1− + ρ M 21− M





2 η 2 4 + O( ) , t2

→ 0.

Substituting in this equation the expression (10.18a) for M , we find equation (10.20a). The sine rule for the triangle D2 implies the identity sin w sin ψ = , ρ/M W which, using the expressions for ψ (equations (10.17b) and (10.18b)), for M (equation (10.18a)) and for W (equation (10.20a)), yields equation (10.20b). Using the identity  ρ −iψ s−1 e 1− = W s−1 ei(s−1)w , M

10. SEVERAL REPRESENTATIONS FOR THE BASIC SUM

105

as well as the expressions for M , μ, W and w obtained earlier, it is possible to compute all the expressions appearing in the rhs of (10.17a):  2  2 i2 η   M s−1 = e(it+σ−1) ln M = e− 2 eO t eO t2 , η 2 2 e−i(s−1)μ = eμt e−i(σ−1)μ = et e−η eO( t) ei(σ−1)( t −) eO( ) ,  2 η 2     2   2 η 2   η 2 s−1 s−1 (it+σ−1) O t +O t2 W = (1 − ρ) e = (1 − ρ)s−1 eO( η) eO t , ρ

ρ

eiw(s−1) = e−tw ei(σ−1)w = e−η 1−ρ e 1−ρ [O()+O( t )] . 

Substituting the above expressions in (10.17a) and using the identities −i

e−ite

= e−it(1−i+O(

2

))

= e−it−t+O(

2

t)

,

in order to simplify the integrand of the integral in equation (10.17a), we find the equation (10.21)   1− ηt it(ρ−1)e−i iπs e−z z s−1 dz e (1 − ρ)s−1 dρ s = t lim , 0 < η < t. e− 2 −i −z it(ρ−1)e →0 1−e 1−e Cηt 0 2  t→0

The transformation ρ = 1 − ut maps this equation to equation (10.12). We note that a detailed analysis in the neighborhood of ρ = 1 shows that similar estimates are also valid near ρ = 1.

APPENDIX A

The Asymptotics of Γ(1 − s) and χ(s) The asymptotic formulae given in theorems 4.1 and 4.4 involve the functions χ(s) and Γ(1 − s). In this appendix, we derive the asymptotics of these functions as t → ∞. We use the following well-known formula for the asymptotics of the Gamma function (see for example Olver [20], page 294): −s s ln s

Γ(s) = e

e



2π s

 12  1+

1 +O 12s



1 s2



where δ > 0. Hence, as t → ∞, Γ(s) =

√

2πe−σ−it es ln[(it)(1+ it )] σ

1+

1 σ 12it(1+ it )

1  (it) 2 1 +

s → ∞,

,

+O 1 σ 2



1 s2

| arg s| ≤ π − δ,



it √    s σ σ2 1 2π iπ (σ+it) it + 2t2 +O ( t3 ) − iπ = √ e 4 e 2 t e−σ−it e t       1 1 1 σ +O 2 +O 2 × 1+ (A.1) 1− 12it t 2it t    √ s− 1 iπs − iπ −it − iσ2 +O( 1 ) 1 1 − 6σ 2 2t 2 2 4 t +O 2 e e e e = 2πt 1+ 12it t       2 √ s− 1 iπs − iπ −it 1 1 iσ 1 − 6σ 2 2 4 +O 2 +O 2 = 2πt e e e 1− 1+ 2t t 12it t    √ s− 1 iπs − iπ −it 1 ic(σ) = 2πt 2 e 2 e 4 e +O 2 1+ , 0 ≤ σ ≤ 1, t → ∞, t t

where c(σ) is defined by 1 σ (1 − σ) − . 2 12 Replacing σ by 1 − σ in (A.1) and taking the complex conjugate of the resulting equation, we find    √ −iπ(1−s) 1 iπ 1 ic(σ) +O 2 e 4 eit 1 − , Γ(1 − s) = 2πt 2 −s e 2 t t (A.2) 0 ≤ σ ≤ 1, t → ∞. c(σ) =

On the other hand, the definition (2.2) of χ(s) implies (2π)s χ(1 − s) =

 iπ   iπs iπ iπs (2π)s (2π)1−s Γ(s) e 2 (1−s) − e− 2 (1−s) = e− 2 + e 2 Γ(s). 2i π 107

A. THE ASYMPTOTICS OF Γ(1 − s) AND χ(s)

108

Replacing in this equation Γ(s) by the rhs of (A.1) we find (A.3) χ(1 − s) = (2π)

1 2 −s

t

s− 12 − iπ −it 4

e

e

   1 ic(σ) +O 2 1+ , t t

0 ≤ σ ≤ 1,

t → ∞.

Equation (A.3) together with the identity (2.13) imply the following asymptotic expression for χ(s):    1 1 iπ 1 ic(σ) , 0 ≤ σ ≤ 1, t → ∞. χ(s) = (2π)s− 2 t 2 −s e 4 eit 1 − +O 2 t t It is of course straightforward to extend the above asymptotic formulae to higher order.

APPENDIX B

Numerical Verifications B.1. Verification of Theorem 3.1 Letting σ = 1/2, the error term in equation (3.2) is given by t = 10 −(10.4 + 5.22i) × 10−5 −(4.00 + 4.19i) × 10−3

2

η=t 3 η = t2

t = 102 (−10.2 + 2.97i) × 10−9 −(1.40 + 1.11i) × 10−5

t = 103 (15.1 − 4.46i) × 10−13 −(7.80 + 9.81i) × 10−8

Note that the error is proportional to t3 /η 3+σ as expected. In order to demonstrate the effect of the higher order terms in (3.2), we also consider the difference between ζ(s) and the first term on the rhs of (3.2), i.e., the difference η [ 2π ] n−s . ζ(s) − n=1

For σ = 1/2, this difference is given by 2

η=t 3 η = t2

t = 10 t = 102 t = 103 −0.291 + 0.274i −0.341 + 0.207i −0.380 + 0.121i 0.266 + 0.0471i 0.127 + 0.020i 0.0360 + 0.0612i

Similarly, the difference between ζ(s) and the first two terms on the rhs of (3.2), i.e., the difference [ η ]  2π  1  η 1−s −s ζ(s) − n − , 1 − s 2π n=1 is given by 2

η=t 3 η = t2

t = 10 −(8.27 + 6.52i) × 10−2 (1.58 − 1.50i) × 10−1

t = 102 −(6.95 + 10.3i) × 10−4 (9.37 − 26.0i) × 10−3

t = 103 −(3.40 + 10.6i) × 10−4 (−4.93 + 3.32i) × 10−3

The above example illustrates that the inclusion of the higher order terms in (3.2) improves the convergence of the asymptotic series considerably. B.2. Verification of Theorem 3.2 The error term in equation (3.22) is given by σ=0 σ = 1/2 σ=1

t = 10 (1.97 − 3.81i) × 10−3 (2.23 − 4.34i) × 10−3 (2.42 − 4.78i) × 10−3

t = 102 (−7.76 + 65.1i) × 10−7 (−2.74 + 23.5i) × 10−7 (−9.41 + 82.5i) × 10−8

t = 103 (5.62 − 3.40i) × 10−9 (6.46 − 3.82i) × 10−10 (7.13 − 4.22i) × 10−11

Note that the error is proportional to t−σ−3 as expected. 109

110

B. NUMERICAL VERIFICATIONS

B.3. Verification of Theorem 4.1 Letting σ = 1/2, the error term in equation (4.2), i.e., the term    1 1 < η < t 3 < ∞, O ηt , −iπs − πt σ−1 2 e Γ(1 − s)e η × √  − At 4 1 O e η2 + ηt2 , t 3 < η < t < ∞, is given by t = 102 (3.04 + 7.27i) × 10−3 (45.4 − 6.75i) × 10−5 7.27 − 1.45i) × 10−1

η = 10 η = t1/4 η = t5/12

t = 104 (8.05 − 2.53i) × 10−6 (8.05 − 2.53i) × 10−6 (−8.69 + 9.53i) × 10−5

t = 106 (84.1 − 5.12i) × 10−10 (−443.6 + 6.91i) × 10−7 (4.38 − 13.7i) × 10−6

In order to illustrate the increased accuracy achieved by including the higher order terms in (4.2), we also display the corresponding table when only the first two sums on the rhs of (4.2) are included: For σ = 1/2, the difference1 [t]

η

[ 2π ] η   1 1 ζ(s) − − χ(s) s n n1−s n=1 n=1

is given by t = 102 (5.55 − 14.1i) × 10−2 (4.76 − 7.53i) × 10−2 −(2.14 + 2.15i) × 10−1

η = 10 η = t1/4 η = t5/12

t = 104 (−14.4 + 7.92i) × 10−3 (−14.4 + 7.92i) × 10−3 (3.09 − 1.91i) × 10−2

t = 106 (−15.9 + 4.40i) × 10−4 −(26.3 + 6.92i) × 10−3 (−5.70 + 10.4i) × 10−3

B.4. Verification of Corollary 4.3 Letting σ = 1/2 and N = 2, we find that the error term in equation (4.24) is given by 7 12

η=t 3 η = t4 η = 10t

t = 10 −(4.08 + 2.47i) × 10−1 (6.92 + 55.7i) × 10−2 (1.40 − 1.12i) × 10−2

t = 102 (−2.39 + 3.50i) × 10−1 (9.93 + 25.7i) × 10−2 (2.24 + 5.28i) × 10−4

t = 103 (−6.42 + 15.9i) × 10−2 −(8.75 + 12.2i) × 10−3 (2.45 + 186.1i) × 10−7

B.5. Verification of Theorem 4.4 Letting σ = 1/2, the error term in equation (4.31), i.e., the term    1 1 < η < t 3 < ∞, O ηt , −iπs − πt σ−1 Γ(1 − s)e 2 η × e √  − At 4 1 O e η2 + ηt2 , t 3 < η < t < ∞, is given by  η=

2πt √ 100

η = 2πt √ η = 200πt

t = 102

t = 104

(−5.96 + 8.83i) ×

10−4

10−3

(3.59 − 10.8i) × (13.7 − 5.21i) × 10−6

(6.44 + 1.30i) ×

t = 106 10−4

(4.64 + 69.7i) × 10−7

10−5

(104.6 + 8.97i) × 10−7 −(14.2 + 6.74i) × 10−7

(2.72 − 28.1i) × (−9.02 + 2.27i) × 10−4

1 This difference is exactly the error in the “approximate functional equation” of Hardy and Littlewood cf. [12].

B.5. VERIFICATION OF THEOREM 4.4

111

√ Remark B.1. In the particular case of η = 2πt, the formula of Siegel (Eq. (32) of [24]) and the corresponding formula of Titchmarsh (Theorem 4.16 of [26]) are equivalent to the formula of our theorem 4.4. This equivalence can easily be checked numerically—for example, for N = 3 all three formulas yield the same numerical values for the error terms. In this regard, we note that Titchmarsh states Siegel’s formula in terms of the function Ψ(a) defined by 2  iw2 a2 5 cos π( a2 − a − 18 ) e 4π + aw e−iπ( 2 − 8 ) Ψ(a) = dw = , 2π ew − 1 cos πa L which is related to our function Φ(τ, u) defined in (4.29) by  1 −iπ( a22 − 58 ) e Ψ(a) = −iΦ −1, a − , a ∈ C. 2

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Memoirs of the American Mathematical Society

9 781470 450984

MEMO/275/1351

Number 1351 • January 2022

ISBN 978-1-4704-5098-4