Johan Andersson

Summation formulae and zeta functions

Department of Mathematics

Stockholm University

2006

Johan Andersson

Summation formulae and zeta functions

4

Doctoral Dissertation 2006Department of mathematicsStockholm UniversitySE-106 91 Stockholm

Typeset by LATEX2ec©2006 by Johan Anderssone-mail: johana@math.su.seISBN 91-7155-284-7Printed by US-AB, Stockholm, 2006.

ABSTRACT

In this thesis we develop the summation formula

∑ad−bc=1

c>0

f

(a bc d

)= “main terms”+

∑m,n 6=0

1π

∫ ∞

−∞

σ2ir(|m|)σ2ir(|n|)F (r;m,n)dr

|nm|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

∑m,n 6=0

ρj,k(m)ρj,k(n)F((

12 − k

)i;m,n

)+∞∑

j=1

∑m,n 6=0

ρj(m)ρj(n)F (κj ;m,n),

where F (r;m,n) is a certain integral transform of f , ρj(n) denote the Fouriercoefficients for the Maass wave forms, and ρj,k(n) denote Fourier coefficients ofholomorphic cusp forms of weight k. We then give some generalisations and ap-plications. We see how the Selberg trace formula and the Eichler-Selberg traceformula can be deduced. We prove some results on moments of the Hurwitz andLerch zeta-function, such as∫ 1

0

∣∣ζ∗( 12 + it, x

)∣∣2dx = log t + γ − log 2π + O(t−5/6

),

and ∫ 1

0

∫ 1

0

∣∣φ∗ (x, y, 12 + it

)∣∣4dxdy = 2 log2 t + O((log t)5/3

),

where ζ∗(s, x) and φ∗(x, y, s) are modified versions of the Hurwitz and Lerch zetafunctions that make the integrals convergent. We also prove some power sumresults. An example of an inequality we prove is that

√n ≤ inf

|zi|≥1max

v=1,...,n2

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤ √n + 1

if n + 1 is prime. We solve a problem posed by P. Erdos completely, and disprovesome conjectures of P. Turan and K. Ramachandra.

PREFACE

I have been interested in number theory ever since high school when I read clas-sics such as Hardy-Wright. My special interest in the zeta function started inthe summer of 1990 when I studied Aleksandar Ivic’s book on the Riemann zetafunction and read selected excerpts from “Reviews in number theory”. I rememberlong hours from that summer trying to prove the Lindelof hypothesis, and workingon the zero density estimates. My first paper on the Hurwitz zeta function camefrom that interest (as well as some ideas I had about Bernoulli polynomials in highschool). The spectral theory of automorphic forms, and its relationship to the Rie-mann zeta function has been an interest of mine since 1994, when I first visitedMatti Jutila in Turku, and he showed a remarkable paper of Yoichi Motohashi tome; “The Riemann zeta function and the non-euclidean Laplacian”. Not only didit contain some beautiful formulae. It also had interesting and daring speculationson what the situation should be like for higher power moments. The next year Ivisited a conference in Cardiff, where Motohashi gave a highly enjoyable talk onthe sixth power moment of the Riemann zeta function. One thing I remember -“All the spectral theory required is in Bump’s Springer lecture notes”. Anyway thesixth power moment might have turned out more difficult than originally thought,but it further sparked my interest, and I daringly dived into Motohashi’s inter-esting, but highly technical Acta paper. My first real break-through in this areacame in 1999, when I discovered a new summation formula for the full modulargroup. Even if I have decided not to include much related to Motohashi’s originaltheory (the fourth power moment), the papers 7-13 are highly influenced by hiswork. Although somewhat delayed, I am pleased to finally present my results inthis thesis.

Johan Andersson

ACKNOWLEDGMENTS

First I would like to thank the zeta function troika Aleksandar Ivic, Yoichi Mo-tohashi and Matti Jutila. Aleksandar Ivic for his valuable comments on earlyversions of this thesis, as well as his text book which introduced me to the zetafunction; Yoichi Motohashi for his work on the Riemann zeta function which hasbeen a lot of inspiration, and for inviting me to conferences in Kyoto and Ober-wolfach; Matti Jutila for first introducing me to Motohashi’s work, and invitingme to conferences in Turku.

Further I would like to thank Kohji Matsumoto and Masanori Katsurada, forbeing the first to express interest in my first work, and showing that someone outthere cares about zeta functions; Dennis Hejhal for his support when I visited himin Minnesota; Jan-Erik Roos for his enthusiasm during my first years as a graduatestudent; My friends at the math department (Anders Olofsson, Jan Snellman +others) for mathematical discussions and a lot of good times; My family: myparents, my sister and especially Winnie and Kevin for giving meaning to lifeoutside of the math department.

And finally I would like to thank my advisor Mikael Passare, for his neverending patience, and for always believing in me.

Johan Andersson

CONTENTS

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Part I - The power sum method 21

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

On some Power sum problems of Turan and Erdos . . . . . . . . . . . . . . 29

Disproof of some conjectures of P Turan . . . . . . . . . . . . . . . . . . . . 41

Disproof of some conjectures of K Ramachandra . . . . . . . . . . . . . . . 47

Part II - Moments of the Hurwitz and Lerch zeta functions 53

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

Mean value properties of the Hurwitz zeta function . . . . . . . . . . . . . . 61

On the fourth power moment of the Hurwitz zeta-function . . . . . . . . . . 67

On the fourth power moment of the Lerch zeta-function . . . . . . . . . . . 79

Part III - A new summation formula on the full modular group and Klooster-man sums 89

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

A note on some Kloosterman sum identities . . . . . . . . . . . . . . . . . . 99

A note on some Kloosterman sum identities II . . . . . . . . . . . . . . . . 107

A summation formula on the full modular group . . . . . . . . . . . . . . . 115

Contents 12

A summation formula over integer matrices . . . . . . . . . . . . . . . . . . 133

A summation formula over integer matrices II . . . . . . . . . . . . . . . . . 145

The summation formula on the modular group implies the Kuznetsov summa-tion formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

The Selberg and Eichler-Selberg trace formulae . . . . . . . . . . . . . . . . 169

INTRODUCTION

This thesis in analytic number theory consists of three parts and altogether thirteenpapers. Each of the parts has a separate introduction. This thesis is by no meanscomplete and is a result of a number compromises. There are additional papers Iwould have liked to include, some of which I refer to as forthcoming papers. Evenif they are mostly complete they would still need some more work to be included inthis thesis. However, even if they would have painted a fuller picture and providedfurther motivation of my main result, the current version does not depend uponthem and it is entirely self contained. Part I and parts of part II of this thesishave also been included in my licentiate thesis [3], and three of the papers havebeen published in mathematical journals ([1], [2] and [4]).

The Riemann zeta-function

The Riemann zeta-function1

ζ(s) =∞∑

n=1

n−s (Re(s) > 1) (0.1)

is perhaps the deepest function in all of mathematics. First of all it is arithmetic

ζ(s) =∏

p prime

11− p−s

, (Re(s) > 1) (0.2)

and has an Euler product. It is symmetric

ζ(s) = χ(s)ζ(1− s) χ(s) = 2sπs−1 sin(π

2s)Γ(1− s) (0.3)

and has a functional equation. Furthermore it is random. A theorem illustratingthis is the Voronin universality theorem (see e.g [8]). If f is a non-vanishinganalytic function on D for a compact subset D ⊂ s ∈ C : 1/2 < Re(s) < 1,there exists a sequence Tk = Tk(D, f) so that

limk→∞

maxs∈D

|ζ(s + iTk)− f(s)| = 0. (0.4)

1 Introduced by Euler [6] and studied by Riemann [10]. General references: Ivic [7] andTitchmarsh [12].

Introduction 14

The depth of the function comes from its simple definition and these three prop-erties. I particular The Euler product implies that the function contains all infor-mation about the primes, and therefore its study belongs in the realms of numbertheory. The most fundamental open problem about the Riemann zeta-function isthe Riemann hypothesis2 that says that ζ(s) = 0 implies that either Re(s) = 1/2 ors is an even negative integer. Of near equal importance are the weaker conjectures

1. The Lindelof hypothesis: |ζ(1/2 + it)| = O(|t|ε), ∀ε > 0.

2. The density hypothesis: N(σ, T ) = O(T 2(1−σ)+ε), ∀ε > 0,

where N(σ, T ) denote the number of zeroes ρ with real part Re(ρ) ≥ σ, and imagi-nary part −T ≤ Im(ρ) ≤ T . It is obvious that the Riemann hypothesis implies thedensity hypothesis and with some complex analysis, it is possible to show that theRiemann hypothesis implies the Lindelof hypothesis, and the Lindelof hypothesisimplies the density hypothesis. Many consequences of the Riemann hypothesis arein fact implied already by the density hypothesis.

Part I - The power sum method

In an attempt to prove the density hypothesis we were lead to study the Turanpower sum method. The method allows us to obtain lower bounds for

maxv=M(n),...,N(n)

∣∣∣∣∣1 +n∑

k=1

zvk

∣∣∣∣∣,and in the first papers on his method Turan stated some conjectures on thesebounds that would indeed imply the density hypothesis. However we prove thatTuran’s conjectures were in fact false. We also prove some other results in thefield, in particular solving a problem of Erdos. Although it does not strictly usethe power sum method, we have also included a disproof of a certain conjecture ofRamachandra. This conjecture would also have had implications on the Riemannzeta function if true. The Voronin universality theorem (0.4) is a crucial part ofthe disproof. All of the above together constitutes part I of our thesis.

Part II - Moments of the Hurwitz and Lerch zeta functions

A problem closely related to the Lindelof hypothesis is the moment problem, whichconsists in estimating

Zk(T ) =∫ T

0

∣∣ζ( 12 + it

)∣∣2kdt. (0.5)

2 See Riemann [10] or Bombieri [5]

Introduction 15

The general conjecture is that

Zk(T ) ∼ ckT logk2T (0.6)

for all integers k ≥ 1. It should here be mentioned that by the method of Bala-subramanian and Ramachandra (see [9]), the lower bound

Zk(T ) ≥ dkT logk2T

for some dk > 0 has been proved. It is also known that the Lindelof conjecture isequivalent to a weak form of the conjecture

Zk(T ) T 1+ε (0.7)

for all integers k ≥ 1, and not even the Riemann hypothesis seems to imply thestrong form of the conjecture (0.6), which has so far only been proved for k = 1, 2.

A way of trying to understand the Riemann zeta-function is to see how itrelates to other zeta functions, such as the closely related Hurwitz zeta-function

ζ(s, x) =∞∑

k=0

(k + x)−s, (Re(s) > 1)

and the Lerch zeta-function

φ(x, y, s) =∞∑

k=0

e(kx)(k + y)−s. (Re(s) > 1)

The Lindelof hypothesis is expected to hold also for the Hurwitz and Lerch zeta-functions. However the Riemann hypothesis is false. This is due to the fact thatwhile the Hurwitz and Lerch zeta-functions admit a functional equation, theydo not have an Euler product. In more generality the Riemann hypothesis isassumed for a class of zeta-functions with Euler product and functional equation(the Selberg class - see Selberg [11]). It is thus interesting to study what differsbetween the Riemann zeta function and the Hurwitz and Lerch zeta functions, inorder to better understand the arithmetic nature of the Riemann zeta function. Inparticular, for the Hurwitz zeta function it is of interest to study the correspondingmoments ∫ 1

0

∫ T

0

∣∣ζ∗( 12 + it, x

)∣∣2kdtdx, (0.8)∫ 1

0

∣∣ζ∗( 12 + it, x

)∣∣2kdx, (0.9)

Introduction 16

and ∫ T

0

∣∣ζ∗( 12 + it, x

)∣∣2kdt, (0.10)

where ζ∗(s, x) = ζ(s, x) − x−s. In Part II we show some new results on theseproblems.

Part III - A new summation formula on the full modular group andKloosterman sums

In the final part of our thesis we consider the following question. What is theappropriate generalization of the classical Poisson summation formula

∞∑n=−∞

f(n) =∞∑

n=−∞f(n)

to the case of the full modular group? There are several generalizations in theliterature. The pre-trace formula, or even the Selberg trace formula has beensuggested. The pre-trace formula allows us to get an expansion of SO(2) bi-invariant functions f on SL(2, R) in terms of modular forms and Eisenstein series.The central result in Part III of the thesis is a proof of such a formula withoutassuming SO(2) bi-invariance. We will obtain a formula of the following type:

∑ad−bc=1

c>0

f

(a bc d

)= “main terms”+

∑m,n 6=0

1π

∫ ∞

−∞

σ2ir(|m|)σ2ir(|n|)F (r;m,n)dr

|nm|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

∑m,n 6=0

ρj,k(m)ρj,k(n)F((

12 − k

)i;m,n

)+∞∑

j=1

∑m,n 6=0

ρj(m)ρj(n)F (κj ;m,n),

where F (r;m,n) is a certain integral transform of f , the ρj(n) denote the Fouriercoefficients for the Maass wave forms, and the ρj,k(n) denote Fourier coefficientsof holomorphic cusp forms of weight k. For an exact version of this formula werefer to Theorem 1 in “A summation formula on the full modular group”. Ourmethod of proof will use the Kloosterman sums

S(m,n; c) =∑

hh≡1 (mod c)0≤h,h<c

e

(mh + nh

c

),

the Bruhat decomposition, the classical Poisson summation formula and the Kuznetsovsummation formula. We also consider the case of integer matrices of determinant

Introduction 17

D, or in other words, the image of SL(2, Z) under Hecke operators TD. We provesome related identities about Kloosterman sums. We show that the Kuznetsovsummation formula as well as the Selberg trace formula and the Eichler-Selbergtrace formula can be proved as consequences. We also indicate how these resultsare closely connected to the explicit formula given by Motohashi for the fourthpower moment of the Riemann zeta function.

BIBLIOGRAPHY

[1] J. Andersson, Mean value properties of the Hurwitz zeta-function, Math.Scand. 71 (1992), no. 2, 295–300.

[2] , On some power sum problems of Turan and Erdos, Acta Math. Hun-gar. 70 (1996), no. 4, 305–316.

[3] , Power sum methods and zeta-functions, Licentiate thesis, StockholmUniversity, 1998.

[4] , Disproof of some conjectures of K. Ramachandra, Hardy-RamanujanJ. 22 (1999), 2–7.

[5] E. Bombieri, Problems of the millenium: The Riemann hypothesis,http://www.maths.ex.ac.uk/˜mwatkins/zeta/riemann.pdf.

[6] L. Euler, Introductio in analysin infinitorum, Bousquet, Lausanne, 1748.

[7] A. Ivic, The Riemann zeta-function, A Wiley-Interscience Publication, JohnWiley & Sons Inc., New York, 1985, The theory of the Riemann zeta-functionwith applications.

[8] A. Laurincikas, Limit theorems for the Riemann zeta-function, Kluwer Aca-demic Publishers Group, Dordrecht, 1996.

[9] K. Ramachandra, On the mean-value and omega-theorems for the Riemannzeta-function, Tata Institute of Fundamental Research Lectures on Mathe-matics and Physics, vol. 85, Published for the Tata Institute of FundamentalResearch, Bombay, 1995.

[10] B. Riemann, Uer die Anzahl der Primzahlen unter einer gegebenen Grosse,Monatsberichte der Berliner Akademie (November 1859), 671–680, also avail-able at http://www.emis.de/classics/Riemann/.

[11] A. Selberg, Old and new conjectures and results about a class of Dirichletseries, Proceedings of the Amalfi Conference on Analytic Number Theory(Maiori, 1989) (Salerno), Univ. Salerno, 1992, pp. 367–385.

Bibliography 19

[12] E. C. Titchmarsh, The theory of the Riemann zeta-function, second ed., TheClarendon Press Oxford University Press, New York, 1986, Edited and witha preface by D. R. Heath-Brown.

PART I - THE POWER SUM METHOD

INTRODUCTION

The density hypothesis and the power sum method

From the explicit formula

∞∑n=1

Λ(n)f(n) =∫ ∞

0

f(x)dx−∑

ζ(ρ)=0

∫ ∞

0

xρ−1f(x)dx, (I.11)

where Λ(n) denotes the Van-Mangholt function, it is easy to see that

∫ H

0

∣∣∣∣∣∞∑

n=1

Φ(n/T )Λ(n)n−1/2−it

∣∣∣∣∣2

dt ∫ 1

0

T 2σ−1+ε(N(σ, 2H) + 1)dσ (I.12)

for a positive test-function Φ ∈ C∞0 (]0,∞[) and log T log H, where N(σ, T )

denotes the number of zeroes of the Riemann zeta function with real part greateror equal to σ, and with imaginary part between −T and T . (Note: the Riemannhypothesis states that N(σ, T ) = 0, for all σ > 1/2.) Can we invert this formula?More precisely, can we find for some H = H(T ) so that log H ∼ log T and suchthat ∫ 1

1/2

T 2σ−1N(σ, T/2)dσ T ε

∫ T

0

∣∣∣∣∣∞∑

n=1

Φ(n/H)Λ(n)n−1/2−it

∣∣∣∣∣2

dt. (I.13)

If this were true we could use a standard mean value formula for Dirichlet-polynomials

∫ T

0

∣∣∣∣∣∣∑k≤T

akkit−1/2

∣∣∣∣∣∣2

dt T∑k<T

|ak|2/k +∑k<T

|ak|2 (I.14)

to obtain the estimate

N(σ, T ) = O(T 2(1−σ)+ε

). (I.15)

This result is called the density hypothesis and can be can be used as a substitutefor the Riemann zeta function in a lot of applications, such as local prime number

Introduction 24

theorems. Both the density hypothesis and the Riemann hypothesis imply

π(x + h)− π(x) ∼ h

log x, (xθ < h < x) (I.16)

for each θ > 1/2.Turan introduced a new method in 1941 (see [10]) in an attempt to prove

the density hypothesis with this idea. However he was never able to prove thedensity hypothesis. Instead he stated two conjectures that would imply the densityhypothesis. The problems with inverting (I.12) comes from the fact that there isno way of knowing that locally the sum over the zeroes in the critical strip do notcancel each other. If we knew that the zeroes in the critical strip (but not on thecritical line) were “isolated”, i.e. N(σ, T +1)−N(σ, T ) = o(log T ), for σ > 1/2, wecould have used Turan’s method. However this has been shown to be equivalentto the even stronger Lindelof hypothesis (see Titchmarsh [9], Chapter 13). In ourpaper “Disproof of some power sum problems of P. Turan” we show that the twoconjectures of Turan are false3. Furthermore it can be shown that it is a futileattempt to try to prove the density hypothesis just by these methods. One wouldneed at least something like

infλ∈Cn,λ1=0

maxx∈[1,C]

∣∣∣∣∣n∑

k=1

eλkx

∣∣∣∣∣ > e−o(n), (C > 1)

to be true, and we manage to prove that there is no such estimate.Even though the density hypothesis was not proved by these methods, the

Turan power sum theory allowed us to obtain non-trivial estimates (and the bestin its time) for the density of the zeroes (and a θ < 1) in (I.16) before moremodern zero-detecting methods were found. This could be done because (I.13)can be proved for positive constants c1, c2 and for each T and some H so thatc1 log T ≤ log H ≤ c2 log T . Even though the Turan theory currently does not giveas good estimates as the state-of the art zero-detecting methods, we suspect thatit will eventually lead to the same-strength estimates (neither better nor worse).

The power sum method

Turan power sum theory has over the years found a number of other applications.In his book [11] P. Turan has 15 chapters for the theory and 40(!) chapters ofapplications. Turan himself thought of his method as his most important contri-bution to mathematics. His book also contains two chapters of open problems. In

3 This result was presented at the Halberstam conference in analytic theory in Urbana Cham-paign, 1995 and a slightly different version of this paper also appeared in our Licentiate thesis[2].

Introduction 25

the paper “On some power sum problem of Turan and Erdos” we solve some ofthem4. One of the problems we solve completely is the following one of P. Erdos.

Problem. Does there exist for each integer 1 ≤ m ≤ n− 1 a c = c(m) so that

max1≤v≤c(m)n,v integer

|zv1 + · · ·+ zv

n|mink=1,...,n |zk|v

≥ m? (I.17)

We show that in fact one can use c(m) = max(1021m, [(2π2m)m/2]). Anotherproblem of P. Turan concerns that of estimating

inf|zi|≥1

maxv=1,...,n2

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣. (I.18)

We also show that if n + 1 is prime then√

n ≤ (I.18) ≤√

n + 1, and that wehave

√n ≤ (I.18) for all integers n.

We would like to state here a similar open problem of Erdos-Newman (notpower sum though) that Erdos did send to us in 1995.

Problem. Is it true that there is an absolute constant c > 0 such that for everychoice of ak = ±1, 0 ≤ k ≤ n one has

max|z|=1

∣∣∣∣∣n∑

k=0

akzk

∣∣∣∣∣ > (1 + c)√

n? (I.19)

For many similar open problems, see Montgomery [6]. None of the results inthis paper have any applications on zeta-functions. (However we use some resultsfrom that theory, e.g. the distribution of consecutive primes.) It should be notedthat the constants 23/84 and 65/42 in Proposition 2 can be improved to 0.2675and 1.535 thanks to improvements on the difference between consecutive primes(Harman-Baker [4]).

Omega results for the Riemann zeta-function in short intervals

As well as considering order estimates of the Riemann zeta-function there is in-terest in finding omega-estimates (that is estimates from below) in short intervals.The method used here is the one by Ramachandra-Balasubramian (see Ramachan-dra’s book [7]), which gives omega-estimates for the 2k-th moment:∫ T+H

T

∣∣ζ( 12 + it

)∣∣2kdt ≥ ck(log H)k2

, for H log log T. (I.20)

4 This paper has also been published as [1] and appeared in our licentiate thesis [2]

Introduction 26

In a series of papers Ramachandra and Balasubramian generalized these resultsto a wide class of Dirichlet series, called the Titchmarsh series. In a paper inAsterisque [8], Ramachandra states some conjectures which essentially would al-low us to localize these results even further. Ramachandra-Balasubramian wroteanother paper [5] which contained further conjectures. In our paper “Disproofof some conjectures of Ramachandra” we disprove these conjectures (2 differentdisproofs) as well as an even weaker conjecture given by Ramachandra in a privatecommunication5. An excerpt from this paper which essentially disprove all theseconjectures, is the following result.Result. Suppose that H, ε > 0, 0 < δ < 1/2. Then there exist an N ,a1 = 1,|ak| ≤ kδ−1 so that

maxt∈[0,H]

∣∣∣∣∣N∑

n=1

annit

∣∣∣∣∣ < ε

We finally remark that this introduction is also taken in a somewhat modified formfrom our Licentiate thesis [2].

5 This paper has been published as [3] and also appeared in our Licentiate thesis

BIBLIOGRAPHY

[1] J. Andersson, On some power sum problems of Turan and Erdos, Acta Math.Hungar. 70 (1996), no. 4, 305–316.

[2] , Power sum methods and zeta-functions, Licentiate thesis, StockholmUniversity, 1998.

[3] , Disproof of some conjectures of K. Ramachandra, Hardy-RamanujanJ. 22 (1999), 2–7.

[4] R. C. Baker and G. Harman, The difference between consecutive primes, Proc.London Math. Soc. (3) 72 (1996), no. 2, 261–280.

[5] R. Balasubramanian and K. Ramachandra, On Riemann zeta-function andallied questions. II, Hardy-Ramanujan J. 18 (1995), 10–22.

[6] H. L. Montgomery, Ten lectures on the interface between analytic numbertheory and harmonic analysis, CBMS Regional Conference Series in Math-ematics, vol. 84, Published for the Conference Board of the MathematicalSciences, Washington, DC, 1994.

[7] K. Ramachandra, On Riemann zeta-function and allied questions, Asterisque(1992), no. 209, 57–72, Journees Arithmetiques, 1991 (Geneva).

[8] , On the mean-value and omega-theorems for the Riemann zeta-function, Tata Institute of Fundamental Research Lectures on Mathematicsand Physics, vol. 85, Published for the Tata Institute of Fundamental Re-search, Bombay, 1995.

[9] E. C. Titchmarsh, The theory of the Riemann zeta-function, second ed., TheClarendon Press Oxford University Press, New York, 1986, Edited and witha preface by D. R. Heath-Brown.

[10] P. Turan, Uber die Verteilung der Primzahlen. I, Acta Univ. Szeged. Sect.Sci. Math. 10 (1941), 81–104.

[11] , On a new method of analysis and its applications, Pure and AppliedMathematics, John Wiley & Sons Inc., New York, 1984, With the assistance of

Bibliography 28

G. Halasz and J. Pintz, With a foreword by Vera T. Sos, A Wiley-IntersciencePublication.

On some Power Sum problems of Turan andErdos

Johan Andersson∗

1 Introduction

In this paper I will show some elementary inequalities, and e.g. solve a problemgiven by Paul Erdos. I will also consider some problems of Paul Turan. Lemma1 was given by Cassels [1], and its proof will follow his approach, although I willuse it in a more general setting. In this paper e(x) will denote e2πix, bxc willdenote the integral part of x, and x will denote the distance between x and itsnearest integer. Furthermore v will always be integer valued.

2 Main theorem

Lemma 1. (Cassels) If zk are complex numbers we have

maxv=1,...,2n+1

Re

(n∑

k=1

zvk

)≥ 0.

Proof. Let ak, k = 1, . . . , 2n be defined by

n∏k=1

(z − zk)(z − zk) =2n∑

k=0

a2n−kzk

and σv =∑n

k=1 zvk + zv

k, Clearly 2 Re(∑n

k=1 zvk) = σv. By the construction the a′ks

are real. Now suppose σv < 0 for v = 1, . . . , 2n + 1. We now apply the Newtonidentities

σv + a1σv−1 + · · · + av−1σ1 + vav = 0, v = 1, 2, . . . , 2n (1)

∗This paper has been published in Acta Math. Hungar. 70 (1996), no. 4, 305–316.

29

and successively get a1 > 0, a2 > 0, . . . , a2n > 0 and with the relation σ2n+1 +a1σ2n + · · · + a2nσ1 = 0 we get σ2n+1 > 0, and a contradiction, hencemaxv=1,...,2n+1 σv ≥ 0.

Theorem 1. If zk are complex numbers and 1 ≤ m ≤ n we have

maxv=1,...,2nm−m(m+1)+1

|∑n

k=1 zvk |√∑m

k=1 |zk|2v≥ 1.

Proof. Put sv =∑n

k=1 zvk , and sa,b,v =

∑bk=a zv

k . We get

|sv|2 = |s1,m,v + sm+1,n,v|2 = |s1,m,v|2 + |sm+1,n,v|2 + 2 Re(s1,m,vsm+1,n,v) =

=m∑

k=1

|zk|2v + |sm+1,n,v|2 + 2 Re

(s1,m,vsm+1,n,v +

∑1≤j<k≤m

zvj zk

v

)=

=m∑

k=1

|zk|2v + |sm+1,n,v|2 + 2 Re B(v),

where B(v) is a pure power sum with nm − m(m+1)2

elements. i. e.

B(v) =

nm−m(m+1)2∑

k=1

wvk. (2)

From Lemma 1 we have an integer v such that 1 ≤ v ≤ 2nm − m(m + 1) + 1,such that Re(B(v)) ≥ 0, and for that v we have |sv|2 =

∑mk=1 |zk|2v+ something

non-negative, and |sv| ≥√∑m

k=1 |zk|2v.

Corollary 1. If 1 ≤ m ≤ n and |zk| ≥ 1 then

maxv=1,...,2nm−m(m+1)+1

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥ √m.

Proof. This follows from the theorem and the inequality√√√√ m∑k=1

|zk|2v ≥√

m mink=1,...,n

|zk|v.

From the two extreme cases in corollary 1 (m = 1 and m = n) we get

30

Corollary 2. If |zk| ≥ 1 then

maxv=1,...,2n−1

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥ 1

which is the classic result by Cassels (see [1] or [8] section 3). We also get

Corollary 3. If |zk| ≥ 1 then

maxv=1,...,n2−n+1

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥ √n

which is a rather interesting new result, which we will use in the followingsections. Note that the only condition is that |zk| ≥ 1. Under the assumption|zk| = 1 similar results have been proved earlier by e.g. Cassels, Newman andSzalay who even proved

max1≤v≤bcnc

∣∣∣∣∣n∑

k=1

bkzvk

∣∣∣∣∣ ≥√√√√(1 − n − 1

bcnc

) n∑k=1

bk, (3)

when bk > 0 and c ≥ 1. (See Turan [8] page 80, or Leenmann-Tijdeman [5] for aweaker result.)

3 Problem 10 of Turan

In his classic book [8] Turan states a number of problems. One of them (Problem10, page 190) is the following:

If minj |zj| = 1 then what is

infzj

max1≤v≤n2,v integer

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣? (4)

We see that Corollary 3 clearly gives a lower estimate for this expression. Inthe following we will show that this estimate is rather strong, in the sense thatfor infinitely many integers, e.g. the primes the upper estimate is very close. Webegin with two lemmas which give upper estimates for (4).

Lemma 2. (H. Montgomery) If p is prime then there exists numbers z1, . . . , zp−1

such that |zk| = 1 and |∑p−1

k=1 zvk | ≤

√p for all integers 1 ≤ v ≤ p2 − p − 1.

Proof. Take zk = χ(k)e(k/p), χ a character mod p of order p− 1, see [8] page 83.See Queffelec [6] for a fuller account.

31

Lemma 3. If q is a prime power then there exists numbers z1, . . . , zq+1 such that|zk| = 1 and |

∑q+1k=1 zv

k | ≤√

q for all integers 1 ≤ v ≤ q2 + q.

Proof. (Here I essentially follow Fabrykowski [3].) Suppose q is a prime power.By the Singer theorem (see [7]), we can find numbers a1, . . . , aq+1, such that theq(q + 1) numbers ai − aj, i 6= j form all nonzero residues modulo q2 + q + 1.Consider zk = e(ak/(q

2 + q + 1)). We get |∑q+1

k=1 zvk |2 = q + 1 +

∑q+1k 6=j e((ak −

aj)v/(q2 + q + 1)) = q +∑q2+q

k=0 e(kv/(q2 + q + 1)) = q + 1−e(v)1−e(v/(q2+q+1))

= q for all

integers v such that 1 ≤ v ≤ q2 + q, and thus we even have |∑q+1

k=1 zvk | =

√q for

those v.

It is now an easy task to state and prove the following proposition.

Proposition 1.

For n+1 prime we have:√

n ≤ infzk∈C,|zk|≥1

maxv=1,...,n2−n+1

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤ (i)

≤ infzk∈C,|zk|≥1

maxv=1,...,n2+n−1

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤ √n + 1.

For n-1 prime power we have:√

n − 2 ≤ infzk∈C,|zk|≥1

maxv=1,...,n2−n

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤ (ii)

≤√

n − 1 ≤√

n ≤ infzk∈C,|zk|≥1

maxv=1,...,n2−n+1

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ .Proof. The inequality

√n − 2 ≤ inf

zk∈C,|zk|≥1max

v=1,...,n2−n

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣in (ii) follows from Corollary 1 with m = n − 2. The inequality

√n ≤ inf

zk∈C,|zk|≥1max

v=1,...,n2−n+1

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣in (i) and (ii) is exactly Corollary 3. The inequality

infzk∈C,|zk|≥1

maxv=1,...,n2+n−1

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤ √n + 1

32

in (i) follows from Lemma 2. The inequality

infzk∈C,|zk|≥1

maxv=1,...,n2−n

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤ √n − 1

in (ii) follows from Lemma 3.

From Proposition 1 (i) we immediately obtain the best interval estimate sofar for (4) and infinitely many integers.

Corollary 4. If n + 1 is a prime we have

√n ≤ inf

zk∈C,|zk|≥1max

v=1,...,n2

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤ √n + 1.

For arbitrary n we can’t use the methods of Corollary 4 which requires pri-mality. But since we know that the primes are rather densely distributed we canuse local prime density results from Iwaniec and Pintz (see [4]). First, howeverwe need a lemma.

Lemma 4. (Erdos and Renyi) There exists numbers zk for k = 1, . . . , n suchthat |zk| = 1 and maxv=1,...,m |

∑nk=1 zv

k | ≤√

6n log(m + 1).

Proof. See Erdos and Renyi [2].

Similar results with explicit zk’s have been proved by Leenmann-Tijdemann(see [5] or [8] page 82).

Proposition 2.

infzk∈C,|zk|≥1

maxv=1,...,bn2−n65/42c

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ =√

n + O(n23/84

√log n

)Proof. We know that there exists a prime p between n−1 and n− 1

2n23/42,∀n ≥ N0

by Iwaniec-Pintz’s theorem (see [4]). Choose z1, . . . , zp+1 fulfilling the conditionin Lemma 3, and zp+2, . . . , zn fulfilling the condition in Lemma 4. We get for1 ≤ v ≤ p2 + p.

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤∣∣∣∣∣p+1∑k=1

zvk

∣∣∣∣∣+∣∣∣∣∣

n∑k=p+2

zvk

∣∣∣∣∣ ≤≤ √

p +√

6(n − p − 1) log(p2 + p + 1) =

=√

p + O(n23/84√

log n)

33

It is enough since bn2 − n65/42c ≤ (n − 12n23/42)2 ≤ p2 + p.

In the other direction we put m = n−2bn65/84c for n ≥ N0, and use Corollary1.

maxv=1,...,2nm−m(m+1)+1

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥ √m =

√n − 2bn65/84c =

√n + O(n23/84)

We also get 2nm − m(m + 1) + 1 = 2n(n − 2bn65/84c) − (n − 2bn65/84c)(n −2bn65/84c + 1) + 1 = n2 − 4bn65/84c2 − n + 2bn65/84c + 1 ≤ n2 − bn65/42c.

Queffelec [6] has showed the ≤ part of the proposition with essentially thesame method. See also Queffelec [6] for an application of these estimates infunctional analysis. Note that we still have not considered (4) for all n’s. In thenext section we will do that (and obtain much worse results than e.g. Corollary4 or Proposition 2), but we will also generalize results in the previous sections.

4 A further result

With Lemma 4 of Erdos and Renyi in mind we might consider the followinggeneralization of (4)

If minj |zj| = 1 then what is

infzj

maxv=1,...,n2m,v integer

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣? (5)

For general integers m first I prove an analogue of Corollary 3:

Theorem 2. If |zk| ≥ 1 then

maxv=1,...,n2m

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥((

n

m

)m!2) 1

2m

.

Proof. ∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣2m

=

(n∑

k=1

zvk

)m( n∑k=1

zvk

)m

= A(v) + B(v) + B(v)

Where A(v) is a power sum consisting of only positive real elements. SinceA(v) + B(v) + B(v) consists of exactly n2m elements and A(v) is non-emptywe must have that B(v) have less than 1

2(n2m − 1) elements. We now find a

lower estimate of the number of terms of A(v). It is clear that for each m-tuple

34

k1, . . . , km of distinct integers ki we have m!2 number of terms contributing toA(v), since each permutation on both sides occur. By elementary combinatoricswe know that we can pick

(nm

)such sets of integers from 1, . . . , n. and since

|zk| ≥ 1 we have A(v) ≥(

nm

)m!2,∀v. We now apply Lemma 1 on B(v) and we

obtain the theorem.

We now consider a well known version of the Stirling formula, which we willuse

n! ≥√

2πnn+ 12 e−n (6)

see e.g. Abramowitz-Stegun, Handbook of mathematical functions, formula 6.1.38.

Proposition 3. If |zk| ≥ 1, and 1 ≤ m ≤ n then

maxv=1,...,n2m

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥√(n −m)me−1

Proof. By Theorem 2 we have

maxv=1,...,n2m

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥≥((

n

m

)m!2) 1

2m

= (n(n − 1) · · · (n −m + 1)m!)1

2m ≥

≥ ((n −m)mm!)1

2m =√

n −m(m!)1

2m ≥

≥ (According to (6)) ≥√

(n −m)me−1(√

2πm)1

2m ≥√

(n −m)me−1.

We now have the means to obtain an interval estimate for (5)

Corollary 5. We have for 1 ≤ m ≤ n√e−1(1 − m

n) ≤ 1√

mninf

zk∈C,|zk|≥1max

v=1,...,n2m

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤√12 log(n)

(1 +

1

n

)Proof. The first inequality is exactly Proposition 3. The second inequality is seenby Lemma 4 and the fact that√

log(n2m + 1) =

√2m log n + log(1 +

1

n2m) ≤√

2m log n +1

n2m≤√

2m log n(1 +1

n)

35

We note that the quotient between the upper and lower estimate is essentiallyindependent of m for small m’s.

5 A problem of Erdos

In this section I will use Corollary 1, Proposition 3 and a classic result by Dirichletin Diophantine approximation to solve the following problem of Erdos, given inTuran [8] as problem 47 on page 196.

Does there exist for each integer 1 ≤ m ≤ n − 1 a c = c(m) so that

max1≤v≤c(m)n,v integer

|zv1 + · · · + zv

n|mink=1,...,n |zk|v

≥ m? (7)

I will start by proving an essential lemma.

Lemma 5. If |zk| ≥ 1, and n ≥ m + 1 then

maxv=1,...,b

“2π2nn−m

”n/2c

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥ m

Proof. We use the following version of Dirichlet’s theorem:

Given real numbers a1, a2, . . . , an, and an A ≥ 2 then we can find

an integer k in the range 1 ≤ k ≤ An such that kai ≤1

A. (8)

See e.g. Turan [8] section 15.

Now put A =√

2π2nn−m

and let zk = |zk|e(θk). By (8) choose a v, 1 ≤ v ≤ An

such that vθk ≤ 1A. We get∣∣∣∣∣

n∑k=1

zvk

∣∣∣∣∣ ≥ Re

(n∑

k=1

zvk

)= Re

(n∑

k=1

e(vθk)|zk|v)

=

=n∑

k=1

cos(2πvθk)|zk|v ≥ (By the relation 1 − x2

2≤ cos x) ≥

≥n∑

k=1

(1 − 1

2(2πvθk)2

)|zk|v ≥

≥n∑

k=1

(1 − 1

2

(2π

A

)2)|zk|v ≥ min

k=1,...,n|zk|vn

(1 − n −m

n

)≥ m.

36

We now state the main result in this section.

Proposition 4. If |zk| ≥ 1, we have for n ≥ m + 1 that

maxv=1,...,bmax(1021m,(2π2m)m/2)nc

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥ m

Proof. We will show this result in three cases.Case 1: m + 1 ≤ n ≤ 32m

Proof. First we put f(x) = ( 2π2xx−m

)x/2. From Lemma 5 we obtain that h(m) =1m

maxn=m+1,...,32m f(n) clearly is sufficient as c(m) in case 1. We get h(m) ≤1m

maxx∈[m+1,32m] f(x). We now notice that f(x)’s only critical point for x >m is a minimum point, hence the maximum must be obtained in the intervalend-points, and we get h(m) ≤ 1

mmax(f(m + 1), f(32m)). Since 1

mf(32m) =

1m

((32/31)2π2)16m ≤ 1021m, and

1

mf(m + 1) =

1

m(2π2(m + 1))

m+12 = (2π2m)m/2 (1 + 1

m)m/2

√2π2(m + 1)

m≤

≤ (2π2m)m/2

, for e.g. m > 1000

Since clearly 1021m is the dominating term for m < 1000 this is enough. Wehave now proved the proposition in Case 1.

Case 2: 32m ≤ n ≤ m2

Proof. We see that it only makes sense for m ≥ 32. Then we have by Proposition3 that

maxv=1,...,n2bm/8c

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥√(n − bm/8c)bm/8ce−1 ≥

≥√

31m3

4

1

8me−1 ≥ m

√331

32e−1 ≥ m.

Since n2bm/8c ≤ mm2 ≤ (2π2m)m/2 when n is in the above interval, we have proved

the proposition in case 2.

Case 3: m2 ≤ n

Proof. By Corollary 1 we get

maxv=1,...,2nm2

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≥ m

Which clearly is sufficient.

37

Since we have proved all three cases we have proved the proposition.

We see that Proposition 4 solves the problem (7), of Erdos affirmative withc(m) = max(1021m, (2π2m)m/2) We also note that we do not really need the pow-erful results of section 4. By using either Lemma 1, Corollary 1 and Dirichlet’stheorem or just Corollary 1 and Dirichlet’s theorem we can solve the problem,although the results one gets will be asymptotically worse than the ones provedabove(they will be of the order (C1m)5m/2, respectively Cm2

2 for constants C1 andC2). Note that for large m, 1021m < (2π2m)m/2. hence it is the term (2π2m)m/2

that will be asymptotically significant. It seems that this is essentially the asymp-totically best possible estimate obtainable with these methods, although for smallm it is easy to obtain better estimates.

6 On Turan’s problem 17 and 18

I will finally present a short solution to two further power sum problems fromTuran’s book [8]. Problem 17 page 191 is the following:

Show that for arbitrarily small ε > 0, there is an m0(ε, n), such that for everyinteger m > m0(ε, n) there exists a system (z∗1 , . . . , z

∗n) with maxj |z∗j | = 1 for

which the inequality

maxm+1≤v≤m+n,v integer

∣∣∣∣∣n∑

k=1

z∗vk

∣∣∣∣∣ ≤ εn (9)

holds. In problem 18 he asks if problem 17 also holds if maxj |z∗j | = 1 is replacedby minj |z∗j | = 1. I will now show a proposition which implies that there in factexists an m0(ε, n) such that the statements in both problems are true.

Proposition 5. For each m ≥ 4πn2ε−n there exists complex numbers zk , with|zk| = 1 such that

maxm+1≤v≤m+n,v integer

∣∣∣∣∣n∑

k=1

zvk

∣∣∣∣∣ ≤ εn

38

Proof. Let zk = e2πi kmn . For m + 1 ≤ v ≤ m + n we get∣∣∣∣∣

n∑k=1

zvk

∣∣∣∣∣ ≤∣∣∣∣∣

n∑k=1

zvk − e2πi k

n

∣∣∣∣∣+∣∣∣∣∣

n∑k=1

e2πi kn

∣∣∣∣∣ =

=

∣∣∣∣∣n∑

k=1

(e2πi v

mkn − e2πi k

n

)∣∣∣∣∣ ≤n∑

k=1

∣∣∣1 − e2πi( vm−1) k

n

∣∣∣ ≤≤ (By the relation |eix − 1| ≤ 2x) ≤ n

∣∣∣∣4π(m + n

m− 1

)∣∣∣∣ ≤ 4πn2

m≤ εn

References

[1] J. W. S. Cassels, On the sums of powers of complex numbers, Acta Math.Acad. Sci. Hungar. 7 (1956), 283–289.

[2] P. Erdos and A. Renyi, A probabilistic approach to problems of Diophantineapproximation, Illinois J. Math. 1 (1957), 303–315.

[3] J. Fabrykowski, A note on sums of powers of complex numbers, Acta Math.Hungar. 62 (1993), no. 3-4, 209–210.

[4] H. Iwaniec and J. Pintz, Primes in short intervals, Monatsh. Math. 98 (1984),no. 2, 115–143.

[5] H. Leenman and R. Tijdeman, Bounds for the maximum modulus of the firstk power sums, Nederl. Akad. Wetensch. Proc. Ser. A 77=Indag. Math. 36(1974), 387–391.

[6] H. Queffelec, Sur un theoreme de Gluskin-Meyer-Pajor, C. R. Acad. Sci. ParisSer. I Math. 317 (1993), no. 2, 155–158.

[7] J. Singer, A theorem in finite projective geometry and some applications tonumber theory, Trans. Amer. Math. Soc. 43 (1938), 377–385.

[8] P. Turan, On a new method of analysis and its applications, Pure and AppliedMathematics, John Wiley & Sons Inc., New York, 1984, With the assistanceof G. Halasz and J. Pintz.

39

Disproof of some conjectures of P. Turan

Johan Andersson∗

1 Introduction

In his first paper on power sum theory [4], P. Turan stated a power sum conjecturewhich implies the density hypothesis of the Riemann zeta-function. In this paperwe will show that a considerably weaker statement is still false. That is if

Claim. For some C > 1 one has that

infλ∈Cn,λ1=0

maxx∈[1,C]

∣∣∣∣∣n∑

k=1

eλkx

∣∣∣∣∣ > e−o(n). (1)

Then

Theorem. Claim is false.

2 Some conjectures

We first state the two original conjectures of Turan [4]. It should be noted thatHalasz in a note in [6], (page 2083) wrote that the conjectures were probablyfalse, but that no disproof had been found.

Conjecture 1. If ω(n) is a positive increasing function with limn→∞ ω(n) = ∞and for some n ≥ C1 there exists a M = M(n) such that the inequality

nω(n) ≤ M(n) ≤ n2

ω(n)2,

then

maxM(1− 1

ω(n))≤v≤M

v integer

|zv1 + · · ·+ zv

n| > e−n2

Mω(n)

∗This result was first presented at the Halberstam conference in analytic number theory inUrbana-Champaign, 1995.

41

for all systems (z1, . . . , zn) satisfying the conditions

z1 = 1, and 1− n2

2M2≤ |zv| ≤ 1. (v = 2, 3, . . . , n) (2)

Conjecture 2. For n ≥ c0, z1 = 1 and |z2| ≤ 1, . . . , |zn| ≤ 1 one has that

maxn3/2(1−n−0.42)≤v≤n3/2

|zv1 + · · ·+ zv

k | > e−n0.09

.

In [4] Turan shows that Conjecture 2 implies the density hypothesis for theRiemann zeta-function. If we use a modification of his method it is possibleto show that the conjectures can be weakened. However we would still needsomething like the existence of a C > 1 so that Claim is true to prove the densityhypothesis by this approach. We have

Proposition 1. Conjecture 1 and Conjecture 2 are false.

Proof. The falsity of the conjectures follows from Theorem and the fact that bothconjectures implies Claim. To prove the first two implications, pick a λ ∈ Cn. Wecan assume that maxk Re(λk) = 0, since otherwise if j is such that maxk Re(λk) =Re(λj), then consider λk = λk − λj, with λ1 and λj interchanged.

1. Conjecture 1 =⇒ Claim.

Choose zk = eλk/(M(1− 1ω(n)

)). First divide zknk=1 into two disjoint sets

zkin′

i=1 and zmjn′′

j=1 such that zkin′

i=1 is exactly the set that fulfills (6).Now consider z ∈ Cn, where zki

= zki, and zmi

= 1 − n2/(2M2). Then fore.g. ω(n) (log n)2 we have

e−o(n) e−n/ω(n)2 ,

e−n2/(Mω(n)) − 2n(1− n2/(2M2))M ,

maxM(1− 1

ω(n))≤v≤M

v integer

|zv1 + · · ·+ zv

n| − |zv1 − zv

1 + · · ·+ zvn − zv

n|,

≤ maxx∈[1,1/(1−1/ω(n))]

∣∣∣∣∣n∑

k=1

eλkx

∣∣∣∣∣, max

x∈[1,C]

∣∣∣∣∣n∑

k=1

eλkx

∣∣∣∣∣,for some C > 0.

42

2. Conjecture 2 =⇒ Claim.

Choose zk = eλk/(n3/2(1−n−0.42)). We obtain

e−o(n) e−n0.09

,

maxn3/2(1−n−0.42)≤v≤n3/2

|zv1 + · · ·+ zv

k |,

≤ maxx∈[1,1/(1−n−0.42)]

∣∣∣∣∣n∑

k=1

eλkx

∣∣∣∣∣, max

x∈[1,C]

∣∣∣∣∣n∑

k=1

eλkx

∣∣∣∣∣,for some C > 0.

3 Proof of theorem

First we remark that in the next section we will only use the principal part ofthe logarithm, and αβ will mean eβ log α. We have the following lemma

Lemma. One has for α and β > 1 real numbers that

∞∑k=−∞

(1 + αki)−β =2π

αΓ(β)

∞∑k=1

(2πk

α

)β−1

e−2πk/α.

Proof. By using the Poisson summation formula

∞∑k=−∞

∫ ∞

−∞f(x)e−ikxdx = 2π

∞∑k=−∞

f(2πk)

on

f(x) =

xβ−1e−x/α, when x > 0,

0, when x ≤ 0;

we get for β > 1 the identity

∞∑k=−∞

Γ(β)(1/α + ik)−β = 2π∞∑

k=1

(2πk)β−1e−2πk/α.

By dividing both sides with Γ(β)αβ we obtain the lemma.

43

Note that the Lemma is a special case of the functional equation for theLerch zeta-function. In fact a proof for the functional equation for the Lerchzeta-function along these lines has been given by Oberhettinger [3].

Proposition 2. One has that if 0 < a ≤ b < 2π, then

n∑k=−n

(1 +

ik

n

)−nx

e−Cn

for each x ∈ [a, b], and some C = C(a, b) > 0.

Proof. By using β = nx, and α = 1n

in the Lemma we obtain

n∑k=−n

(1 +

ki

n

)−nx

= A + B, (3)

where

A =∑|k|>n

(1 +

ki

n

)−nx

,

and

B =2πn

Γ(nx)

∞∑k=0

(2πkn)nx−1e−2πkn.

It is easy to see that

A n2−na/2 (4)

for x ≥ a. The dominating term will be the first, which will have absolute value2−x/2. To estimate B we use the Stirling formula

Γ(λ) ∼√

2πλλ+1/2e−λ.

We get

B n(nx)−nx−1/2enx

∞∑k=0

(2πkn)nx−1e−2πkn

√

n/x∞∑

k=0

(2πk

xe1−2πk/x

)nx

√

n/x

(2π

xe1−2π/x

)−nx

√

n/b

(2π

be1−2πb

)−nb

,

(5)

44

for x ≤ b, since the terms tends to zero fast enough. From equations (3), (4) and(5), we see that we can choose

C(a, b) = min

(b

(log

(b

2π

)+ 1− 2π

b

),a log 2

2

)− ε, (6)

which is a constant > 0 for some ε > 0, since the function

f(x) = log

(2π

x

)+

2π

x− 1

has the unique maximum 0 at x = 2π.

Proof of Theorem. That Claim is false is a direct consequence of Proposition 2.Choose for λ ∈ C2n+1,

λ2k+1 = − 1

Clog

(1− ki

n

),

and

λ2k = − 1

Clog

(1 +

ki

n

).

4 Summary

The results of our Theorem shed new light on Turan’s achievement. The powersum result

maxm+1≤v≤m+n

v integer

∣∣∣∣∣n∑

k=1

bkzk

∣∣∣∣∣ ≥ 1.007

(n

4e(m + n)

)n

mink=1,...,n

|b1 + · · ·+ bk| (7)

for |z1| = 1 ≥ |z2| ≥ · · · ≥ |zn|. This result, which is a refined version of Turan’ssecond main theorem (see Turan [5](with constant 4e instead of 8e), or Kolesnik-Straus [1]), seems more powerful than ever when we consider the much morespecial case when bk = 1 and the maximum with respect to v is considered forv in the interval [m, n + m] instead of just the integers m, . . . , n + m, for e.g.n = m and we still obtain results of decreasing exponential order). It should alsobe noted that already Makai [2] showed that 4e is in general the best possibleconstant in (7). Even though the situation might seem disappointing, note thatconjectures 1 to 2 and the Claim deal with conditions on the maximum norm,maxk |zk| = 1. Claim should be true for the minimum norm (Re(λk) > 0) and itis quite likely that we have

45

Conjecture 3. If C > 1 one has that

infλ∈Cn,Re(λk)≥0

maxx∈[1,C]

∣∣∣∣∣n∑

k=1

eλkx

∣∣∣∣∣ > e−o(n). (8)

It is possible to show that the corresponding non-pure power sum problem(i.e. when bk = 1 do not need to be true), then the analogue of Conjecture 3 isfalse. However in the pure power sum case this seems likely (to us). It would not(to our knowledge) have any consequences to the theory of the Riemann zeta-function. Nevertheless it would be of interest to investigate it further. Anotherproblem is to see to what extent Claim fails. That is

Problem. Given C > 1. Find the smallest constant B = B(C) so that

infλ∈Cn,λ1=0

maxx∈[1,C]

∣∣∣∣∣n∑

k=1

eλkx

∣∣∣∣∣ e−(B+ε)n, (9)

for all ε > 0.

From (7) it is easy to see that B ≤ log(4eC/(C − 1)). We also see thatequation (6) can give us a lower bound for B.

References

[1] G. Kolesnik and E. G. Straus, On the sum of powers of complex numbers,Studies in pure mathematics - To the memory of Paul Turan, Birkhauser,Basel-Boston, Mass., 1983, pp. 427–442.

[2] E. Makai, On a minimum problem II, Acta. Math. Hung. 15 (1964), no. 1–2,63–66.

[3] F. Oberhettinger, Note on the Lerch zeta-function, Pacific J. Math. 6 (1956),117–120.

[4] P. Turan, Uber die Verteilung der Primzahlen (1), Acta. Sci. Math. Szeged10 (1941), 81–104.

[5] , On a new method of analysis and its applications, John Wiley &Sons, New York, 1984.

[6] , Collected papers of Paul Turan, Akademiai Kiado (Publishing Houseof the Hungarian Academy of Sciences), Budapest, 1990.

46

Disproof of some conjectures of K. Ramachandra

Johan Andersson∗

1 Introduction

In a recent paper [9] K. Ramachandra states some conjectures, and gives conse-quences in the theory of the Riemann zeta function. In this paper we will willpresent two different disproofs of them. The first will be an elementary appli-cation of the Szasz-Muntz theorem. The second will depend on a version of theVoronin universality theorem, and is also slightly stronger in the sense that itdisproves a weaker conjecture. An elementary (but more complicated) disproofhas been given by Rusza-Lazkovich [11].

2 Disproof of some conjectures

2.1 The conjectures

We will first state the three conjectures as given by Ramachandra [9], and Ramachandra-Balasubramian [5]:

Conjecture 1. For all N ≥ H ≥ 1000 and all N−tuples a1 = 1, a2, . . . , aN ofcomplex numbers we have

1

H

∫ H

0

∣∣∣∣∣N∑

n=1

annit

∣∣∣∣∣ dt ≥ 10−1000.

Conjecture 2. For all N ≥ H ≥ 1000 and all N−tuples a1 = 1, a2, . . . , aN ofcomplex numbers we have when M = H(log H)−2 that

1

H

∫ H

0

∣∣∣∣∣N∑

n=1

annit

∣∣∣∣∣2

dt ≥ (log H)−1000

M∑n=1

|an|2.

∗This paper has been published in Hardy-Ramanujan J. 22 (1999), 2–7.

47

Conjecture 3. There exist a constant c > 0 such that

∫ T

0

∣∣∣∣∣N∑

n=1

annit

∣∣∣∣∣2

dt ≥ c∑n≤cT

|an|2.

2.2 The Szasz-Muntz theorem

To disprove these conjectures, we first consider the following classic result of Szasz

Lemma 1. (Szasz) If we have that

∞∑n=1

1 + 2 Re(λn)

1 + |λn|2= +∞,

where Re(λn) ≥ 0 then the set of finite linear combinations of xλn is dense inL2(0, 1).

Proof. See Szasz [12], theorem A.

We will now state a theorem that will effectively disprove the above conjec-tures:

Theorem 1. For each D ≥ 0 and ε > 0 there exists an N ≥ 0 and complexnumbers a2, . . . , aN , such that

∫ D

0

∣∣∣∣∣1 +N∑

n=2

annit

∣∣∣∣∣2

dt ≤ ε.

Proof. Since −1 ∈ L2(0, 1) and

∞∑n=2

1 + 2 Re(−i log n)

1 + | − i log n|2= +∞ (1)

we have by the lemma that for each δ > 0 there exists an N > 0 and complexnumbers a2, . . . , aN such that

∫ 1

0

∣∣∣∣∣1 +N∑

n=2

anx−i log n

∣∣∣∣∣2

dx < δ.

48

We obtain

δ >

∫ 1

0

∣∣∣∣∣1 +N∑

n=2

anx−i log n

∣∣∣∣∣2

dx ≥∫ 1

e−D

∣∣∣∣∣1 +N∑

n=2

anx−i log n

∣∣∣∣∣2

dx =

= (Substituting t = − log x) =∫ D

0

e−t

∣∣∣∣∣1 +N∑

n=2

annit

∣∣∣∣∣2

dt ≥ e−D

∫ D

0

∣∣∣∣∣1 +N∑

n=2

annit

∣∣∣∣∣2

dt.

By choosing δ = e−Dε we obtain the theorem.

It is now an easy task to falsify the conjectures.

Proposition. Conjectures 1, 2 and 3 are false

Proof. For Conjecture 1, choose ε < 10−2000H and D = H in Theorem 1 andapply the Cauchy-Schwarz inequality

∫ H

0

∣∣∣∣∣N∑

n=1

annit

∣∣∣∣∣ dt ≤√

H

√√√√∫ H

0

∣∣∣∣∣N∑

n=1

annit

∣∣∣∣∣2

dt.

For Conjecture 2, choose ε < (log H)−1000H, and D = H in Theorem. To disproveConjecture 3, choose e.g. a1 = T = 1, and ε = c/2.

2.3 The Voronin theorem

In private correspondence, Ramachandra asked whether the conjectures hold un-der the additional growth assumption |ak| (Hk)100. Ramachandra and Bal-asubramian have proved conjectures 1 and 2 under the this and the additionalfurther assumption that N < exp(exp(cH)). This shows that N must be verylarge compared to H for the conjectures to be false. However, they are in factstill false, although their proof requires a deeper result. A version of the Voroninuniversality theorem for the Riemann zeta-function. We will state the theoremthat shows that the conjectures are still false below:

Theorem 2. Suppose that H, ε > 0, 0 < δ < 12. Then there exists |ak| ≤ kδ−1

such that

maxt∈[0,H]

∣∣∣∣∣1 +N∑

n=2

annit

∣∣∣∣∣ dt < ε.

The idea goes as follows: We use the following version of the Voronin univer-sality theorem

49

Theorem 3. (Voronin-Bagchi) For any compact subset K of the complex numberssuch that x ∈ K =⇒ 1

2< Re(x) < 1, non vanishing analytic function f on K

and ε > 0, we have a real t such that |ζ(z + it)− f(z)| ≤ ε, for all z ∈ K

Proof. See Bagchi [1].

and the following version of the approximate functional equation for the Rie-mann zeta-function

∣∣∣∣∣ζ(σ + it)−N∑

k=1

k−σ−it

∣∣∣∣∣ ≤ CNσ−1, for e.g. t < N < 2t and σ ≥ 1

2.

Proof. (of Theorem 2) Choose T0 so that when T > T0 then∣∣∣∣∣ζ(1− δ + it)−∑k<2T

kδ−1−it

∣∣∣∣∣ < ε

3

when T ≤ t ≤ T + H. Now choose T > T0 so that

|ζ(1− δ + it)| < ε

3

for T ≤ t ≤ T + H (This is possible by applying Theorem 3 to f(z) = ε/3, andK = [1− δ, 1− δ + iH]). By using the triangle inequality we get∣∣∣∣∣∑

k<2T

kδ−1−it

∣∣∣∣∣ < ε

for all T < t < T + H. Now we can choose N = T and an = nδ−1−iT in thetheorem.

3 Summary

In [7] K. Ramachandra stated a similar conjecture to conjectures 1 and 2. Al-though it was more general in the sense that it considered Dirichlet series of formA(s) = 1 +

∑anλ

−sn , for certain λn generalizing λn = n it was much weaker as it

had three further conditions. H depended on N , |an| were bounded from aboveand A was bounded in certain regions in the complex plane. Under these addi-tional assumptions, A(s) is called a Titschmarsh-series and for these, analoguesof conjectures 1 and 2, and similar conjectures were proved in Ramachandra [8], and Balasubramian and Ramachandra [4], [3] and [2]. It would certainly be

50

interesting to see if all these additional assumptions are needed, or if a certainsubset of them implies the truth of conjectures 1 and 2. For a disproof of theLp(0, D), p > 2 version of conjectures 1 and 2, or for considering much moresparse sequences than log n in (1), we need a stronger version of the Lemma,the reader is referred to the literature on the theory of completeness of complexexponentials, see e.g. the classic by Levinson [6] or Redheffer [10] for a morerecent survey. It should be noticed that problems similar to conjectures 1 and 2are also studied in Turan power sum theory (although there it is essentially theλn’s which vary, instead of the an’s), see Turan [13] for a thorough treatment.

References

[1] B. Bagchi, A joint universality theorem for dirichlet l-functions. (english),Math. Z. 181 (1982), no. 3, 319–334.

[2] R. Balasubramanian and K. Ramachandra, Proof of some conjectures on themean-value of Titchmarsh series. I, Hardy-Ramanujan J. 13 (1990), 1–20.

[3] , Proof of some conjectures on the mean-value of Titchmarsh series.II, Hardy-Ramanujan J. 14 (1991), 1–20.

[4] , Proof of some conjectures on the mean-value of Titchmarsh series.III, Proc. Indian Acad. Sci. Math. Sci. 102 (1992), no. 2, 83–91.

[5] , On Riemann zeta-function and allied questions. II, Hardy-Ramanujan J. 18 (1995), 10–22.

[6] N. Levinson, Gap and Density Theorems, American Mathematical Society,New York, 1940, American Mathematical Society Colloquium Publications,v. 26.

[7] K. Ramachandra, Progress towards a conjecture on the mean-value of Titch-marsh series-i, Recent progress in Analytic number theory (H. Halberstamand C. Hooley, eds.), vol. 1, Academic Press, London, New York, Toronto,Sydney, San Fransisco, 1981, pp. 303–318.

[8] , Proof of some conjectures on the mean-value of Titchmarsh se-ries with applications to Titchmarsh’s phenomenon, Hardy-Ramanujan J.13 (1990), 21–27.

[9] , On Riemann zeta-function and allied questions, Asterisque (1992),no. 209, 57–72, Journees Arithmetiques, 1991 (Geneva).

51

[10] R. M. Redheffer, Completeness of sets of complex exponentials, Advances inMath. 24 (1977), no. 1, 1–62.

[11] M. Rusza, I. Lazkovich, Sums of periodic functions and a problem of Ra-machandra, Tech. report, Mathematical institute of the Hungarian Academyof Sciences, 1996.

[12] 0. Szasz, Uber die Approximation stetiger Funktionen durch lineare Aggregatevon Potenzen, Math Ann. 77 (1916), 482–496.

[13] P. Turan, On a new method of analysis and its applications, Pure and Ap-plied Mathematics, John Wiley & Sons Inc., New York, 1984, With theassistance of G. Halasz and J. Pintz, With a foreword by Vera T. Sos, AWiley-Interscience Publication.

52

PART II - MOMENTS OF THE HURWITZ ANDLERCH ZETA FUNCTIONS

INTRODUCTION

It is conjectured that for the Riemann zeta-function one has

Zk(T ) =∫ T

0

∣∣ζ( 12 + it

)∣∣2kdt ∼ ckT logk2

T. (II.21)

This has so far only been proved for k = 1, 2, although the lower bound has beenproved for a general k ≥ 1 by the method of Balasubramanian and Ramachandra(see [5]). A way to understand the Riemann zeta-function better is to see how itrelates to other zeta functions, such as the Hurwitz zeta-function defined by

ζ(s, x) =∞∑

k=0

(k + x)−s,

and the Lerch zeta-function defined by

φ(x, y, s) =∞∑

k=0

e(kx)(k + y)−s,

for Re(s) > 1 and by analytic continuation elsewhere. We may thus ask: Whatcan be said about the moments∫ T

0

∣∣ζ( 12 + it, x

)∣∣2kdt, (II.22)

and ∫ T

0

∣∣φ(x, y, 12 + it

)∣∣2kdt? (II.23)

This problems turns out to be difficult, since the behavior will be different de-pending on whether x and y are rational or not. This is due to the fact that whilethe Hurwitz and Lerch zeta-functions admit a functional equation, they do nothave an Euler product, and they are in general non-arithmetic. For the specialcase of rational x and y, we can express them in terms of finite sums of DirichletL-functions, and they will inherit arithmetical properties.

Introduction 56

The mean square of the Hurwitz zeta function

As we have a parameter x in the Hurwitz zeta function we may consider momentswith respect to x instead of t. We will start with the mean square. In a paperfrom 1956 [4], M Mikolas used the functional equation (the Fourier-expansion)

ζ(s, x) =2Γ(1− s)(2π)1−s

(sin

πs

2

∞∑k=1

sin 2πkx

k1−s+ cos

πs

2

∞∑k=1

cos 2πkx

k1−s

), (II.24)

valid for Re(s) < 0 and x ∈ [0, 1], and the Parseval identity to prove the followingformula for the inner product of two Hurwitz zeta-functions:∫ 1

0

ζ(z, x)ζ(w, x)dx = 2(2π)z+w−2Γ(1− z)Γ(1− w) cos(π

2(z − w)

)ζ(2− z − w),

(II.25)

for Re(z),Re(w),Re(z+w) < 1. By then setting w = z he obtained a mean-squareformula for Re(z) < 1/2. However on the critical line Re(z) = 1/2, the integraldiverges. This lead Koksma and Lekkerkerker [3] to instead consider the integral∫ 1

0

∣∣ζ∗( 12 + it, x

)∣∣2dx, (II.26)

where ζ∗(s, x) = ζ(s, x)− x−s, and by using the approximate functional equation

ζ∗(s, x) =∑

k<2 Re(s)

(k + x)−s + O(1), (Re(s) > 0) (II.27)

for s = 12+it, squaring and using explicit estimates they obtained (II.26)=O(log T ).

When we wrote our paper “Mean value properties of the Hurwitz zeta function”the best result we knew of at the time was an O(1) error term. However WenPeng Zhang [6] had recently proved that (II.26) = log t + γ − log 2π + O

(t−θ)

forθ = 7/36 − ε. Furthermore he had conjectured that one can choose θ = 1/4. Inour paper we used analytic continuation of Mikolas formula as well as analyticcontinuation with respect z of ∫ 1

0

ζ∗(s, x)x−zdx, (II.28)

to obtain∫ 1

0

∣∣ζ∗( 12 + it, x

)∣∣2dx = γ − log 2π + Re

(Γ′(

12 + it

)Γ(

12 + it

) − 2∞∑

k=1

ζ(

12 + it + k

)− 1

12 + it + k

).

(II.29)

Introduction 57

This showed that we can could chose θ = 47/56, by the best order estimate ofthe Riemann zeta-function known at the time we wrote the paper. This especiallyproved Wen Peng Zhang’s conjecture. By an estimate of Huxley [1] this can beimproved to θ = 481/570. Of course the Lindelof hypothesis gives θ = 1 − ε,for each ε > 0. For a more detailed history of the problem as well as furtherdevelopments see e.g. Katsurada-Matsumoto [2].

The Fourth power moment

In analogy with the Riemann zeta function case it might seem natural to assumethat ∫ 1

0

∣∣ζ∗( 12 + it, x

)∣∣4dx ∼ c log4 t. (II.30)

In our paper “On the fourth power moment of the Hurwitz zeta function” we showthat a formula like (II.30) cannot be true. In fact the moment∫ 1

0

∣∣ζ∗( 12 + it, x

)∣∣4dx

will be rather chaotic in the same sense as the absolute values of the Riemann zetafunction in the critical strip. On one hand we will be able to prove that∫ 1

0

∣∣ζ∗( 12 + it, x

)∣∣4dx = Ω(

exp(

c(log t)1/4

log log t

)).

On the other hand we prove that its mean value

1T

∫ T

0

∫ 1

0

∣∣ζ∗( 12 + it, x

)∣∣4dxdt = 2 log2 T + O((log T )5/3

),

will grow just like a log power of t. We notice that the log power here is 2 and not4 as in the case of the Riemann zeta function. For rational x = p/q with (p, q) = 1we will have

1T

∫ T

0

∣∣∣∣ζ∗(12

+ it,p

q

)∣∣∣∣4dx ∼ cq log4 T

though. Similarly in our paper “On he fourth power moment of the Lerch zetafunction” we show the corresponding results

1T

∫ T

0

∫ 1

0

∣∣φ∗(x, y, 12 + it

)∣∣4dxdt = 2 log2 T + O((log T )5/3

), (0 ≤ y ≤ 1),

(II.31)

Introduction 58

and

1T

∫ T

0

∫ 1

0

∣∣φ∗(x, y, 12 + it

)∣∣4dydt = 2 log2 T + O((log T )5/3

). (0 ≤ x ≤ 1),

(II.32)

and indicate that for rational values of x, y

1T

∫ T

0

∣∣∣∣φ∗(p

q,h

k,12

+ it

)∣∣∣∣4dt ∼ ch,k,p,q log4 T. (II.33)

Here we use the definition

φ∗(x, y, s) = φ(x, y, s)− y−s − (2π)s−1Γ(1− s)×

×(

e

(1− s

4− yx

)xs−1 + e

(s− 1

4+ y(1− x)

)(1− x)s−1

), (II.34)

which is chosen so that φ does not tend to infinity when x, y → 0, 1. For the Lerchzeta function we are able to prove that∫ 1

0

∫ 1

0

∣∣φ∗ (x, y, 12 + it

)∣∣4dxdy = 2 log2 t + O((log t)5/3

),

which is a natural generalization to (II.29).

Higher power moments

In general we expect the following conjectures to be true

1T

∫ T

0

∣∣∣∣ζ∗(12

+ it,p

q

)∣∣∣∣2k

dx ∼ cp,q,k logk2T,

1T

∫ T

0

∣∣∣∣φ∗(p

q,m

n,12

+ it

)∣∣∣∣2k

dt ∼ cm,n,p,q,k logk2T,

1T

∫ T

0

∫ 1

0

∣∣ζ∗( 12 + it, x

)∣∣2kdxdt ∼ k! logk T,

1T

∫ T

0

∫ 1

0

∣∣φ∗(x, y, 12 + it

)∣∣2kdxdt ∼ k! logk T, (0 ≤ y ≤ 1),

1T

∫ T

0

∫ 1

0

∣∣φ∗(x, y, 12 + it

)∣∣2kdydt ∼ k! logk T, (0 ≤ x ≤ 1),

and ∫ 1

0

∫ 1

0

∣∣φ∗ (x, y, 12 + it

)∣∣2kdxdy ∼ k! logk t.

We will return to this topic in the forthcoming work.

BIBLIOGRAPHY

[1] M. N. Huxley, Exponential sums and the Riemann zeta function. IV, Proc.London Math. Soc. (3) 66 (1993), no. 1, 1–40.

[2] M. Katsurada and K. Matsumoto, Explicit formulas and asymptotic expansionsfor certain mean square of Hurwitz zeta-functions. I, Math. Scand. 78 (1996),no. 2, 161–177.

[3] J. F. Koksma and C. G. Lekkerkerker, A mean-value theorem for ζ(s, w), Ned-erl. Akad. Wetensch. Proc. Ser. A. 55 = Indagationes Math. 14 (1952), 446–452.

[4] M. Miklos, Transformation und orhogonalitat bei ζ(s, u). Verallgemeinungder Riemannschen Funktionalgleichung von ζ(s), Acta. Sci. Mat. Szeged. 17(1956), 143–164.

[5] K. Ramachandra, On the mean-value and omega-theorems for the Riemannzeta-function, Tata Institute of Fundamental Research Lectures on Mathe-matics and Physics, vol. 85, Published for the Tata Institute of FundamentalResearch, Bombay, 1995.

[6] W. P. Zhang, On the Hurwitz zeta-function, Illinois J. Math. 35 (1991), no. 4,569–576.

Mean value properties of the Hurwitzzeta-function

Johan Andersson∗

1 Introduction

The Lindelof hypothesis for the Hurwitz zeta-function states that ζ(1/2+ it, x) =O(tε) for each fixed x and ε > 0. When x = 1 we have the usual hypothesis. Thehypothesis is far from being proved but in 1952 Koksma and Lekkerkerker provedthe estimate ∫ 1

0

|ζ∗(12

+ it, x)|2dx = O(log t),

where ζ∗(s, x) = ζ(s, x + 1) and since then the result has been sharpened. Theresult is very similar to Lindelof’s hypothesis, but it states that the mean square issmall, not the function for fixed x. In this paper I will continue in Lekkerkerker’stradition and calculate some integrals of which∫ 1

0

|ζ∗(12

+ it, x)|2dx = log t + γ − log 2π + O(t−4756

+ε) (ε > 0)

is an improvement on former estimates. Especially I will obtain a better error-term depending on the Lindelof hypothesis. Thus the hypothesis implies a bettermean square formula.

2 Integrals involving the Hurwitz zeta-function

From Hurwitz formula for the Hurwitz zeta-function

ζ(s, x) =2Γ(1− s)

(2π)1−s

(sin

πs

2

∞∑k=1

sin 2πkx

k1−s+ cos

πs

2

∞∑k=1

cos 2πkx

k1−s

),

Re(s) < 0 ∧ x ∈ [0, 1] (1)

∗This paper has been published in Math. Scand. 71 (1992), no. 2, 295–300.

61

we see directly with Parseval’s identity (see Miklos Mikolas [5], [4],[6]):∫ 1

0

ζ(z, x)ζ(w, x)dx = (2)

2(2π)z+w−2Γ(1− z)Γ(1− w) cos(π

2(z − w))ζ(2− z − w)

max(Re(z), Re(w), Re(z + w)) < 1

The formula holds initially for max(Re(z), Re(w)) < 0, but I will show laterthat it holds for the extended region. We have that ζ(s, x) = ζ∗(s, x) + x−s, andζ∗(s, x) is continuous w.r.t. x for x ∈ [0, 1] and thus∫ 1

0

ζ∗(z, x)ζ∗(w, x)dx =

∫ 1

0

(ζ(z, x)− x−z)(ζ(w, x)− x−w)dx = (3)∫ 1

0

(ζ(z, x)ζ(w, x) + x−(z+w) − ζ(z, x)x−w − ζ(w, x)x−z)dx =∫ 1

0

(ζ(z, x)ζ(w, x)− x−(z+w) − x−wζ∗(z, x)− x−zζ∗(w, x))dx =∫ 1

0

(ζ(z, x)ζ(w, x)− x−(z+w))dx−∫ 1

0

(ζ∗(z, x)x−w + ζ∗(w, x)x−z)dx

max(Re(w), Re(z)) < 1

I will now deduce an expression for the last two integrals

∫ 1

0

ζ∗(z, x)x−wdx = (partial integration) =

(we use∂ζ∗

∂x(s, x) = −sζ∗(s + 1, x))[

ζ∗(z, x)x1−w

1− w

]1

0

+z

1− w

∫ 1

0

ζ∗(z + 1, x)x1−wdx =

ζ(z)− 1

1− w+

z

1− w

∫ 1

0

ζ∗(z + 1, x)x1−wdx

Consider

sn(z, w) =n∑

k=0

(z)k

(1− w)k+1

(ζ(z + k)− 1),

we see that

∫ 1

0

ζ∗(z, x)x−wdx = sn(z, w) + Rn

where Rn =(z)n+1

(1− w)n+1

∫ 1

0

ζ∗(z + n + 1, x)xn+1−wdx

62

after n + 1 partial integrations. We have∫ 1

0ζ∗(n + z)xn−wdx = O((1/2)n),

since ζ∗(z+n, x) ∼ (x+1)−z−n when n →∞. We also have limn→∞(1−w)n

(z)n

(z)n+1

(1−w)n+1=

1, hence Rn → 0, and sn(z, w) converges when n →∞, and

∫ 1

0

ζ∗(z, x)x−wdx =∞∑

k=0

(z)k

(1− w)k+1

(ζ(z + k)− 1) (4)

z 6∈ Z− ∪ 0, 1 ∧ Re(w) < 1

We combine formula (3) and (4) and get∫ 1

0

ζ∗(z, x)ζ∗(w, x)dx = (5)

2(2π)z+w−2Γ(1− z)Γ(1− w) cos(π

2(z − w))ζ(2− z − w)+

+1

1− z − w−

∞∑k=0

((z)k

(1− w)k+1

(ζ(z + k)− 1) +(w)k

(1− z)k+1

(ζ(w + k)− 1)

)z /∈ Z and w /∈ Z ∧ z + w 6= 1

I will now show (by analytic continuation) that the equality really holds in theregion stated. We know that the integral in (5) is analytic w.r.t. z and w for allz, w /∈ Z (Since the function under the integral-sign is analytic w.r.t. z and w anduniformly continuous, w.r.t x for z, w 6= 1. I will now show that the right handside of the equality (5) is analytic w.r.t. z and w. The first product is clearlyanalytic since its factors are analytic. The expression 1

1−z−wis also analytic when

z + w 6= 1. We therefore only have to consider the last sum. First we notice thatby symmetry it is enough to prove that

∞∑k=0

(z)k

(1− w)k+1

(ζ(z + k)− 1)

is analytic for w and z. We have

∞∑k=0

(z)k

(1− w)k+1

(ζ(z + k)− 1) =M−1∑k=0

(z)k

(1− w)k+1

(ζ(z + k)− 1)+

(z)M

(1− w)M+1

∞∑k=0

(z + M)k

(2− w + M)k

(ζ(k + M + z)− 1)

Clearly the first term in this expression is analytic since it is a finite sum ofanalytic functions (when z, w /∈ Z). We see that it is enough to prove that the last

63

sum is analytic, since it is multiplied by an analytic function. Let w ∈ w : |w−w0| < 1 and z ∈ z : |z − z0| < 1. Choose M > |min(Re(1− w0), Re(z0))|+ 3.We see that

∣∣∣∣∣∞∑

k=0

(z + M)k

(2− w + M)k

(ζ(k + M + z)− 1)

∣∣∣∣∣ ≤∞∑

k=0

(|z + M |+ 1)k

(|2− w + M | − 1)k

(ζ(2 + k)− 1)

which converges. By Weierstrass M-test the sum converges uniformly with z andw in the given neighborhoods and is thus analytic. We already have the equality(5) for max(Re(z), Re(w)) < 0 and by uniqueness of analytic continuation we get(5) in the region stated.

From (5) we see when z = σ + it and z = σ − it∫ 1

0

|ζ∗(σ + it, x)|2dx = (6)

2(2π)2σ−2|Γ(1− σ − it)|2 cosh(πt)ζ(2− 2σ)+

+1

2σ − 1− 2 Re

(∞∑

k=0

(σ + it)k

(1− σ + it)k+1

(ζ(σ + it + k)− 1)

)¬(σ ∈ Z ∧ t = 0)

From (6) we see (with Stirling’s formula)∫ 1

0

|ζ∗(σ + it, x)|2dx =1

2σ − 1+ (2π)2σ−1ζ(2− 2σ)t1−2σ− (7)

−2

tIm(ζ(σ + it)) + O

(1

t

), σ > 0

From (3), (4) and (5) we also see that∫ 1

0

(ζ(z, x)ζ(w, x)− x−z−w)dx = (8)

2(2π)(z+w−2)Γ(1− z)Γ(1− w) cos(π

2(z − w))ζ(2− z − w)− 1

1− z − w

max(Re(z), Re(w)) < 1 ∧ z + w 6= 1

From (8) we see that the convergence region in (2) holds. If we put w := 1−s− ε

64

and z := s− ε we get∫ 1

0

(ζ(s− ε, x)ζ(1− s− ε, x)− x2ε−1)dx = (according to 3) =

2(2π)(−2ε−1)Γ(1− s + ε)Γ(s + ε) cos(π

2(1− 2s))ζ(1 + 2ε)− 1

2ε=

= (Laurent series development) =

1

π((1− 2 log(2π)ε + O(ε2)))(Γ(1− s) + Γ′(1− s)ε + O(ε2))

(Γ(s) + Γ′(s)ε + O(ε2)) sin(πs)(1

2ε+ γ + O(ε))− 1

2ε=

(1

2πΓ(s)Γ(1− s) sin(πs)− 1

2)1

ε+

+sin(πs)

π((γ − log 2π)Γ(1− s)Γ(s)+

+12(Γ(1− s)Γ′(s) + Γ′(1− s)Γ(s))) + O(ε) =

= (according to the reflexion formula for the Gamma-function) =

γ − log 2π + 12

(Γ′(s)

Γ(s)+

Γ′(1− s)

Γ(1− s)

)+ O(ε) =

γ − log 2π + 12(Ψ(s) + Ψ(1− s)) + O(ε)

We let ε tend to zero and get∫ 1

0

(ζ(s, x)ζ(1− s, x)− x−1)dx =

γ − log 2π + 12(Ψ(s) + Ψ(1− s)). Re(s) ∈ (0, 1)

(9)

We now consider a special case. First when w + z = 1. From formula (9), (3)and (4) we get∫ 1

0

ζ∗(s, x)ζ∗(1− s, x)dx = γ − log 2π + 12(Ψ(1− s) + Ψ(s))−

∞∑k=0

(ζ(s + k)− 1

s + k+

ζ(1− s + k)− 1

1− s + k

). s /∈ Z (10)

First we only have the formula for Re(s) ∈ (0, 1), but by the same argument as in(5) we know that the formula holds in the region stated. (In fact the integral andthe sum is just a special case of the sum and integral discussed in the proof of (5).We also need that Ψ(s) is an analytic function and then by analytic continuation

65

the equality is valid). When s = 1/2 + it we get∫ 1

0

|ζ∗(12

+ it, x)|2dx = γ − log 2π + Re

(Ψ(1

2+ it)− 2

∞∑k=0

ζ(1/2 + it + k)− 1

1/2 + it + k

).

(11)

Directly from (11) we see (since Ψ(s) = log(s)+O(|1s|) and ζ(1/2+it) = O(t

956

+ε)(see [2]))∫ 1

0

|ζ∗(12

+ it, x)|2dx = log t + γ − log 2π − 2

tIm(ζ(1

2+ it)) + O(

1

t) (12)

= log t + γ − log 2π + O(t−4756

+ε). (ε > 0)

We see that the truth of the Lindelof hypothesis would imply the error-termO(tε−1) The estimate (12) is however far better than the previously best estimate:∫ 1

0

|ζ∗(12

+ it, x)|2dx = log t + O(1).

See [7] V.V. Rane, and also [3], and [1] for former estimates.

References

[1] R. Balasubramian, A note on Hurwitz zeta-function, Ann.Acad.Sci. Fenn. AI Math. 4 (1979), 41–44.

[2] E. Bombieri and H. Iwaniec, On the order of ζ(1/2+ it), Ann. Sc. Norm. Sup.Pisa. 13 (1986), 449–372.

[3] J. F. Koksma and C. G. Lekkerkerker, A mean-value theorem for ζ(s, w),Nederl. Akad. Wetensch. Proc. Ser. A. 55 = Indagationes Math. 14 (1952),446–452.

[4] M. Miklos, Transformation und orhogonalitat bei ζ(s, u). Verallgemeinungder Riemannschen Funktionalgleichung von ζ(s), Acta. Sci. Mat. Szeged. 17(1956), 143–164.

[5] , Integral formulae of arithmetical characteristics relating to the zeta-function of Hurwitz, Publ. Math. Debrecen 5 (1957), 44–53.

[6] , Uber die Charakterisirung der Hurwitzschen Zetafunktion mittelsFunktional-gleichungen, Acta. Sci. Math. Szeged. 19 (1958), 247–250.

[7] V. V. Rane, On Hurwitz zeta-function, Math. Ann 264 (1983), 147 – 151.

66

On the fourth power moment of the Hurwitz zetafunction

Johan Andersson∗

Abstract

It is well known that∫ 1

0

∣∣ζ∗(12 + it, x

)∣∣2dx = log x + γ − log 2π + O(t−5/6

),

where ζ∗(s, x) = ζ(s, x)− x−s and ζ(s, x) denote the Hurwitz zeta function. See e.g.Andersson [3]. In this paper we prove the less known fact that∫ 1

0

∣∣ζ∗(12 + it, x

)∣∣4dx = Ω

(exp

(c(log t)1/4

log log t

)), (1)

for some c > 0. In particular this means that there is no corresponding formula forhigher power moments. This result was first presented at the Kubelius conference inPalanga 1996, and first appeared in print in the introduction of our licentiate thesis[2]. We also prove that on the average with respect to t we have

1T

∫ T

0

∫ 1

0

∣∣ζ∗(12 + it, x

)∣∣4dxdt = 2 log2 T + O((log T )5/3

),

whereas for rational parameters x = p/q we have

1T

∫ T

0

∣∣∣∣ζ∗(12

+ it,p

q

)∣∣∣∣4dt = cq log4 T + O(log3 T

)whenever (p, q) = 1.

Contents

1 Introduction 68

2 Omega estimates for moments of the Hurwitz zeta function 69

3 The fourth power moment of the Hurwitz zeta function 71

∗Department of Mathematics, Stockholm University, johana@math.su.se

67

1 Introduction

The moment problem is an important problem in the theory of the the Riemann zetafunction. The mean square

1

T

∫ T

0

∣∣ζ(12

+ it)∣∣2dt = log T + 2γ − 1 + O

(T−1/2

)(2)

for the Riemann zeta function has been used as a motivation to study the mean square ofthe Hurwitz zeta function. In our paper [3] we considered this problem, and we proved∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣2dx = log x + γ − log 2π − 2t−1 Im

(ζ(

12

+ it))

+ O(t−1), (3)

where ζ∗(s, x) = ζ(s, x)−x−s, is defined so that the integral is convergent. For the Riemannzeta function we have a good formula for the fourth power moment1:

1

T

∫ T

0

∣∣ζ(12

+ it)∣∣4dt ∼ c4 log4 T. (4)

It is natural to consider the corresponding problem for the Hurwitz zeta function. Inanalogy with the Riemann zeta function case it might seem natural to assume that∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dx ∼ c log4 t. (5)

In this paper we will show that a formula like (5) cannot be true. In fact the moment∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dx

will be rather chaotic in the same sense as the absolute values of the Riemann zeta functionin the critical strip. On one hand we will be able to prove that∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dx = Ω

(exp

(c(log t)1/4

log log t

)).

On the other hand we prove that its mean value with respective to t

1

T

∫ T

0

∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dxdt = 2 log2 T + O

((log T )5/3

),

will grow just like a log power of t. We notice that the log power here is 2 and not 4 as inthe case of the Riemann zeta function. This comes from the fact that the Riemann zetafunction by its nature is arithmetic and has an Euler product, whereas the Hurwitz zetafunction in general is not arithmetic. In the special case of x being rational it will havearithmetic properties though, since it can be expressed as sums of Dirichlet L-functions,which do have Euler products2. This will allow us to prove that

1

T

∫ T

0

∣∣∣∣ζ∗(1

2+ it,

p

q

)∣∣∣∣4dt ∼ cq log4 T,

whenever (p, q) = 1.

1A result of Ingham [5].2For a further discussion of this phenomenon see our forthcoming paper [1].

68

2 Omega estimates for moments of the Hurwitz zeta

function

Lemma 1. There exists constants C1, C2 > 0 such that if κ ≥ 1 and t ≥ 2 are real numbers,then ∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣2κ

dx > C1

(∣∣∣∣ζ (1

2+

1

2κ− it

)∣∣∣∣2κ

− C2

)(log t)−2κ−1.

Proof. The case κ = 1 follows trivially from (3). Hence we can assume κ > 1. Theapproximate functional equation for the Hurwitz zeta function (see e.g. Rane [9]) saysthat

ζ∗(

12

+ it, x)

=∑

1≤n≤X

(n + x)−1/2−it

+ χ(

12

+ it) ∑

1≤k≤Y

k−1/2+ite(−kx) + O(|X|−1/2 + |Y |−1/2

), (2πXY = t) (6)

where χ(s) is the factor in the functional equation for the Riemann zeta function (inparticular we have that |χ(1/2 + it)| = 1). By choosing Y = t this gives us

ζ∗(

12

+ it, x)

= χ(

12

+ it) ∑

1≤k≤t

k−1/2+ite(−kx) + O(1). (7)

Another estimate we will use is∣∣∣∣∣ ∑1≤k≤t

e(kx)k−λ

∣∣∣∣∣ min((1− x)λ−1 + xλ−1, t1−λ

). (0 < x, λ < 1)

In particular this implies that∫ 1

0

∣∣∣∣∣ ∑1≤k≤t

e(kx)k−λ

∣∣∣∣∣dx ≤ Cλ, (0 < λ < 1) (8)

as well as

∫ 1

0

∣∣∣∣∣ ∑1≤k≤t

e(kx)k−λ

∣∣∣∣∣1/(1−λ)

dx log t. (2 ≤ t, 0 < λ < 1)

(9)

From the approximate functional equation for the Riemann zeta function (Titchmarsh [10],Theorem 4.11) we have that

ζ

(1

2+

1

2κ− it

)=∑

1≤k≤t

k−1/2−1/(2κ)−it + O(1), (κ > 0)

69

which in view of the Parseval identity equals

χ(

12

+ it) ∫ 1

0

(∑1≤k≤t

e(kx)k−1/(2κ)

)(∑1≤k≤t

e(−kx)k−1/2−it

)dx + O(1).

Applying the identity (7) we can rewrite this as

χ(

12

+ it) ∫ 1

0

(∑1≤k≤t

e(kx)k−1/(2κ)

)(ζ∗(

12

+ it, x)

+ O(1))dx + O(1).

By using equation (8) we can remove the O(1) term within the integral and we obtain

ζ

(1

2+

1

2κ− it

)= χ

(12

+ it) ∫ 1

0

(∑1≤k≤t

e(kx)k−1/(2κ)

)ζ∗(

12

+ it, x)dx + O(1).

By using the Holder inequality [4]∫ b

a

|f(x)g(x)|dx ≤(∫ b

a

|f(x)|p)1/p(∫ b

a

|g(x)|qdx

)1/q

,

(1

p+

1

q= 1

)with

a = 0, b = 1, p =2κ

2κ + 1, and q = 2κ,

together with the fact that |χ(1/2 + it)| = 1, we get that∣∣∣∣ζ (1

2+

1

2κ− it

)+ O(1)

∣∣∣∣

∫ 1

0

∣∣∣∣∣ ∑1≤k≤t

e(kx)k−1/2−1/(2κ)

∣∣∣∣∣2κ/(2κ+1)

dx

(2κ+1)/(2κ)(∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣2κ

dx

)1/(2κ)

.

With (9) this implies that∣∣∣∣ζ (1

2+

1

2κ− it

)+ O(1)

∣∣∣∣ (log t)(2κ+1)/(2κ)

(∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣2κ

dx

)1/(2κ)

.

The Lemma follows by raising this inequality to the 2κ’th power.

Theorem 1. There exist constants C1, C2 > 0 such that whenever

C1 log log T ≤ H ≤ T,

and κ > 1, then

maxT≤t≤T+H

∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣2κ

dx > exp

(C2(log H)(κ−1)/(2κ)

log log H

).

70

Proof. This follows from Lemma 1 and Theorem 3.3.1 (for the special case of z = 1) inRamachandra [6].

Theorem 1 should be compared to the Omega estimates of Ramachandra and Sankara-narayanan [7] ∣∣∣∣ζ∗(1

2+ it,

p

q

)∣∣∣∣ = Ω

(exp

(Cp,q

√log t

log log t

)),

for the Hurwitz zeta function with a rational parameter. We also remark that from thetheorem it follows that there exist no formula like (3) for any integer or fractional momentstrictly greater than 2. In fact for any κ > 1 we can choose a sequence tk∞k=1 such that∫ 1

0

∣∣ζ∗(12

+ itk, x)∣∣2κ

dx > (log tk)N (k > k0(N))

will grow faster than any polynomial in log tk.

3 The fourth power moment of the Hurwitz zeta func-

tion

From Theorem 1 we have the following consequence in the special case of κ = 2 that

Corollary. There exist constants C1, C2 > 0 such that whenever

C1 log log T ≤ H ≤ T,

then

maxT≤t≤T+H

∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dx > exp

(C2(log H)1/4

log log H

).

Since it is impossible to get a nice asymptotic formula for the fourth power momentwhen we integrate with respect to x we may ask whether it is possible to get a niceasymptotic formula when we integrate with respect to t. In the case of x being rationalwe show that

Theorem 2. Let (p, q) be relatively prime integers. Then

1

T

∫ T

0

∣∣∣∣ζ∗(1

2+ it,

p

q

)∣∣∣∣4dt = cq log4 T + O(log3 T

),

where

cq = q−1∏P |q

P prime

(1− 1/P )−2(1− 1/P 2)−1.

71

Proof. We have that the Hurwitz zeta-function can be expressed in terms of DirichletL-functions

ζ

(s,

p

q

)=

1

φ(q)

∑χ (mod q)

qsχ(p)L(s, χ).

From this we obtain

1

T

∫ T

0

∣∣∣∣ζ∗(1

2+ it,

p

q

)∣∣∣∣4dt =∑

χ1,χ2,χ3,χ4 (mod q)

χ3(p)χ4(p)q2χ1(p)χ2(p)

T (φ(q))4×

×∫ T

0

L(

12

+ it, χ1

)L(

12

+ it, χ1

)L(

12− it, χ3

)L(

12− it, χ4

)dt (10)

By the fact that3

1

T

∫ T

0

∣∣L(12

+ it, χ1

)∣∣2∣∣L(12

+ it, χ2

)∣∣2dt = O(log2 T

)whenever χ1 6= χ2 or χ1 6= χ2, we see by the Cauchy-Schwarz inequality that the contribu-tion coming from the non diagonal case is

1

T

∫ T

0

∣∣L(12

+ it, χ1

)L(

12

+ it, χ1

)L(

12− it, χ3

)L(

12− it, χ4

)∣∣dt = O(log3 T

)unless χ1 = χ2 = χ3 = χ4, or χ1 = χ2 = χ3 = χ4. Hence we have that (10) equals

∑χ (mod q)

2q2

T (φ(q))4

∫ T

0

∣∣L(12

+ it, χ)∣∣4dt + O

(log3 T

).

The result follows from the following Theorem of Rane [8].

∑χ (mod q)

1

φ(q)

∫ T

0

∣∣L(12

+ it, χ)∣∣4dt =

1

2π2

∏P |q

P prime

1− 1/P

1− 1/P 2T log4(Tq)

+ O(log3(qT )(log log 3q)5

),

and the fact that

φ(q) = q∏P |q

P prime

(1− 1/P ).

3This is easily proved with e.g. the approximate functional equation of Dirichlet L-series and and asomewhat generalized version of the Ramanujan identity.

72

We see that the method depends heavily on arithmetic properties, and it is not possibleto generalize it to arbitrary x. To see how the Hurwitz zeta-function behaves for arbitraryx we will instead consider the fourth power moment when we integrate with respect toboth x and t. We will prove

Theorem 3. One has that

1

T

∫ T

0

∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dxdt = 2(log T )2 + O

((log T )5/3

).

Theorem 3 can certainly be improved if we use more careful estimates. We believethat by using the approximate functional equation (6) with X Y

√t, and not just

carelessly throwing away terms into the error term, we should be able to prove that

1

H

∫ T+H

T

∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dxdt = P2(log T ) + o(1),

(T 1/2+ε < H < T 1−ε

)and by using more advanced estimates from the theory of exponential sums we mightimprove the result to an even shorter interval [T, T + H], for H = T θ and some θ < 1/2.For now we will be content with the present version of the theorem since we are at thispoint in time mainly concerned with the leading term.

Proof of Theorem 3

We will first prove some Lemmas.

Lemma 2. Suppose that 0 < j, k < n are integers such that k, n− k 6= j. Then∣∣∣∣∣∫ B

A

(j(n− j)

k(n− k)

)it

dt

∣∣∣∣∣ ≤ n2

|n− j − k| |j − k|.

Proof. By symmetry we may assume that k, j ≤ n/2, since otherwise we can considerk → n − k and/or j → n − j. Likewise we may assume k < j since otherwise we canconsider (j, k) → (k, j).

Assume that 0 < y < x < 1/2. We define

f(x) = log(x(1− x)), and obtain f ′(x) =1− 2x

(1− x)x.

Since f(x) is increasing on the interval (0, 1/2] we have that

f(x)− f(y) > f

(x + y

2

)− f(y). (11)

We have by the mean value theorem that

f

(x + y

2

)− f(y) = f ′(ξ)

(x + y

2− y

),

73

for some y < ξ < (x + y)/2. Since f ′ > 0 is decreasing on the interval (0, 1/2) this impliesthat

f

(x + y

2

)− f(y) ≥ f ′

(x + y

2

)(x + y

2− y

).

By simplifying this and combining it with (11) we obtain

f(x)− f(y) ≥ 2(x− y)(1− x− y)

(x + y)(2− (x + y)).

Since the denominator is positive and less than or equal to 1 for 0 < x, y < 1 this gives us

f(x)− f(y) ≥ 2(x− y)(1− x− y). (12)

We remark that the integral can be written as∫ B

A

(j(n− j)

k(n− k)

)it

dt =

∫ B

A

exp

(it

(f

(j

n

)− f

(k

n

)))dt,

and by the inequality ∣∣∣∣∫ B

A

eiαtdt

∣∣∣∣ =

∣∣∣∣eiαB − eiαA

iα

∣∣∣∣ ≤ 2

|α|,

valid for real α we have that∣∣∣∣∣∫ B

A

(j(n− j)

k(n− k)

)it

dt

∣∣∣∣∣ ≤ 2∣∣f( jn

)− f

(kn

)∣∣ ,and by (12) this gives us

≤ n2

|n− j − k| |j − k|.

Lemma 3. Suppose that B > A. Then

1

B − A

∫ B

A

∫ 1

0

∣∣∣∣∣M∑

n=1

n−1/2+ite(−nx)

∣∣∣∣∣4

dxdt = 2

(M∑

j=1

j−1

)2

−M∑

j=1

j−2 + O

(M log2 M

B − A

).

Proof. By squaring the Dirichlet polynomial

M∑n=1

n−1/2+ite(−nx),

74

we obtain that(M∑

n=1

n−1/2+ite(−nx)

)2

=2M∑n=2

( ∑1≤j,n−j≤M

(j(n− j))−1/2+it

)e(−nx).

With the Parseval identity we obtain

∫ 1

0

∣∣∣∣∣M∑

n=1

n−1/2+ite(−nx)

∣∣∣∣∣4

dx =

=2M∑n=2

( ∑1≤j,n−j≤M

(j(n− j))−1/2+it

)( ∑1≤k,n−k≤M

(k(n− k))−1/2−it

). (13)

By identifying the diagonal term we get the main term and by using Lemma 2 on the nondiagonal terms we get

∫ B

A

∫ 1

0

∣∣∣∣∣M∑

n=1

n−1/2+ite(−nx)

∣∣∣∣∣4

dxdt = (B − A)

2

(M∑

j=1

j−1

)2

−M∑

j=1

j−2

+ O

2M∑k=1

∑1≤j,n−j≤M1≤k,n−k≤M

n−k,k 6=j

(jk(n− j)(n− k))−1/2 n2

|n− j − k||j − k|

.

By summing the elements in the error terms this can be estimated by

(B − A)

2

(M∑

j=1

j−1

)2

−M∑

j=1

j−2

+ O(M log2 M

).

Proof of Theorem 2. By dyadic division it is sufficient to prove the result for

1

T

∫ 2T

T

∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dxdt.

Let us first assume that

T ≤ t ≤ 2T.

By using the approximate functional equation (6) with

Y T3√

log T, and X =

t

2πY 3√

log T ,

75

we obtain

ζ∗(

12

+ it, x)

=∑

1≤n≤X

(n + x)−1/2−it + χ(

12

+ it) ∑

1≤k≤Y

k−1/2+ite(−kx) + O(X−1/2

).

By estimating the first sum trivially by absolute values, we find that the first term can beestimated by

√X, and we obtain

ζ∗(

12

+ it, x)

= χ(

12

+ it) ∑

1≤k≤Y

k−1/2+ite(−kx) + O(

6√

log T). (14)

From Lemma 3 we have that

1

T

∫ 2T

T

∫ 1

0

∣∣∣∣∣ ∑1≤k≤Y

k−1/2+ite(−kx)

∣∣∣∣∣4

dxdt = 2 log2 T + O((log T )5/3

). (15)

Combining the fact that |χ(1/2 + it)| = 1, with the equations (14) and (15) finally givesus

1

T

∫ 2T

T

∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dxdt = 2 log2 T + O

((log T )5/3

).

References

[1] J. Andersson, Higher power moments of the Hurwitz and Lerch zeta functions, forth-coming.

[2] , Power sum methods and zeta-functions, Licentiate thesis, Stockholm Univer-sity, 1998.

[3] , Mean value properties of the Hurwitz zeta-function, Math. Scand. 71 (1992),no. 2, 295–300.

[4] O. Holder, Uber einen Mittelwertsatz., Gottingen Nachr., 38-47, 1889.

[5] A. E. Ingham, Mean value theorems in the theory of the Riemann zeta-function, proc.lond. math. soc,(2), 27 (1926),273-300.

[6] K. Ramachandra, On the mean-value and omega-theorems for the Riemann zeta-function, Published for the Tata Institute of Fundamental Research, Bombay, 1995.

[7] K. Ramachandra and A. Sankaranarayanan, Omega-theorems for the Hurwitz zeta-function, Arch. Math. (Basel) 53 (1989), no. 5, 469–481.

[8] V. V. Rane, A note on the mean value of L-series, Proc. Indian Acad. Sci. Math. Sci.90 (1981), no. 3, 273–286.

76

[9] , On Hurwitz zeta function, Math. Ann. 264 (1983), no. 2, 147–151.

[10] E. C. Titchmarsh, The theory of the Riemann zeta-function, second ed., The Claren-don Press Oxford University Press, New York, 1986, Edited and with a preface by D.R. Heath-Brown.

77

On the fourth power moment of the Lerch zeta function

Johan Andersson∗

Abstract

It is well known that∫ 1

0

∣∣ζ∗(12 + it, x

)∣∣2dx = log t + γ − log 2π + O(t−5/6

),

where ζ∗(s, x) = ζ(s, x) − x−s and ζ(s, x) denotes the Hurwitz zeta function. In arecent paper [2] we proved that it is not possible to extend this result to higher powermoments for the Hurwitz zeta function. In this paper we show that when we considerthe Lerch zeta function we have an analogous result for the fourth power moment∫ 1

0

∫ 1

0

∣∣φ∗ (x, y, 12 + it

)∣∣4dxdy = 2 log2 t + O((log t)5/3

),

where

φ∗(x, y, s) = φ(x, y, s)− y−s − (2π)s−1Γ(1− s)×

×(

e

(1− s

4− yx

)xs−1 + e

(s− 1

4+ y(1− x)

)(1− x)s−1

),

is defined so that the integral converges. This result was first presented at theKubelius conference in Palanga, 1996, but this is the first time it appears in print.

1 Introduction

In our paper [4] we discussed the problem of estimating∫ 1

0

∣∣ζ∗ (12

+ it, x)∣∣2dx, (1)

where ζ∗(s, x) = ζ(s, x)− x−s is defined so that the integral converges. Here

ζ(s, x) =∞∑

k=0

(k + x)−s, (Re(s) > 1)

∗Department of Mathematics, Stockholm University, johana@math.su.se

79

denotes the Hurwitz zeta function. This has been studied in several papers, starting withKoksma-Lekkerkerker [7]. For an excellent historic account of this problem, see Katsurada-Matsumoto [6]. Although it is very difficult to estimate ζ(1/2 + it, a) for fixed a, we haveobtained very sharp estimates for (1). In our paper [4] we obtained∫ 1

0

∣∣ζ∗ (12

+ it, x)∣∣2dx = log t + γ − log 2π + t−1 Im

(ζ(

12

+ it))

+ O(t−1).

It would be natural to continue these studies with the integral∫ 1

0

∣∣ζ∗ (12

+ it, x)∣∣4dx.

However, in our paper [2] we showed that there exists no similar formula for this expression1.In fact we showed that∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dx = Ω

(exp

(c(log t)1/4

log log t

)). (2)

We believe that the right problem to study, if we want to obtain an analogous formula to(3), is that of estimating ∫ 1

0

∫ 1

0

∣∣φ∗ (x, y, 12

+ it)∣∣4dxdy. (3)

Here

φ(x, y, s) =∞∑

n=0

e(nx)(n + y)−s (Re(s) > 1)

is the Lerch zeta function (For its theory see Laurincikas-Garunkstis [8]), and

φ∗(x, y, s) = φ(x, y, s)− y−s − (2π)s−1Γ(1− s)×

×(

e

(1− s

4− yx

)xs−1 + e

(s− 1

4+ y(1− x)

)(1− x)s−1

), (4)

is defined so that the integral converges. Although we will not be able to get as sharp aresult as (3) we will obtain the the quite satisfactory estimate∫ 1

0

∫ 1

0

∣∣φ∗ (x, y, 12

+ it)∣∣4dxdy = 2 log2 t + O

((log t)5/3

). (5)

We will also extend our result

1

T

∫ T

0

∫ 1

0

∣∣ζ∗(12

+ it, x)∣∣4dxdt = 2 log2 T 2 + O

((log T )5/3

),

1This was also presented at the Kubelius conference, Palanga 1996. It first appeared in print in a lessrefined form in the introduction our licentiate thesis [3].

80

from our paper [2] to the case of the Lerch zeta function. We will prove that

1

T

∫ T

0

∫ 1

0

∣∣φ∗(x, y, 12

+ it)∣∣4dxdt = 2 log2 T + O

((log T )5/3

), (0 ≤ y ≤ 1), (6)

as well as

1

T

∫ T

0

∫ 1

0

∣∣φ∗(x, y, 12

+ it)∣∣4dydt = 2 log2 T + O

((log T )5/3

). (0 ≤ x ≤ 1), (7)

Notice that the log power in these three problems (5), (6) and (7) is 2, and not 4 as is thecase2

1

T

∫ T

0

∣∣ζ(12

+ it)∣∣4dt ∼ c4 log4 T (8)

of the Riemann zeta function. This comes from the fact that the Riemann zeta functionby its nature is arithmetic and has an Euler product, whereas neither the Hurwitz zetafunction nor the Lerch zeta function in general are arithmetic3. For a further discussion ofthis phenomenon see our forthcoming paper [1].

2 Preliminaries

We will use the notation

xN =

x, 1 ≤ x < N + 1,

0, otherwise.(9)

With this notation we have that

φ∗N(x, y, s) =∑

1≤k≤N

(k + x)−1/2−ite(ky), (10)

=∞∑

k=−∞

k + x−sN e(ky). (0 ≤ x, y < 1)

We obtain

(φ∗N(x, y, s))2 =∞∑

k1,k2=−∞

k1 + x−sN k2 + x−s

N e((k1 + k2)y),

and with the Parseval identity we get∫ 1

0

(φ∗N(x, y, s))2(φ∗N(x,−y, z))2dy =

∞∑k1,k2,k3,k4=−∞k1+k2=k3+k4

k1 + x−sN k2 + x−s

N k3 + x−zN k4 + x−z

N .

2A result of Ingham [5].3In the special case of x, y being rational, the Lerch zeta function will have arithmetic properties. In

particular it can be expressed as sums of Dirichlet L-functions, which do have Euler products. Thus thecorresponding fourth power moment will have a log T term of the fourth order.

81

By integrating this with respect to x we have∫ 1

0

∫ 1

0

(φ∗N(x, y, s))2(φ∗N(x,−y, z))2dydx =

∞∑k1,k2,k3,k4=−∞k1+k2=k3+k4

∫ 1

0

k1 + x−sN k2 + x−s

N k3 + x−zN k4 + x−z

N dx.

With

k1 = n + [x], k2 = m + [x], k3 = n + m + [x], and k4 = [x],

where [x] as usual denote the integer part of x, we obtain a proof for the following Lemma.

Lemma 1. Let s, w be complex variables, and let ΦN be defined by (10). One has theidentity∫ 1

0

∫ 1

0

(φ∗N(x, y, s))2(φ∗N(x,−y, z))2dydx =

∞∑n,m=−∞

∫ N+1

1

n + x−sN m + x−s

N n + m + x−zN x−zdx.

By specializing this to

s = σ + it, and z = σ − it, (11)

we obtain

Lemma 2. Let ΦN be defined by (10), and σ, t be real numbers. One has the identity∫ 1

0

∫ 1

0

|φ∗N(x, y, σ + it)|4dydx =

∞∑n,m=−∞

∫ N+1

1

n + x−σ−itN m + x−σ−it

N n + m + x−σ+itN x−σ+itdx.

The most important case is the critical line σ = 1/2. We will first prove a result thatallows us to estimate the terms arising from Lemma 2.

Lemma 3. Suppose that m, n, n + m 6= 0 are integers, and t is a real number. Then thereexist a constant C > 0 such that∣∣∣∣∫ N+1

1

n + x−1/2−itN m + x−1/2−it

N n + m + x−1/2+itN x−1/2+itdx

∣∣∣∣ ≤ CN

(|t|+ 1)|mn|.

Proof. By symmetry we may assume that m, n ≥ 1, since otherwise we will get an integralof such type with the substitution y = x + min(m, n). With the choice of functions

f(x) =(n + x)(m + x)

x(n + m + x),

= 1 +nm

x(n + m + x), (12)

82

and

g(x) =1

x(n + m + x),

we can now write the integral as ∫ B

A

(f(x))it−1/2g(x)dx,

for some 1 ≤ A ≤ B ≤ N + 1. By (12) we see that f(x) is decreasing and that f ′(x) < 0,hence we can write the integral as∫ B

A

f ′(x)(f(x))it−1/2 g(x)

f ′(x)dx.

We do partial integration and we find that the integral equals[(f(x))it+1/2

it + 1/2

g(x)

f ′(x)

]B

A

−∫ B

A

(f(x))it+1/2

it + 1/2

(d

dx

g(x)

f ′(x)

)dx. (13)

We have that

g(x)

f ′(x)= − x(n + m + x)

mn(m + n + 2x), (14)

and

d

dx

g(x)

f ′(x)= −

((m + n + x)2 + x2

mn(m + n + 2x)2

).

We see that the last integral equals

1

it + 1/2

∫ B

A

(f(x))it

√(n + x)(m + x)

(m + n + x)x

((m + n + x)2 + x2

mn(m + n + 2x)2

)dx

By estimating the integral by its absolute values, this term can be estimated by

CN

(|t|+ 1)|mn|.

Finally, by (14) we see that the first two terms in (13) can also be estimated by

CN

(|t|+ 1)|mn|.

83

Lemma 4. Let ΦN be defined by (10). One has the estimate∫ 1

0

∫ 1

0

∣∣φ∗N(x, y, 12

+ it)∣∣4dydx =

N∑n=1

4

n

(log(n + 1) + log

(1− n

N + 1

))− 3N

N + 1+ O

(N log2 N

|t|+ 1

).

Proof. By Lemma 2 we have that∫ 1

0

∫ 1

0

∣∣φ∗N(x, y, 12

+ it)∣∣4dydx =

∞∑n,m=−∞

∫ N+1

1

n + x−1/2−itN m + x−1/2−it

N n + m + x−1/2+itN x−1/2+itdx. (15)

From the diagonal terms, that is when

n = 0, or m = 0,

we obtain the terms

2∞∑

n=−∞

∫ N+1

1

n + x−1N x−2dx−

∫ N+1

1

x−2dx.

which equals

4N∑

n=1

∫ N+1−n

1

dx

(n + x)x− 3

∫ N+1

1

dx

x2. (16)

We have that

1

(n + x)x=

1

n

(1

x− 1

n + x

)(17)

and we have that equation (16) equals

N∑n=1

4

n(log(N + 1− n) + log(n + 1)− log(N + 1))− 3N

N + 1. (18)

It is clear that the integral in (15) vanishes when |m| > N or |n| > N , hence it is sufficientto consider the case |m|, |n| ≤ N and from Lemma 3 we obtain the error term

N∑n,m=−Nn,m6=0

O

(N

(|t|+ 1)|mn|

)= O

(N log2 N

|t|+ 1

).

84

3 The main results

We are now ready to combine our lemmas to obtain

Theorem 1. One has that∫ 1

0

∫ 1

0

∣∣φ∗ (x, y, 12

+ it)∣∣4dxdy = 2 log2 t + O

((log t)5/3

).

Proof. By using the definition of φ∗N(x, y, s), and φ∗(x, y, s), equations (10) and (4), theapproximate functional equation for the Lerch zeta function can be written as

φ∗(x, y, 1

2+ it

)= φ∗X

(x, y, 1

2+ it

)+ χ

(12

+ it)φ∗Y(y, 1− x, 1

2− it

)+ O

(|X|−1/2 + |Y |−1/2

), (2πXY = t > 0)

where χ(s) is the factor in the functional equation for the Riemann zeta function (inparticular we have that |χ(1/2 + it)| = 1). By using it with

X t3√

log t, and Y =

t

2πX 3√

log t,

and by estimating the second term by its absolute values, we find that

φ∗(x, y, 1

2+ it

)= φ∗X

(x, y, 1

2+ it

)+ O

((log t)1/6

). (19)

Since

N∑n=1

log(n)

n= 1

2(log N)2 + O(log N),

we get by Lemma 4 that∫ 1

0

∫ 1

0

∣∣φ∗X(x, y, 12

+ it)∣∣4dydx = 2 log t + O

((log t)5/3

).

From this and (19) the Theorem follows.

Theorem 2. One has the estimates

1

T

∫ T

0

∫ 1

0

∣∣φ∗(x, y, 12

+ it)∣∣4dxdt = 2 log2 T + O

((log T )5/3

), (0 ≤ y ≤ 1)

and

1

T

∫ T

0

∫ 1

0

∣∣φ∗(x, y, 12

+ it)∣∣4dydt = 2 log2 T + O

((log T )5/3

). (0 ≤ x ≤ 1)

85

Proof. First we notice that by the approximate functional equation of the Lerch zeta-function it is sufficient to prove the first of the estimates. The proof is then identical to theproof of the corresponding result for the Hurwitz zeta function, Theorem 2 in our paper[2], so we will skip the proof here. We will just remark that instead of Lemma 3 from thatpaper ∣∣∣∣∣

∫ B

A

(j(n− j)

k(n− k)

)it

dt

∣∣∣∣∣ ≤ n2

|n− j − k| |j − k|

(j, k, n, n− k > 0

k, n− k 6= j

)(20)

we will need the corresponding inequality∣∣∣∣∣∫ B

A

((j + y)(n− j + y)

(k + y)(n− k + y)

)it

dt

∣∣∣∣∣ ≤ (n + 2y)2

|n− j − k| |j − k|

j, k, n, n− k > 0k, n− k 6= j0 ≤ y ≤ 1

(21)

This follows by the same proof though, since the assumption that n, j, k are integers isactually not needed in the proof of the Lemma, and the case (21) follows by letting (j, k, n)be (j + y, k + y, n + 2y) in (20).

4 Final discussion

In our talk at the Kubelius conference in Palanga, 1996 we were using the approximatefunctional equation with

X = Y =

√t

2π. (22)

This has the advantage that if done carefully, and we estimate all the occurring terms weshould obtain a better error term, since Lemma 3 gives a better error term for shorterranges of N . There are certain new terms appearing though, and the proof will necessarilybe a bit more technical. At the moment we are content with the present version of thetheorem. It should also be mentioned that we have the following

1

T

∫ T

0

∣∣∣∣φ∗(p

q,h

k,1

2+ it

)∣∣∣∣4dt ∼ ch,k,p,q log4 T. (23)

In the special case of h = p = 0 this reduces to the fourth power moment of the Riemannzeta function (8). In the case of h = 0 we have that φ∗(h/k, p/q, 1/2 + it) = ζ∗(1/2+it, p/q)and this follows from Theorem 2 from our paper [2]. The proof for the general case can beconstructed along the same lines as that proof by expressing the Lerch zeta functions interms of Dirichlet L-functions.

We may finally ask. Can we obtain a similar formula as the one in Theorem 1 for higherpower moments ∫ 1

0

∫ 1

0

∣∣φ∗ (x, y, 12

+ it)∣∣2k

dxdy?

A reason to doubt it is that the higher moment case fails for the Hurwitz zeta function.However our best bet is that it will not fail in this case.

86

Conjecture. One has that∫ 1

0

∫ 1

0

∣∣φ∗ (x, y, 12

+ it)∣∣2k

dxdy ∼ k! (log t)k.

We will discuss more about this conjecture in [1]. The truth of the conjecture dependson just how dense the rational numbers are among the real numbers. By the method ofRamachandra-Balasubramanian [9] it is namely possible to show that for rational param-eters of x, y and the integral with respect to t from 0 to T the integral can be estimatedfrom below by cT logk2

T , and not just cT log2k T . We remark that although we believein the Conjecture it should indeed be a very difficult conjecture to prove. The Lindelofhypothesis for the Riemann zeta function as well as the Lindelof conjecture for the Lerchzeta function would follow from the Conjecture.

References

[1] J. Andersson, Higher power moments of the Hurwitz and Lerch zeta functions, forth-coming.

[2] , On the fourth power moment of the Hurwitz zeta function, 2006.

[3] , Power sum methods and zeta-functions, Licentiate thesis, Stockholm Univer-sity, 1998.

[4] , Mean value properties of the Hurwitz zeta function, Math. Scand. 71 (1992),no. 2, 295–300.

[5] A. E. Ingham, Mean value theorems in the theory of the Riemann zeta-function, proc.lond. math. soc,(2), 27 (1926),273-300.

[6] M. Katsurada and K. Matsumoto, Explicit formulas and asymptotic expansions forcertain mean square of Hurwitz zeta-functions, Proc. Japan Acad. Ser. A Math. Sci.69 (1993), no. 8, 303–307.

[7] J. F. Koksma and C. G. Lekkerkerker, A mean-value theorem for ζ(s, w), Nederl. Akad.Wetensch. Proc. Ser. A. 55 = Indagationes Math. 14 (1952), 446–452.

[8] A. Laurincikas and R. Garunkstis, The Lerch zeta-function, Kluwer Academic Publish-ers, Dordrecht, 2002.

[9] K. Ramachandra, On the mean-value and omega-theorems for the Riemann zeta-function, Tata Institute of Fundamental Research Lectures on Mathematics andPhysics, vol. 85, Published for the Tata Institute of Fundamental Research, Bombay,1995.

87

PART III - A NEW SUMMATION FORMULA ON THEFULL MODULAR GROUP AND KLOOSTERMAN

SUMS

INTRODUCTION

A summation formula for the modular group

The classical Poisson summation formula∞∑

n=−∞f(n) =

∞∑n=−∞

f(n).

is a fundamental tool in analytic number theory The importance of this theoremsuggests that one should study generalizations of this result. In particular we wishto consider analogues for discrete groups such as the full modular group. Can wefind some similar expansion of ∑

γ∈PSL(2,Z)

f(γ)?

When f is SO(2) bi-invariant an expansion in terms of spectral theory can ob-tained by the pre-trace formula, and indeed Terras ([11], p. 265) choose to callthis the “Poisson summation formula for SL(2, Z)”. The condition that f is SO(2)bi-invariant seems too restrictive though, and indeed the condition can quite easilybe reduced to SO(2) right (or left)-invariant (for details see our forthcoming paper[1]). The condition that f is SO(2) right (or left)- invariant still seems to restric-tive. The summation makes sense whenever the sum is convergent. In particularf ∈ C∞

0 (PSL(2, R)) should be a sufficient condition. In part III of our thesis wedevelop such a formula. We generalize it to matrices of fixed determinant, andshow how it relates to the Kuznetsov summation formula as well as the Selbergand Eichler-Selberg trace formulae.

It turns out that this question is closely related to the expansion of the fourthpower moment of the Riemann zeta function∫ ∞

−∞g(t)

∣∣ζ( 12 + it

)∣∣4dt,

given by Yoichi Motohashi in his important paper [8] in terms of spectral theory,as well as to the additive divisor problem given by Kuznetsov [6] and Motohashi[9]. This was indeed our first approach to this subject and we will describe thisrelation in the next section.

Introduction 92

Summation formulae and functional equations

A general idea in the analytic theory of numbers is that whenever you have asummation formula, you should have a functional equation, and vice verse (seee.g Ferrar [4]). For example, the classical functional equation of the Riemann zetafunction is proved by the classical Poisson summation formula (after some trick tosolve the convergence problems). The existence of a Poisson summation formulaand a functional equation are in fact equivalent. We can also derive the Poissonsummation formula from the functional equation6. The following table providessome other examples:

summation formula Functional equation

∑f(n) =

∑f(n)

(Poisson summation formula)

ζ(s) = χ(s)ζ(1− s)

∑χ(n)f(n) = τ(χ)χ(−1)

N

∑χ(n)f

(nN

)(twisted Poisson summation formula)

L(s, χ) = χε(s)τ(χ)Ns L(1− s, χ)

∑d(n)f(n) =

∑d(n)f(n)

(Voronoi summation formula)

ζ2(s) = χ2(s)ζ2(1− s)

Selberg trace formula Selberg zeta∑f(n + α, m + β) =

∑f(n, m)e(−nα−mβ)

(shifted double Poisson summation formula)

two Hurwitz zeta

The way of going back and forth between these is by using Mellin inversions

∞∑n=1

anf(n) =1

2πi

∫ c+∞i

c−∞i

Mf(s)A(s)ds, A(n) =∞∑

n=1

ann−s,

where

Mf(s) =∫ ∞

0

xs−1f(x)dx

is the Mellin transform of f . We should note that even though the functional equa-tion and the summation formula are equivalent, the summation formula sometimesgives better results. For example, sometimes it is better to use an approximate

6 Actually the functional equation for the Riemann zeta-function corresponds to the Poissonsummation formula for the Fourier-cosine transform.

Introduction 93

version of the functional equation rather than the full functional equation for theRiemann zeta-function. This is most simply proved by the Poisson summationformula directly. In Motohashi’s series of papers and book (see e.g [10], page 165),he essentially discovers, even though it is in a certain sense hidden, some formulaeof the type

ζ(u)ζ(v)ζ(w)ζ(z) = “main term”+∑j

αjHj( 12 (u + v + w + z − 1))Hj( 1

2 (u + z − v − w + 1))×

×Hj( 12 (v + z − u− w + 1))F (u, z, v, w;κj) + “similar terms”, (III.35)

where F is a certain hypergeometric function, and

Hj(s) =∞∑

n=1

tj(n)n−s (III.36)

are Dirichlet-series (Hecke L-series corresponding to Maass wave forms - see [10],page 104). In a strict sense (III.35) is false. The right hand side is divergent.However, if we integrate the left hand side over test functions of certain restrictedclass we have equality. Our general analogy suggests that this is a sort of functionalequation that should have a corresponding summation formula. If we formallyapply four dimensional Mellin-transformation a quadruple sum can be written as

∞∑a,b,c,d=1

f

(a bc d

)=

1(2πi)4

∫ 2+∞i

2−∞i

∫ 2+∞i

2−∞i

∫ 2+∞i

2−∞i

∫ 2+∞i

2−∞i

ζ(u)ζ(v)ζ(w)ζ(z)Mf

(u vw z

)dudvdwdz,

(III.37)

where Mf( u vw z ) now denotes the four-dimensional Mellin transform. By using

the expression (III.35) for the product of four zeta-function, and using the linearchange of variables

s1 = 12 (u + v + w + z − 1),

s2 = 12 (u + z − v − w + 1),

s3 = 12 (v + z − u− w + 1),

we find that the right hand side in equation (III.37) can be re-written as

1(2πi)3

∫ 1/2+∞i

1/2−∞i

∫ 1/2+∞i

1/2−∞i

∫ 7/2+∞i

7/2−∞i

(“main term”+

+∞∑

j=1

αjHj(s1)Hj(s2)Hj(s3)F (u, z, v, w;κj) + “similar terms”)

ds1ds2ds3,

Introduction 94

and we should be able, by using (III.36) and once again using Mellin inversion, torecover

∞∑a,b,c,d=1

f

(a bc d

)= “main term”+

∞∑j=1

∞∑m,n,D=1

αjtj(n)tj(m)tj(D)f(κj ;m,n,D)+

+ “other similar terms” + “certain terms of less importance”, (III.38)

where f is some transform of f , and the kernel-function is some (possibly com-plicated) hypergeometric function. If we look at the papers of Motohashi [9] andKuznetsov [6] on the additive divisor problem, we see that we can in fact get asimilar formula for fixed D for sums∑

ad−bc=D

f

(a bc d

)where the summation over D in the right hand side of (III.38) should be sup-pressed. This of course corresponds to the change of summation order

∞∑a,b,c,d=−∞

f

(a bc d

)=

∞∑D=−∞

∑ad−bc=D

f

(a bc d

).

Note that Motohashi calls the dissection D < 0, D = 0, D > 0, the “Atkinsondissection” after Atkinson’s paper [2] on the mean square of the Riemann zetafunction. We will emphasize that in this case the summation formula we will findin Part III in our thesis will be new, but the corresponding functional equation isknown.

The summation formula - A simplified approach

We originally thought of the idea to use the results coming from the Fourth powermoment in the spring of 1999. We were looking at some quadruple exponentialsums coming from a problem involving the Lerch zeta function, when we realizedthat we had a Mellin Barnes integral involving four zeta functions, which we couldtreat with Motohashi’s method. It was clear to us that there had to be simplerproofs. A first simplification was given at a seminar in Turku in the fall of 1999and is presented in the paper “A summation formula over integer matrices II”. Inthe spring of 2000 we discovered a further simplification and we will here describethe main idea in the proof as given in the paper “A summation formula on thefull modular group”. The idea relies on the Bruhat decomposition of the Big cellof PSL(2, Z), or in other words the fact that any element in PSL(2, Z), which isnot the identity can be written in a unique way as(

1 m0 1

)(h ∗c h

)(1 n0 1

)

Introduction 95

whenever m,n, h, h, c are integers such that

c > 0, and 0 ≤ h, h < c.

The idea is then to proceed by using the double Poisson summation formula, andthen the classical Kloosterman sums

S(m,n; c) =∑

hh≡1 (mod c)0≤h,h<c

e

(mh + nh

c

)

appear in a natural way. In fact for f belonging to a test function class onPSL(2, R) we obtain∑

ad−bc=1c>0

f

(a bc d

)=

∞∑m,n=−∞

∞∑c=1

S(m;n, c)fc(m,n),

for fc(m,n) a certain integral transform of f . The case when m,n 6= 0 can now betreated with the Kuznetsov summation formula [5]. This will allow us to obtain ageneral summation formula of the following type∑

ad−bc=1c>0

f

(a bc d

)= “main terms”+

∑m,n 6=0

1π

∫ ∞

−∞

σ2ir(|m|)σ2ir(|n|)F (r;m,n)dr

|nm|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

∑m,n 6=0

ρj,k(m)ρj,k(n)F((

12 − k

)i;m,n

)+∞∑

j=1

∑m,n 6=0

ρj(m)ρj(n)F (κj ;m,n).

(III.39)

Here F (r;m,n) is a certain integral transform of f , the ρj(n) denote the Fouriercoefficients for the Maass wave forms, and the ρj,k(n) denote Fourier coefficientsof holomorphic cusp forms of weight k. For an exact version of this formula werefer to Theorem 1 in “A summation formula on the full modular group”.

The summation formula and Hecke operators

In order to relate this formula to matrices of determinant D in order to apply it onquadruple exponential sums such as in (III.38), we need to apply the Hecke opera-tors in some way or another. We will choose to do so directly on the Kloostermansums and by doing so the generalized Kloosterman sums

S(D,m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

)

Introduction 96

will appear. We will need a Kuznetsov summation formula for these generalizedKloosterman sums. A formula of this kind has been given by Bykovsky-Kuznetsov-Vinogradov [3]. The way they prove this is by using the identity

S(D,m, n; c) =∑

d|(D,m,c)

d S

(mD

d2, n;

c

d

),

which they attribute to Heath-Brown. However, a clear and simple proof of thisidentity seems to be lacking in the literature, and thus we chose to write the paper“A note on some Kloosterman sum identities” on this subject. We also remarkthat this identity gives a simple and elementary proof of the Selberg identity. Thisis a subject which has been of some interest to other researchers, e.g. Matthes [7]who gave a different elementary proof.

Applications

There are several applications of the summation formula. As already indicatedit is related to the additive divisor problem and the additive circle problem, aswell as the Fourth power moment. We have chosen not to include this in ourthesis but will instead return to this topic in forthcoming papers. Instead we willconcentrate on showing that it can be used to prove the Selberg trace formula andthe Eichler-Selberg trace formula in a unified way. We also see that it is in factequivalent with the Kuznetsov summation formula, since the summation formulacan be used to prove the Kuznetsov summation formula in a simple way.

BIBLIOGRAPHY

[1] J. Andersson, A summation formula on the modular group and the pre-traceformula, forthcoming.

[2] F. V. Atkinson, The mean-value of the Riemann zeta function, Acta Math.81 (1949), 353–376.

[3] V. Bykovsky, N. Kuznetsov, and A. Vinogradov, Generalized summation for-mula for inhomogeneous convolution, Automorphic functions and their appli-cations (Khabarovsk, 1988), Acad. Sci. USSR Inst. Appl. Math., Khabarovsk,1990, pp. 18–63.

[4] W. L. Ferrar, Summation formulae and their relation to Dirichlet series, Com-positio Math. (1935), no. 1, 344–360.

[5] N. V. Kuznetsov, The Petersson conjecture for cusp forms of weight zero andthe Linnik conjecture. Sums of Kloosterman sums, Mat. Sb. (N.S.) 111(153)(1980), no. 3, 334–383, 479.

[6] , Convolution of Fourier coefficients of Eisenstein-Maass series, Zap.Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI) 129 (1983), 43–84, Automorphic functions and number theory. I. MR 86h:11039

[7] R. Matthes, An elementary proof of a formula of Kuznecov for Kloostermansums, Resultate Math. 18 (1990), no. 1-2, 120–124.

[8] Y. Motohashi, An explicit formula for the fourth power mean of the Riemannzeta-function, Acta Math. 170 (1993), no. 2, 181–220.

[9] , The binary additive divisor problem, Ann. Sci. Ecole Norm. Sup. (4)27 (1994), no. 5, 529–572.

[10] , Spectral theory of the Riemann zeta-function, Cambridge UniversityPress, Cambridge, 1997.

[11] A. Terras, Harmonic analysis on symmetric spaces and applications. I,Springer-Verlag, New York, 1985.

A note on some Kloosterman sum identities

Johan Andersson∗

Abstract

In this paper we will give an elementary proof of the identity

S(D,m, n; c) =∑

d|(D,m,c)

d S

(mD

d2, n;

c

d

),

where

S(D,m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

)

are certain generalized Kloosterman sums and S(m,n; c) = S(1,m, n; c)denote the classical Kloosterman sums. This identity was first statedin Bykovsky-Kuznetsov-Vinogradov [4]. Heath-Brown [5] had previouslyproved a closely related result. In particular this will give us a new proofof the Selberg identity

S(m,n; c) =∑

d|(m,n,c)

d S(mn

d2, 1;

c

d

).

The identity was given by Selberg [12] without proof. Kuznetsov [9] pub-lished the first proof using his summation formula. Matthes [10] has alsoprovided an alternative elementary proof.

1 Kloosterman sums

Kloosterman sums

S(m, n; c) =∑

hh≡1 (mod c)0≤h,h<c

e

(mh + nh

c

) (e(x) = e2πix

)(1)

∗Department of Mathematics, Stockholm University, johana@math.su.se

99

have had a lot of applications in analytic number theory ever since Kloosterman’spaper [8]. An important estimate is Weil’s [13] algebro-geometric estimate

|S(m, n; c)| ≤ d(c)√

c · (m, n, c), (2)

where (m, n, c) is the greatest common divisor, and d(n) =∑

d|n 1 denotes thedivisor function. This estimate follows from his result on the Riemann hypothesisfor curves. In the seventies Kuznetsov [9] developed a summation formula1 thathas since then been of prime importance, and is an essential point in Motohashi’swork [11] on the Riemann zeta function. Kuznetsov obtained a summation for-mula of the type

∞∑c=1

Φ(c)S(m, n; c) =∞∑

j=1

ρj(n)ρj(m)Φm,n(κj) + “other similar terms”, (3)

where Φm,n is a certain integral transform of Φ, ρj(n) are certain Fourier co-efficients of Maass wave forms, and the “other similar terms” is the spectralcontribution coming from the holomorphic cusp forms and the Eisenstein series.For an introduction to “Kloostermania”, the theory of Kloosterman sums andtheir applications, see Huxley [6].

2 Main results

Theorem 1. (Bykovsky-Kuznetsov-Vinogradov identity) One has the identity

S(D, m, n; c) =∑

d|(D,m,c)

dS

(mD

d2, n;

c

d

),

where

S(D, m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

)

are generalized Kloosterman sums.

Proof. Whenever

hk ≡ D (mod c), 0 ≤ h, k < c,

1Kuznetsov [9] and Bruggeman [3] independently developed the closely related trace formula.

100

we have with

p =hk −D

c,

that (h pc k

)is an integer matrix of determinant D. From the theory of Hecke-operators (seee.g Motohashi [11] pages 97-98), we know that it can be written uniquely as aproduct (

h pc k

)=

(a b0 d

)·(

x ql y

),

where

ad = D, 0 ≤ b < d,

and the second factor (x ql y

)is an element in SL(2, Z). We get that(

h pc k

)=

(a b0 d

)(x ql y

),

=

(bl + ax aq + by

dl dy

),

which gives us

l =c

d, y =

k

d, and a =

D

d.

Let us also assume that x is defined so that 0 ≤ x < l and x ≡ y (mod l). Thisimplies that

e

(mh + nk

c

)= e

(m(bl + ax) + ndx

c

).

101

Summing over hk ≡ D (mod c) will now correspond to summing over d|(c, D),xx ≡ 1 (mod c/d), 0 < x, x < c/d and b = 0, . . . , d− 1. We have that

S(D, m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

),

=∑

d|(c,D)

∑xx≡1 (mod c/d)

0≤x,x<c/d

d−1∑b=0

e

(m(bl + ax) + ndx

c

),

=∑

d|(c,D)

∑xx≡1 (mod c/d)

0≤x,x<c/d

e

(max + ndx

c

) d−1∑b=0

e

(mbl

c

).

By applying the identity

d−1∑b=0

e

(mbl

c

)=

d−1∑b=0

e

(mb

d

)=

d, d|m,

0, otherwise,l =

c

d,

we deduce that

S(D, m, n; c) =∑

d|(c,D,m)

d∑

xx≡1 (mod c/d)0≤x,x<c/d

e

(max + ndx

c

),

=∑

d|(c,D,m)

dS

(mD

d2, n,

c

d

).

Theorem 1 was stated in Bykovsky-Kuznetsov-Vinogradov [4] (see page 48,equation (2.29)). It is not proved in their paper though. Instead they referto Heath-Brown [5] for proof. Heath-Brown never proves Theorem 1, but thecorresponding identity for

u(n; q, r) =∑ab=n

S(r, a, b; q).

See [5], equation (51) page 411. He does not explicitly define S(D, m, n; c). Hisproof applies in this setting as well though, and it is similar to our proof. It shouldalso be mentioned that M.N. Huxley [7] found a similar argument independentlyof the author. Neither Heath-Brown [5] nor Huxley [7] mentions Hecke operators.We believe they are a key point if we would like to generalize the theorem to moregeneral Kloosterman sums associated with arithmetical groups.

102

Bykovsky-Kuznetsov-Vinogradov [4] used Theorem 1 to prove a generalizedKuznetsov summation formula (compare with (3)):

∞∑c=1

Φ(c)S(D, m, n; c) =∞∑

j=1

tj(D)ρj(n)ρj(m)ΦD,m,n(κj) + “other similar terms”

(4)

where

tj(D) =ρj(D)

ρj(1)

are the eigenvalues of the Hecke operator TD acting on the Maass wave form. Theidentity in Theorem 1 will occur in the same setting in our forthcoming work [2],where we develop a summation formula over integer matrices with determinantD. In fact we needed a formula like (4) as a Lemma for our other work and haddeveloped an independent proof of (4) when K. Matsumoto made an interestingremark at our seminar [1] at the Millennial conference of number theory. Theresult had previously been proved by Bykovsky-Kuznetsov-Vinogradov [4]!

Theorem 2. One has the identity

S(n1, n2, n3; c) = S(nσ(1), nσ(2), nσ(3); c)

where S(n1, n2, n3; c) is defined as in Theorem 1 and σ ∈ S3 is any permutationacting on 1, 2, 3.

Proof. From the definition of the generalized Kloosterman sums

S(D, m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

)

it is clear that S(D, m, n; c) = S(D, n, m; c). Hence it is sufficient to prove thatS(D, m, n; c) = S(m, D, n; c). But the right hand side of the identity in Theorem1

S(D, m, n; c) =∑

d|(D,m,c)

dS

(mD

d2, n;

c

d

),

is symmetric in D, m, hence so is the left hand side.

Even though Bykovsky-Kuznetsov-Vinogradov [4] gives the identity in Theo-rem 1 they never mention the symmetry of Theorem 2. One reason why we havechosen the notation S(D, m, n; c)2 is to point out this symmetry. It is not at allobvious from its definition.

2Bykovsky-Kuznetsov-Vinogradov [4] used the notation Sc(m,n;D).

103

Corollary. (The Selberg identity) We have that

S(m, n; c) =∑

d|(m,n,c)

dS(mn

d2, 1;

c

d

).

Proof. We have

S(m, n; c) = S(1, m, n; c),

which by using Theorem 2 equals

= S(n, m, 1; c),

which by using Theorem 1 equals

=∑

d|(m,n,c)

dS(mn

d2, 1;

c

d

).

This is the classical Selberg identity. It was given by Selberg [12] withoutproof, and Kuznetsov [9] published the first proof using his summation formula.Matthes [10] has given an elementary proof, by division into cases and usingmultiplicative properties of the Kloosterman sums. It is also of interest to notethat Kuznetsov’s method of proof used the Hecke operators and his summationformula. Matthes gave an elementary proof, without the summation formula andwithout Hecke operators. We give an elementary proof but we use the notion ofHecke operators.

3 Acknowledgments

We would like to thank Martin Huxley for his interest in the problem, and forencouraging us to publish our results.

References

[1] J. Andersson. A Poisson summation formula for SL(2, Z). Conferenceabstract, Millenial conference in number theory, Urbana Champaign,http://atlas-conferences.com/c/a/e/w/11.htm, 2000.

[2] A summation formula over integer matrices, 2006.

104

[3] R. W. Bruggeman. Fourier coefficients of cusp forms. Invent. Math., 45(1):1–18, 1978.

[4] V. Bykovsky, N. Kuznetsov, and A. Vinogradov. Generalized summationformula for inhomogeneous convolution. In Automorphic functions and theirapplications (Khabarovsk, 1988), pages 18–63. Acad. Sci. USSR Inst. Appl.Math., Khabarovsk, 1990.

[5] D. R. Heath-Brown. The fourth power moment of the Riemann zeta function.Proc. London Math. Soc. (3), 38(3):385–422, 1979.

[6] M. N. Huxley. Introduction to Kloostermania. In Elementary and analytictheory of numbers (Warsaw, 1982), pages 217–306. PWN, Warsaw, 1985.

[7] . Three parameter kloosterman sum, private communication. 2002.

[8] H. D. Kloosterman. On the representation of numbers in the form ax2 +by2 + cz2 + dt2. Acta Math., 49:407–464, 1926.

[9] N. V. Kuznetsov. The Petersson conjecture for cusp forms of weight zeroand the Linnik conjecture. Sums of Kloosterman sums. Mat. Sb. (N.S.),111(153)(3):334–383, 479, 1980.

[10] R. Matthes. An elementary proof of a formula of Kuznecov for Kloostermansums. Resultate Math., 18(1-2):120–124, 1990.

[11] Y. Motohashi. Spectral theory of the Riemann zeta-function. CambridgeUniversity Press, Cambridge, 1997.

[12] A. Selberg. Uber die Fourierkoeffizienten elliptischer Modulformen negativerDimension. In C.R. Neuvieme Congres Math Scandinaves, pages 320–322.Helsinki, 1938.

[13] A. Weil. On some exponential sums. Proc. Nat. Acad. Sci. U. S. A., 34:204–207, 1948.

105

A note on some Kloosterman sum identities II

Johan Andersson∗

Abstract

In a previous paper [1] we gave new proof of Kloosterman sum identitiesof Bykovsky-Kuznetsov-Vinogradov and Selberg. In this paper we willgeneralize the identities to Kloosterman sums and generalized Kloostermansums over Hecke congruence subgroups Γ0(N) . Let c be such that N |cand χ a Dirichlet character modulo N . We will define the generalizedKloosterman sum associated to Γ0(N) with character χ to be

Sχ(D,m, n; c) =∑

ad≡D (mod c)0≤a,d<c

χ(k)e(

ma + nd

c

).

Here Sχ(m,n; c) = Sχ(1,m, n; c) denotes the classical Kloosterman sumassociated to Γ0(N) and character χ. We will prove the identities

Sχ(D,m, n; c) =∑

d|(D,m,c)

χ(d)dSχ

(mD

d2, n;

c

d

),

and

Sχ(m,n; c) = χ(n)∑

d|(m,n,c)

χ(d)dSχ

(mn

d2, 1;

c

d

). ((n, N) = 1)

The first identity is a generalization of an identity by Bykovsky-Kuznetsov-Vinogradov [3]. The last identity is a generalization of an identity of Sel-berg [7].

Contents

1 Kloosterman sums 108

∗Department of Mathematics, Stockholm University, johana@math.su.se

107

2 Main results 109

1 Kloosterman sums

The classical Kloosterman sums

S(m, n; c) =∑

hh≡1 (mod c)0≤h,h<c

e

(mh + nh

c

), (e(x) = e2πix) (1)

appear naturally from the Bruhat decomposition (double coset decomposition)of the full modular group. To understand the Kloosterman sums from a grouptheoretical perspective it is more convenient to write them as

S(n, m; c) =∑

( a bc d )∈PSL(2,Z)

0≤a,d<c

e

(ma + nd

c

). (2)

The Hecke congruence subgroups Γ0(N) is the subgroup of the full modular groupPSL(2, Z) given by

Γ0(N) =

(a bc d

)∈ PSL(2, Z) : c ≡ 0 (mod N)

. (3)

The Kloosterman sums associated to a character modulo N , and to the groupΓ0(N), can now be defined similarly to (2) as (see Venkov [8] pp. 32)

Sχ(n, m; c) =∑

( a bc d )∈Γ0(N)

0≤a,d<c

χ(d)e

(ma + nd

c

). (4)

Similarly as in our paper [1] we will also consider generalized Kloosterman sumswhere Γ0(N) is replaced by its image under Hecke operators. We define

Sχ(D, n, m; c) =∑

ad−bc=D0≤a,d<c

χ(d)e

(ma + nd

c

). (c|N) (5)

We have that Sχ(1, n, m; c) = Sχ(n, m; c), and the Kloosterman sum defined by(4) is thus a proper generalization of the classical Kloosterman sum associatedwith Γ0(N).

108

2 Main results

Theorem 1. (generalized Bykovsky-Kuznetsov-Vinogradov identity) We havethat

Sχ(D, m, n; c) =∑

d|(D,m,c)

χ(d) Sχ

(mD

d2, n;

c

d

),

where Sχ(D, m, n; c) defined by (5) are generalized Kloosterman sums.

Proof. By the definition (5)

Sχ(D, n, m; c) =∑

ad−bc=D0≤a,d<c

χ(d)e

(ma + nd

c

).

we have that (a bc d

)is an integer matrix of determinant D. From the theory of Hecke-operators (seee.g Motohashi [6] pages 97-98), we know that it can be written uniquely as aproduct (

a bc d

)=

(α β0 δ

)·(

x ql y

),

where

αδ = D, 0 ≤ β < δ,

and (x ql y

)is an element in SL(2, Z). We get that(

a bc d

)=

(βl + αx αq + βy

δl δy

),

which gives us

α =D

δ, d = δy, and l =

c

δ

109

Let us also assume that x is defined so that 0 ≤ x < l and x ≡ y (mod l). Thisimplies that

e

(ma + nd

c

)= e

(m(βl + αx) + nδx

c

).

Summing over ad − bc = D hence corresponds to summing over δ|(c, D), xx ≡1 (mod c/δ), and β = 0, . . . , δ − 1. We get

Sχ(D, n, m; c) =∑

ad−bc=D0≤a,d<c

χ(d)e

(ma + nd

c

),

=∑

δ|(c,D)

∑xx≡1 (mod c/δ)

0≤x,x<c/δ

χ(δx)d−1∑β=0

e

(m(βl + αx) + nδx

c

),

which by the multiplicative properties χ(ab) = χ(a)χ(b) and e(a + b) = e(a)e(b)equals

=∑

δ|(c,D)

χ(δ)∑

xx≡1 (mod c/δ)0≤x,x<c/d

χ(x)e

(mαx + nδx

c

) d−1∑β=0

e

(mβl

c

).

By applying the identity

δ−1∑β=0

e

(mβl

c

)=

δ−1∑β=0

e

(mβ

δ

)=

δ, δ|m,

0, otherwise,l =

c

δ,

this implies that

Sχ(D, m, n; c) =∑

δ|(c,D,m)

χ(δ) δ∑

xx≡1 (mod c/δ)0≤x,x<c/δ

e

(max + nδx

c

),

which by the fact that α = D/δ and the definition of the Kloosterman sumsequals

=∑

δ|(c,D,m)

χ(δ) δ Sχ

(mD

δ2, n,

c

δ

).

110

Remark 1. The proper way of doing the proof would of course be to use theHecke operators of the congruence subgroups, instead of referring to the Heckeoperators of the full modular group. In this case the Hecke operators are thesame (see e.g. Atkin-Lehner [2]), and that is why it still works.

Similarly as in the case of classical Kloosterman sum we have a Kuznetsovsummation formula (see e.g Venkov [8] chapter 5).

∞∑c=1

Φ(c)Sχ(m, n; c) =∞∑

j=1

ρj,χ(n)ρj,χ(m)Φm,n(κj) + “other terms”,

where Φm,n is a certain integral transform of Φ, and ρj(n) are certain Fouriercoefficients of Maass wave forms. The “other terms” are the corresponding termscoming from the holomorphic cusp forms and the Eisenstein series. In the sameway as in Bykovsky-Kuznetsov-Vinogradov [3], it will now be possible to developa generalized Kuznetsov summation formula.

∞∑c=1

Φ(c)S(D, m, n; c) =∞∑

j=1

tj,χ(D)ρj,χ(n)ρj,χ(m)Φm,n(κj) + “other terms”

where tj,χ(D) is the eigenvalue for the Hecke operator acting on the Maass waveform. We will return to this topic in a later paper. The identity in Theorem 1will also allow us to develop the Selberg identity for these Kloosterman sums.

Theorem 2. One has the following identities

(i) Sχ(D, m, n; c) = Sχ(m, D, n; c),

(ii) Sχ(D, m, n; c) = χ(D)Sχ(D, n, m; c) whenever (D, N) = 1.

Proof. To prove (i) we notice that the right hand side of the identity in Theorem1

Sχ(D, m, n; c) =∑

d|(D,m,c)

χ(d)dSχ

(mD

d2, n;

c

d

), (6)

is symmetric in D, m, hence so is the left hand side. To prove (ii) We use thedefinitions of the generalized Kloosterman sums

Sχ(D, m, n; c) =∑

hk≡D (mod c)

χ(k)e

(mh + nk

c

), (7)

Since (D, N) = 1, since N |c, and since hk ≡ D (mod N) we get that

χ(hk) = χ(D), and χ(k) = χ(D)χ(h),

111

and finally

Sχ(D, m, n; c) = χ(D)∑

hk≡D (mod c)

χ(h)e

(mh + nk

c

).

Corollary. (generalized Selberg identity) We have that if (n, N) = 1 1 then

Sχ(m, n; c) = χ(n)∑

d|(m,n,c)

χ(d)dSχ

(mn

d2, 1;

c

d

). (8)

Proof. By the fact that Sχ(m, n, c) = Sχ(1, m, n; c) we have

Sχ(m, n; c) = Sχ(1, m, n; c),

which by using Theorem 2 (ii) equals

= Sχ(1, n, m; c),

which by using Theorem 2 (i) equals

= Sχ(n, 1, m; c),

which by using Theorem 2 (ii) equals

= χ(n)Sχ(n, m, 1; c).

which by using Theorem 1 equals

= χ(n)∑

d|(m,n,c)

χ(d)dSχ

(mn

d2, 1;

c

d

).

This is the analogue of the Selberg identity. In the case of the classical Kloost-erman sums it was given by Selberg [7] without proof, and proved by Kuznetsov[4] with alternative elementary proofs given in Matthes [5] and Andersson [1].

1can this condition be removed and we can still find a similar formula? It can be replacedby (m,N) = 1. But in general?

112

References

[1] J. Andersson. A note on some Kloosterman sum identities, 2006.

[2] A. O. L. Atkin and J. Lehner. Hecke operators on Γ0(m). Math. Ann.,185:134–160, 1970.

[3] V. Bykovsky, N. Kuznetsov, and A. Vinogradov. Generalized summationformula for inhomogeneous convolution. In Automorphic functions and theirapplications (Khabarovsk, 1988), pages 18–63. Acad. Sci. USSR Inst. Appl.Math., Khabarovsk, 1990.

[4] N. V. Kuznetsov. The Petersson conjecture for cusp forms of weight zeroand the Linnik conjecture. Sums of Kloosterman sums. Mat. Sb. (N.S.),111(153)(3):334–383, 479, 1980.

[5] R. Matthes. An elementary proof of a formula of Kuznecov for Kloostermansums. Resultate Math., 18(1-2):120–124, 1990.

[6] Y. Motohashi. Spectral theory of the Riemann zeta-function. CambridgeUniversity Press, Cambridge, 1997.

[7] A. Selberg. Uber die Fourierkoeffizienten elliptischer Modulformen negativerDimension. In C.R. Neuvieme Congres Math Scandinaves, pages 320–322.Helsingfors, 1938.

[8] A. B. Venkov. Spectral theory of automorphic functions and its applications.Kluwer Academic Publishers Group, Dordrecht, 1990. Translated from theRussian by N. B. Lebedinskaya.

113

A summation formula on the full modular group

Johan Andersson∗

Abstract

In this paper we will develop a new type of formula that will allow us to expanda sum over PSL(2, Z) for test functions f defined on PSL(2, R)∑

“a bc d

”∈PSL(2,Z)

f

(a bc d

)

into spectral objects such as Fourier coefficients for Maass wave forms, holomorphiccusp forms and Eisenstein series. A similar result, the pre-trace formula has previ-ously been known in the special case when f is SO(2) bi-invariant. Our proof is asfollows: Use the Bruhat decomposition to express the sum as a sum of Kloostermansums and then apply the Kuznetsov summation formula.

Contents

1 Introduction 1151.1 The Poisson summation formula . . . . . . . . . . . . . . . . . . . . . . . . 1151.2 The theory of modular forms . . . . . . . . . . . . . . . . . . . . . . . . . . 1161.3 Kloosterman sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

2 Main result 119

3 Some remarks 123

4 Appendix - Integral transforms 124

1 Introduction

1.1 The Poisson summation formula

It is fairly trivial and straightforward to obtain an analogue of the classical Poisson sum-mation formula ∑

n∈Z

f(n) =∑n∈Z

f(n)

∗Department of Mathematics, Stockholm University, johana@math.su.se

115

from Fourier analysis in the case of finite abelian groups, where we have similarly∑g∈G

f(g) =∑g∈G

f(g), (1)

where the dual group G denotes the group of the characters on G, that is functionsρ : G → C such that ρ(g1)ρ(g2) = ρ(g1 + g2). The case of general locally compact abeliangroups can also be treated although it will be a bit more complicated since the dual groupwill be non compact. However in the case of non abelian groups the relevant representationtheory is non trivial. The concept of a dual group does not exist anymore. In this paperwe will consider the non abelian locally compact group PSL(2, Z), the full modular group,and develop an equation where the left hand side is∑

( a bc d )∈PSL(2,Z)

f

(a bc d

).

What will the right hand side be? It will not have a group theoretical structure but wewill see that it can be expressed in terms of the spectral theory of the full modular group.It will come from the classical holomorphic modular forms as well as the Eisenstein seriesand the more mysterious Maass wave forms [11].

1.2 The theory of modular forms

We will start with some definitions, and a short review of the theory of modular forms1.The Poincare upper half plane is defined as

H = z = x + iy : x ∈ R, y ∈ R+

An element g in PSL(2, R) acts on H by Moebius transformations(a bc d

)· z =

az + b

cz + d,

and maps the Poincare upper half plane on itself. A function f defined on H is said to bemodular if

f

((a bc d

)· z)

= f(z).

We have that

F = x + iy : −1 < x ≤ 1, x2 + y2 ≥ 1

is a fundamental domain of the full modular group. That means that for each elementz ∈ H there exist a unique element γ ∈ PSL(2, Z) such that γz ∈ F . We define thePetersson inner product as

〈f, g〉 =

∫F

f(z)g(z)dxdy

y2.

1See e.g. Iwaniec [8] or Motohashi [12].

116

A modular form is a modular function that furthermore is an eigenfunction of the hyper-bolic Laplacian

∆ = y2

(∂2

∂x2+

∂2

∂y2

).

It is clear that any modular function is periodic in z with period 1, since(1 n0 1

)· z = z + n,

and hence every modular function f has a Fourier series in x

f(x + iy) =∞∑

n=−∞

fn(y)e(nx).(e(x) = e2πix

)We say that f is a cusp form if it is a modular form and its constant in the Fourier seriesvanishes, that is f0(y) = 0. The cusp forms for the full modular group are called Maasswave forms, and they are enumerable. We will use the notation

Ψj(x + iy) =∞∑

n=−∞

ρj(n)e(nx)Kiκj(|2πny|), ∆Ψj =

(1

4+ κ2

j

)Ψj(z) (2)

where Kiκj(x) denote the K-Bessel functions and Ψj∞j=1 is an orthonormalized basis of

Maass wave forms and the norm comes from the Petersson inner product. The Eisensteinseries

E(z; s) =y1−s +Γ(s− 1/2)ζ(2s− 1)

Γ(s)ζ(2s)

+2π2√y

Γ(s)ζ(2s)

∑n6=0

|n|s−1/2σ1−2s(|n|)Ks−1/2(2π|n|y)e(nx)

are the only non cusp modular forms. A fundamental result (Maass [11], Selberg [15]) isthat the modular forms span the space of modular functions. We have

f(z) =∞∑

j=1

〈f, Ψj〉Ψj(z) +1

π

∫ ∞

−∞

⟨f, E

(·; 1

2+ ir

)⟩E(z; 1

2+ ir

)dr.

Similarly we will let

Ψj,kθ(k)j=1 , (3)

be an orthonormal basis of holomorphic cusp forms of weight k. We will let ρj,k(n) be thecoefficients in its power series

Ψj,k(z) =∞∑

n=1

ρj,k(n)zn.

117

The cusp forms of weight k fulfill the transformation equation

f

((a bc d

)· z)

= (cz + d)2kf(z).

and the Petersson inner product will be

〈f, g〉 =

∫F

f(z)g(z)y2k−1dxdy.

1.3 Kloosterman sums

Kloosterman sums

S(m, n; c) =∑

hh≡1 (mod c)0≤h,h<c

e

(mh + nh

c

)(4)

have had a lot of applications in analytic number theory ever since Kloosterman’s pa-per [9]. From their definition we immediately deduce some elementary properties of theKloosterman sums such as the symmetry

S(m, n; c) = S(n, m; c), (5)

as well as the facts that

S(m, 0; c) = cc(m), (6)

and

S(0, 0; c) = φ(c), (7)

where

cc(m) =∑

(h,c)=10≤h<c

e

(mh

c

)(8)

denote the Ramanujan sums, and φ(c) denotes Euler’s phi function. An important nontrivial estimate of the Kloosterman sums is Weil’s [18] algebro-geometric estimate

|S(m, n; c)| ≤ d(c)√

c · (m, n, c), (9)

where (m, n, c) is the greatest common divisor, and d(n) =∑

d|n 1 denotes the divisorfunction. It follows from his result on the Riemann hypothesis for curves. In the seventiesKuznetsov [10] developed an important summation formula2 that has since then been auseful tool in the analytic theory of numbers3. He developed the summation formula

2Kuznetsov [10] and Bruggeman [5] independently developed the closely related trace formula.3For example it is an essential point in Motohashi’s [12] work on the fourth power moment of the

Riemann zeta function.

118

Lemma 1. (Kuznetsov summation formula) Suppose that Φ ∈ C3(0,∞) is a function onR+, and that for some δ > 0 and v = 0, 1, 2, 3 it satisfies

∣∣Φ(v)(t)∣∣ t−v

t32+δ + t−δ

. (10)

Then

∞∑c=1

S(m, n; c)Φ(c) =∞∑

j=1

ρj(m)ρj(n)Φm,n(κj) +1

π

∫ ∞

−∞

Φm,n(r)σ2ir(|m|)σ2ir(|n|)dr

|mn|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

ρj,k(m)ρj,k(n)Φm,n

(12− ki

),

where

Φm,n(r) =

∫ ∞

0

Mr

(mn

t2

)Φ(t)dt,

and Mr is defined by

Mr(x) =

iπ

2 sinh πr(J2ri(4π

√x)− J−2ri(4π

√x)) , x ≥ 0,

2 cosh πrK2ri(4π√−x), x < 0.

(11)

Proof. After the change of variables

Φ(t) =1

tφ

(4π

√|mn|t

)(12)

it reduces to the classical Kuznetsov summation formula (This version of the formula istaken from Motohashi [12], Theorem 2.3 and Theorem 2.5).

2 Main result

We have now provided the necessary tools to state and prove our summation formula.

Theorem 1. Let f be a function on PSL(2, R), fulfilling the condition∣∣∣∣ ∂n

∂ξn

∂m

∂ηm

∂v

∂tvf

(ξt ∗t ηt

)∣∣∣∣ t−v

(t2+δ + t−1−δ)(ξ2 + η2 + 1)1+δ,

n, m, v = 0, 1, 2, 3,t > 0.

(13)

Let ∗ denote the number making the determinant of the matrix to be 1. One then has the

119

identity

∑ad−bc=1

c>0

f

(a bc d

)= −

∞∑c=1

φ(c)

∫ ∞

−∞

∫ ∞

−∞f

(cξ ∗c cη

)dξdη

+∞∑

n=−∞

∞∑c=1

cc(n)

∫ ∞

−∞

(f

(cn ∗c cξ

)+ f

(cξ ∗c cn

))dξ

+∞∑

j=1

∑m,n6=0

ρj(m)ρj(n)F (κj; m, n) +∑

m,n6=0

1

π

∫ ∞

−∞

σ2ir(|m|)σ2ir(|n|)F (r; m, n)dr

|nm|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

∑m,n6=0

ρj,k(m)ρj,k(n)F((

12− k)i; m, n

)where

F (r; m, n) =

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(mξ + nη)Mr

(mn

t2

)f

(tξ ∗t tη

)dξdηdt, (14)

Mr is defined by (11), K2ri(x) and J2ri(x) are the K-Bessel and J-Bessel-functions respec-tively, ρj(n) and ρj,k(n) are defined by (2), and (3) are Fourier coefficients of Maass waveforms and holomorphic modular forms respectively, cc(n) are the Ramanujan sums definedby (8) and φ is the Euler phi function. The integral transform can also be given in Iwasawacoordinates

F (r; m, n) =

∫g∈PSL(2,R)

e(mx)e(−(my + n) cot θ)Mr

(mny

sin2 θ

) f(g)√

y

sin θdg, (15)

where

g =

(1 x0 1

)·(√

y 00 1√

y

)·(

cos θ sin θ− sin θ cos θ

), (16)

and the integral is with respect to the Haar measure of the group.

Remark 1. Even if we have not bothered to get the best possible test function class (13)in Theorem 1, it is in fact a rather wide class of test functions, and contains e.g. the class

C∞0 (PSL(2, R))

of all smooth functions with compact support.

We will now give the simplest proof we have found4 (assuming the Kuznetsov summa-tion formula) for our main result. We will use results on integral transforms and Besselfunctions proved in the Appendix.

4This proof was presented at the millennial conference on number theory, Urbana-Champaign, May2000 see conference abstract [1]. A different and a bit more involved proof had been presented in Turkuin October 1999. We will return to that in [3].

120

Proof. We will now use the Bruhat decomposition, a simple change of summation trick,and the Kuznetsov summation formula to prove our theorem. By letting a = mc + h, andd = nc + h, with 0 ≤ h, h < c, we obtain∑

ad−bc=1c>0

f

(a bc d

)=

∞∑c=1

∞∑m,n=−∞

∑hh≡1 (mod c)

0≤h,h<c

f

(mc + h ∗

c nc + h

)

=∞∑

c=1

∑hh≡1 (mod c)

0≤h,h<c

∞∑m,n=−∞

fc

(h

c+ m,

h

c+ n

),

where

ft(ξ, η) = f

(tξ ∗t tη

). (17)

Condition (13) for ξ and η assures us that we can use the shifted double Poisson summationformula (see e.g Zygmund [19] (13.3)), and we get

∞∑c=1

∑hh≡1 (mod c)

0≤h,h<c

∞∑m,n=−∞

e

(mh + nh

c

)fc(m, n), (18)

where

ft(m, n) =

∫ ∞

−∞

∫ ∞

−∞e(mξ + nη)ft(ξ, η)dξdη (19)

is the inverse two dimensional Fourier transform of ft(ξ, η). Lemma 10 ensures that thissum is absolutely convergent. We can change the order of summation in (18) and we obtain

∞∑m,n=−∞

∞∑c=1

∑hh≡1 (mod c)

0≤h,h<c

e

(mh + nh

c

)fc(m, n) =∞∑

m,n=−∞

∞∑c=1

S(m, n; c)fc(m, n), (20)

and the Kloosterman sums appear. Since f fulfills (13), we have by Lemma 5 in theAppendix that the function Φ(t) = ft(m, n) is of the type (10). Hence we can applyLemma 1, the Kuznetsov summation formula and we find that the contribution comingfrom the terms where mn 6= 0 is equal to

∑m,n6=0

[∞∑

j=1

ρj(m)ρj(n)F (κj; m, n)

+1

π

∫ ∞

−∞

σ2ir(|m|)σ2ir(|n|)F (r; m, n)dr

|mn|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

ρj,k(m)ρj,k(n)F((

12− k)i; m, n

)]. (21)

121

From the estimate for the Fourier coefficients of cusp forms∑K/2<κj≤K

|ρj(m)|2 K2 + d3(m) log(2m)√

m, (22)

θ(k)∑j=1

|ρj,k(m)|2 kd3(m)√

m log(2m), (23)

valid uniformly for k,K, m ≥ 1 (see Motohashi [12] Lemma 2.4, and equation (2.2.10)),and the fact that

|F (r; m, n)| |mn|−94 (1 + |r|)−

52 ,∣∣F((1

2+ k)i; m, n

)∣∣ |mn|−2k−73 ,

when f satisfies (13) (see the Appendix Lemma 12 and Lemma 13), we see that thesummation and summation-integration in (21) is absolutely convergent, and hence we canchange the summation order, and we thus obtain the spectral part of Theorem 1. Wewill now consider the contribution coming from m = 0 or n = 0. We get from (20) thecontribution ∑

m=0 or n=0

∞∑c=1

S(m, n; c)fc(m, n), (24)

By equations (5),(6) and (7) this equals

∞∑n=−∞

∞∑c=1

cc(n)(fc(0, n) + fc(0, n)

)−

∞∑c=1

φ(c)fc(0, 0). (25)

By applying the Poisson summation formula on the first term in (25) again, and usingequations (17) and (19) it equals

∞∑n=−∞

∞∑c=1

cc(n)

∫ ∞

−∞

(f

(cn ∗c cξ

)+ f

(cξ ∗c cn

))dξ

−∞∑

c=1

φ(c)

∫ ∞

−∞

∫ ∞

−∞f

(cξ ∗c cη

)dξdη

which account for the second and first term in the theorem.To show how to express the integral transform in Iwasawa coordinates we start with

the integral transform (14)

F (r; m, n) =

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(mξ + nη)Mr

(mn

t2

)f

(tξ ∗t tη

)dξdηdt. (26)

By using the explicit expression of a group element g ∈ SL(2, R), by equation (16) intoIwasawa coordinates, we obtain

g =

(1 x0 1

)·(√

y 00 1√

y

)·(

cos θ sin θ− sin θ cos θ

)=

(√y cos θ − x sin θ√

ysin θ

√y + x cos θ√

y

− sin θ√y

cos θ√y

).

122

Using the change of variables(ξt ∗t ηt

)=

(√y cos θ − x sin θ√

ysin θ

√y + x cos θ√

y

− sin θ√y

cos θ√y

),

in equation (26), we get the equations

ξ = −y cot θ + x, t = −sin θ√

y, and η = − cot θ.

We have the Jacobian∣∣∣∣∣∣∂η∂θ

∂t∂θ

∂ξ∂θ

∂η∂y

∂t∂y

∂ξ∂y

∂η∂x

∂t∂x

∂ξ∂x

∣∣∣∣∣∣ =

∣∣∣∣∣∣∣− 1

sin2 θ− cos θ√

y− y

sin2 θ

0 sin θ2y3/2 − cot θ

0 0 1

∣∣∣∣∣∣∣ = − 1

2y3/2 sin θ,

since the matrix is upper triangular and the determinant equals the product of the diagonalelements. We get

dξdηdt = − dθdxdy

2y3/2 sin θ,

and we see that

F (r; m, n) =

∫ ∞

−∞

∫ ∞

0

∫ 2π

0

e(m(−y cot θ + x)− n cot θ)Mr

(mny

sin2 θ

)×

× f

(√y cos θ − x sin θ√

ysin θ

√y + x cos θ√

y

− sin θ√y

cos θ√y

)dθdxdy

2y3/2 sin θ. (27)

Using the fact that the Haar measure on the group PSL(2, R) can be given by∫g∈PSL(2,R)

f(g)dg =1

2

∫ 2π

0

∫ ∞

−∞

∫ ∞

0

f(g)dydxdθ

y2,

we can write equation (27) as

F (r; m, n) =

∫g∈PSL(2,R)

e(mx)e(−(my + n) cot θ)Mr

(mny

sin2 θ

) f(g)√

y

sin θdg,

where g is given in Iwasawa coordinates by equation (16).

3 Some remarks

The method used in this papers also generalizes to sums over congruence subgroups andcharacters, and the Kuznetsov summation formula has been generalized to that setting by

123

Proskurin ([13], [14]). It should also be noted that from the inversion formulae for theTitchmarsh transforms

Lf (t) =

∫ ∞

0

(J2it(y)− J−2it(y))

2i sinh πt

f(y)dy

y

f(y) =

∫ ∞

−∞Lf (t)

(J2it(y)− J−2it(y))

2 sin πittanh(πt)tdt,

and Kontorevich-Lebedev transforms

Lf (t) =

∫ ∞

0

Kit(y)f(y)dy

y

f(y) =

∫ ∞

−∞Lf (t)Kit(y)

sinh(πt)

π2tdt,

see Iwaniec [8], page 228-231, our main summation formula can be inverted. When f isSO(2) invariant our formula will reduce to the spectral expansion of a modular function,and when f is SO(2) bi-invariant it will reduce to the pre-trace formula. Even if the proofis fairly straightforward, it tends to become a bit technical (a certain tricky integral) so wewill defer the proof to a forthcoming paper [2].

4 Appendix - Integral transforms

In this section we will prove the estimates for integral transforms that are needed in thepaper. We will first calculate the Mellin transform and Mellin-Barnes integral of the kernelfunction Mr as defined by (11).

Lemma 2. One has the following Mellin-transform and Mellin-Barnes representations ofthe Kernel function Mr

(i)

∫ ∞

0

Mr(ε · x)xs−1dx = (2π)−2sΓ(s + ri)Γ(s− ri)

cos πs ε = 1cosh πr ε = −1

,

(|Im (r)| < Re(s) < 3/4)

(ii) Mr(x) =1

2πi

∫ c+∞

c−∞(2π)−2sΓ(s + ri)Γ(s− ri)

cos πs x > 0cosh πr x < 0

|x|−sds

(|Im(r)| < c < 1/2)

Proof. To calculate the Mellin-transform of Mr we will need the Mellin transforms for theJ and K Bessel functions (see [6], 6.8 (1) and 6.8 (22))∫ ∞

0

Kv(x)xs−1dx = 2s−2Γ

(s + v

2

)Γ

(s− v

2

), |Re(v)| < Re(s) (28)∫ ∞

0

Jv(x)xs−1dx = 2s−1 Γ((s + v)/2)

Γ((2 + v − s)/2), −Re(v) < Re(s) < 3/2.

124

By the functional equation for the gamma-function

Γ(s)Γ(1− s) =π

sin πs,

we have that ∫ ∞

0

Jv(x)xs−1dx = 2s−1 Γ((s + v)/2)

Γ((2 + v − s)/2)

=1

π2s−1Γ

(s + v

2

)Γ

(s− v

2

)sin(π

2(s− v)

).

(29)

With the substitution y = 4π√

x we get∫ ∞

0

Mr(±x)xs−1dy = 2(2π)−2s

∫ ∞

0

Mr

(± y2

(4π)2

)y2s−1dy,

and by the definition of Mr(x), (29) and (28) we get∫ ∞

0

Mr(x)xs−1dx =(2π)−2s i

2 sinh πrΓ(s + ri)Γ(s− ri)(sin(π(s− ri))− sin(π(s + ri))),∫ ∞

0

Mr(−x)xs−1dx =(2π)−2s cosh πr

πΓ(s + ri)Γ(s− ri),

so by the fact that cosh πr = cos πri the lemma is clear for Mr(−x). Finally, the fact that

i(sin(π(s− ri))− sin(π(s + ri)))

2 sinh πr=

2i cos(πs) sin(−πri)

2 sinh πr= cos(πs),

concludes the calculation of the Mellin-transform. The Mellin-Barnes integral follows fromMellin-inversion, see Titchmarsh[16], Theorem 30. The restriction c < 1/2 in (ii) comesfrom the fact that the Mellin transform of the function should tend to 0 which is one ofthe conditions of that theorem.

Lemma 3. Let F (x) be a real differentiable function such that F ′(x) is monotone andF ′(x) ≥ m > 0 or F ′(x) ≤ −m < 0 for 0 ≤ x ≤ b. Then∣∣∣∣∫ b

a

eiF (x)dx

∣∣∣∣ ≤ 4

m

Proof. This is taken from Ivic [7], Lemma 2.1.

Lemma 4. Let F (x) be a real, twice-differentiable function in [a, b] such that F ′′(x) ≥m > 0 or F ′′(x) ≤ −m < 0. Then ∣∣∣∣∫ b

a

eiF (x)dx

∣∣∣∣ ≤ 8√m

Proof. This is taken from Ivic [7], Lemma 2.2.

125

Lemma 5. Suppose that f fulfills condition (13). Then Φ(t) = ft(m, n), where ft(m, n) isdefined by (17) fulfills condition (10)

Proof. We have that

Φ(t) =

∫ ∞

−∞

∫ ∞

−∞e(nx + my)f

(tx ∗t ty

)dxdy

and we get

Φ(v)(t) =

∫ ∞

−∞

∫ ∞

−∞e(nx + my)

∂v

∂tvf

(tx ∗t ty

)dxdy, (v = 0, 1, 2, 3)

Using (13) this can be estimated by∫ ∞

−∞

∫ ∞

−∞

t−v

(t2+δ + t−1−δ)

1

(x2 + y2 + 1)1+δdxdy ≤ Cδ

t−v

(t2+δ + t−1−δ), v = 0, 1, 2, 3,

which fulfill condition (10).

In Lemmas 6 and 9 we will first prove the estimate

|Jn(x)| x2

|n|7/3,

which will be used to provide estimates for certain integral transforms appearing in ourmain summation formula.

Lemma 6. Suppose that n ≥ 1, and x is real. Then

|Jn(x)| ≤ 18

πn1/3

Proof. By the Bessel representation (see Guo–Wang [17] 7.5 (20) or Bessel [4])

Jn(z) =1

2π

∫ π

−π

cos(z sin θ − nθ)dθ, (30)

we have that

Jn(z) =1

2π

∫ π

−π

cos(z sin θ − nθ)dθ =1

2πRe

(∫ π

−π

eiF (x)dx

), (31)

where

F (x) = z sin x− nx. (32)

Suppose that n/2 > z. Then we have by the triangle inequality that

F ′(x) = z cos x− n ≤ −n

2,

126

so we can use Lemma 3, and (31) to get∣∣∣∣∫ π

−π

cos(z sin θ − nθ)dθ

∣∣∣∣ ≤ 8

n. (33)

If n/2 ≤ z we divide the integral

∫ π

−π

eiF (x)dx =

∫ −π+n−1/3

−π

+

∫ −n−13

−π+n−1/3

+

∫ n−1/3

−n−1/3

+

∫ π−n−1/3

n−1/3

+

∫ π

π−n−1/3

eiF (x)dx.

For the case

n−1/3 ≤ x ≤ π − n−1/3, (34)

we have by the inequality

x

2≤ sin x,

(0 ≤ x ≤ π

2

)and the fact that sin x is an increasing function for 0 ≤ x ≤ π

2, that

F ′′(x) = −z sin x ≤ −n2/3

4. (35)

From the identity

sin(π − x) = sin(x),

we have the inequality (35) in the whole interval (34), and by Lemma 4 we get∣∣∣∣∣∫ π−n−1/3

n−1/3

eiF (x)dx

∣∣∣∣∣ ≤ 16

n1/3. (36)

Similarly we also get ∣∣∣∣∣∫ −n−1/3

−π+n−1/3

eiF (x)dx

∣∣∣∣∣ ≤ 16

n1/3. (37)

Using (31), (36), (37) and the triangle inequality, and adding the terms we get∣∣∣∣∫ π

−π

cos(z sin θ − nθ)dθ

∣∣∣∣≤∣∣∣∣∫ π

−π

eiF (x)dx

∣∣∣∣ =

∣∣∣∣∣∣∫ −π+n−1/3

−π

+

∫ −n−13

−π+n−1/3

+

∫ n−1/3

−n−1/3

+

∫ π−n−1/3

n−1/3

+

∫ π

π−n−1/3

eiF (x)dx

∣∣∣∣∣∣≤ 1

n1/3+

16

n1/3+

2

n1/3+

16

n1/3+

1

n1/3=

36

n1/3.

127

Lemma 7. One has that uniformly for |n| ≥ 3 and real x,

|Jn(x)| x2

|n|7/3.

Lemma 8. One has that when k is an integer, then

Mi(1/2−k)(x) = π(−1)kJ2k−1

(4π√

x), (x > 0)

Proof. Since x > 0 it follows from (11)

Mi(1/2−k)(x) =πi

2 sinh π(i(1/2− k))

(J2k−1(4π

√x)− J1−2k(4π

√x)),

by applying the identities (see Guo–Wang [17], 7.2 (8)),

Jn(x) = (−1)nJ−n(x),

and

sinh π(i(1/2− k))

i= sin

(π(

12− k))

= cos(πk)

= (−1)k.

Lemma 9. One has that when k ≥ 2 is an integer, then

∣∣Mi(1/2−k)(x)∣∣ x2

|k|7/3.

Proof. This follows from Lemma 7 and Lemma 8.

Proof. By the identity (see Guo–Wang [17], 7.2 (8))

Jn(x) = (−1)nJ−n(x), (38)

it is sufficient to consider n ≥ 3. Applying the identity (see Guo–Wang [17], 7.23 (14))

Jn(x) =x

2n(Jn−1(x) + Jn+1(x)) (39)

two times on Jn(x) gives us

|Jn(x)| x2

n2

2∑j=−2

|Jn+j(x)|, (40)

and the result then follows from Lemma 6.

128

Lemma 10. Suppose that f fulfills condition (13). Then

∞∑c=1

∑hh≡1 (mod c)

0≤h,h<c

∞∑m,n=−∞

∣∣fc(m, n)∣∣ < ∞.

Proof. By using partial integration twice with respect to x and y we get

fc(m, n) =

∫ ∞

−∞

∫ ∞

−∞e(mξ + nη)f

(cξ ∗c cη

)dξdη

= (mn)−2

∫ ∞

−∞

∫ ∞

−∞e(mξ + nη)

∂2

∂ξ2

∂2

∂η2f

(cξ ∗c cη

)dξdη

By (13) this gives us

|fc(m, n)| c−2−δ(mn)−2

If either m or n equals zero, we similarly do partial integration with respect to one variable,and we get

|fc(m, 0)| c−2−δm−2, |fc(0, n)| c−2−δn−2.

When both m = 0 and n = 0, we immediately get

|fc(0, 0)| c−2−δ,

and this concludes our proof.

Lemma 11. One has that∣∣∣∣∫ t

0

yMr

(nm

y2

)dy

∣∣∣∣ t1/2(|mn|3/4 + t3/2

)(1 + |r|)−3/2,∣∣∣∣∫ x

0

∫ t

0

yMr

(nm

y2

)dydx

∣∣∣∣ x3/2(|mn|3/4 + x3/2

)(1 + |r|)−5/2.

Proof. We will use the Mellin-Barnes representation, Lemma 2 with c = 1/4. We get

yMr

(nm

y2

)=

1

2πi

∫ c+∞i

c−∞i

Γ(s + ri)Γ(s− ri)

cos πs, nm > 0cosh πr, nm < 0

(2π)−2s|nm|−sy2s+1ds.

Integrating this with respect to y gives us∫ t

0

yMr

(nm

y2

)dy =

1

2πi

∫ c+∞i

c−∞i

Γ(s + ri)Γ(s− ri)

2s + 2

cos πs, nm > 0cosh πr, nm < 0

(2π)−2s|nm|−st2s+2ds. (41)

129

Integrating this again with respect to t we get∫ x

0

∫ t

0

yMr

(nm

y2

)dydx =

1

2πi

∫ c+∞i

c−∞i

Γ(s + ri)Γ(s− ri)

(2s + 2)(2s + 3)

cos πs, nm > 0cosh πr, nm < 0

(2π)−2s|nm|−sx2s+3ds. (42)

Now moving the integration line to c = −3/4 we will pick up the residues coming froms = ±ir. In order to estimate these residue terms we will use the following variant of theStirling formula(see e.g. Ivic [7], A.34)

|Γ(σ + it)| |t|σ−1/2e−π2|t| σ0 ≤ σ ≤ σ1, 0 < t0 ≤ |t| (43)

For (41), the residue part coming from s = ±ir can be estimated by

O(t2(1 + |r|)−3/2

).

Using (43) again, the remaining integral can be estimated by

O(t1/2(1 + |r|)−3/2|mn|3/4

).

Similarly for (42) the residue part coming from s = ±ir can be estimated by

O(x3(1 + |r|)−5/2

),

and the remaining integral can be estimated by

O(x3/2(1 + |r|)−5/2|mn|3/4

).

Lemma 12. One has that if f fulfills condition (13) then the integral transformF (r; m, n) defined by (14) fulfills

|F (r; m, n)| |mn|−9/4(1 + |r|)−5/2.

Proof. We will use partial integration with respect to x, y, t and differentiate

t−1f

(tx ∗t ty

)three times with respect to x, y and two times with respect to t, and do the correspondingintegration of

e(nx + my)tMr

(nm

t2

).

Condition (13) enables us to do this and ensures that the “limit terms” will disappear. Weget

F (r; m, n) =1

(nm)3

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(nξ + mη)H(t)G(ξ, η, t)dξdηdt, (44)

130

where

G(ξ, η, t) =∂8

∂ξ3∂η3∂t2t−1f

(tξ ∗t tη

),

and

H(t) =

∫ t

0

∫ x

0

yMr

(nm

y2

)dydx.

By Lemma 11 we get

|H(t)| t3/2(|mn|3/4 + t3/2

)(1 + |r|)−5/2. (45)

By the fact that f fulfill condition (13), we have that

|G(x, y, t)| t−3

(t2+δ + t−1−δ)

1

(x2 + y2 + 1)1+δ. (46)

By combining (44), (45) and (46), the proof is completed.

Lemma 13. One has that if f fulfills condition (13) then the integral transform F (r; m, n)defined by (14) fulfills ∣∣F((1

2− k)i; m, n

)∣∣ |mn|−2k−7/3. (47)

Proof. We will use partial integration with respect to x, y and differentiate

f

(tx ∗t ty

)(48)

two times with respect to each variable and do the corresponding integration of

e(nx + my)Mi(1/2−k)

(nmD

t2

). (49)

Condition (13) ensures that we can do this and that the “limit terms” disappear. By thenapplying Lemma 9 the result follows.

References

[1] J. Andersson. A Poisson summation formula for SL(2, Z). Conference ab-stract, Millenial conference in number theory, Urbana Champaign, http://atlas-conferences.com/c/a/e/w/11.htm, 2000.

[2] A summation formula on the modular group and the pre-trace formula, forth-coming.

[3] . A summation formula over integer matrices II, 2006.

131

[4] F. W. Bessel. Analytishe Auflosung der Keplerschen Aufgabe. Berliner Abh. 1816-17,pages 49–55, 1819.

[5] R. W. Bruggeman. Fourier coefficients of cusp forms. Invent. Math., 45(1):1–18, 1978.

[6] A. Erdelyi, W. Magnus, F. Oberhettinger, and F. G. Tricomi. Tables of integraltransforms. Vol. I. McGraw-Hill Book Company, Inc., New York-Toronto-London,1954. Based, in part, on notes left by Harry Bateman.

[7] A. Ivic. The Riemann zeta-function. John Wiley & Sons Inc., New York, 1985. Thetheory of the Riemann zeta-function with applications.

[8] H. Iwaniec. Introduction to the spectral theory of automorphic forms. RevistaMatematica Iberoamericana, Madrid, 1995.

[9] H. D. Kloosterman. On the representation of numbers in the form ax2+by2+cz2+dt2.Acta Math., 49:407–464, 1926.

[10] N. V. Kuznecov. The Petersson conjecture for cusp forms of weight zero and theLinnik conjecture. Sums of Kloosterman sums. Mat. Sb. (N.S.), 111(153)(3):334–383,479, 1980.

[11] H. Maass. Uber eine neue Art von nichtanalytischen automorphen Funktionen unddie Bestimmung Dirichletscher Reihen durch Funktionalgleichungen. Math. Ann.,121:141–183, 1949.

[12] Y. Motohashi. Spectral theory of the Riemann zeta-function. Cambridge UniversityPress, Cambridge, 1997.

[13] N. V. Proskurin. The summation formulas for general Kloosterman sums. Zap.Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI), 82:103–135, 167, 1979.Studies in number theory, 5.

[14] On general Kloosterman sums. Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat.Inst. Steklov. (POMI), 302(Anal. Teor. Chisel i Teor. Funkts. 19):107–134, 200, 2003.

[15] A. Selberg. Harmonic analysis and discontinuous groups in weakly symmetric Rie-mannian spaces with applications to Dirichlet series. J. Indian Math. Soc. (N.S.),20:47–87, 1956.

[16] E. C. Titchmarsh. Introduction to the theory of Fourier integrals. Chelsea PublishingCo., New York, third edition, 1986.

[17] Z. X. Wang and D. R. Guo. Special functions. World Scientific Publishing Co. Inc.,Teaneck, NJ, 1989. Translated from the Chinese by Guo and X. J. Xia.

[18] A. Weil. On some exponential sums. Proc. Nat. Acad. Sci. U. S. A., 34:204–207, 1948.

[19] A. Zygmund. Trigonometric series: Vols. I, II. Cambridge University Press, London,1968.

132

A summation formula over integer matrices

Johan Andersson∗

AbstractIn a previous paper [4] we proved a summation formula for the full modular group.

In this paper we will generalize the result to integer matrices of determinant D. Wedevelop a formula for ∑

ad−bc=Dc>0

f

(a bc d

)

for test functions f . We are able express the sum into spectral objects coming fromthe full modular group. The proof is the same as in [4], but instead of using theclassical Kuznetsov summation formula we use a generalized summation formula ofBykovsky-Kuznetsov-Vinogradov [7] for the generalized Kloosterman sums

S(D,m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

).

Contents

1 Kloosterman sums 1331.1 Elementary properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1331.2 The Kuznetsov summation formula . . . . . . . . . . . . . . . . . . . . . . 134

2 Main result 136

3 Some remarks 139

4 Appendix - Integral transforms 139

1 Kloosterman sums

1.1 Elementary properties

Kloosterman sums

S(m, n; c) =∑

hh≡1 (mod c)0≤h,h<c

e

(mh + nh

c

)(1)

∗Department of Mathematics, Stockholm University, johana@math.su.se

133

is an important part of number theory. In our paper [1] we introduced the generalizedKloosterman sums

S(D, m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

), (2)

and proved the identity of Bykovsky-Kuznetsov-Vinogradov [7]

S(D, m, n; c) =∑

d|(D,m,c)

dS

(mD

d2, n;

c

d

). (3)

We proved the symmetry

S(n1, n2, n3; c) = S(nσ(1), nσ(2), nσ(3); c) (4)

where σ ∈ S3 is any permutation acting on 1, 2, 3. Furthermore, for the classical Kloost-erman sums we have the trivial identities

S(m, 0; c) = cc(m), (5)

as well as

S(0, 0; c) = cc(0) = φ(c), (6)

where

cc(m) =∑

(h,c)=10≤h<c

e

(mh

c

)(7)

denote the Ramanujan sums, and φ(c) denote Euler’s phi function.

1.2 The Kuznetsov summation formula

We will now state a generalized version of the classical Kuznetsov summation formula

Lemma 1. (Bykovski-Kuznetsov-Vinogradov1 summation formula) Suppose that Φ ∈ C3(0,∞)is a function on R+, which for some δ > 0 and v = 0, 1, 2, 3 fulfills

∣∣Φ(v)(t)∣∣ t−v

t32+δ + t−δ

. (8)

1The formula was initially developed independently by the author, but it was pointed out by K. Mat-sumoto at the millennial conference on number theory, Urbana-Champaign, May 2000, that the sameversion of the Kuznetsov summation formula, with the generalized Kloosterman sums S(D,m, n; c), hadbeen previously proved by Bykovski-Kuznetsov-Vinogradov [7].

134

Then

∞∑c=1

S(D, m, n; c)Φ(c) =∞∑

j=1

αjtj(D)tj(m)tj(n)ΦD,m,n(κj)

+1

π

∫ ∞

−∞

ΦD,m,n(r)σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)dr

|nmD|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

αj,ktj,k(D)tj,k(m)tj,k(n)ΦD,m,n

(12− ki

),

where S(D, m, n; c) are the generalized Kloosterman sums defined by equation (2).

ΦD,m,n(r) =

∫ ∞

0

Mr

(Dmn

t2

)Φ(t)dt,

with Mr defined by

Mr(x) =

iπ

2 sinh πr(J2ri(4π

√x)− J−2ri(4π

√x)) , x ≥ 0,

2 cosh πrK2ri(4π√−x), x < 0;

(9)

K2ri(x) and J2ri(x) are the K-Bessel and J-Bessel-functions respectively; ρj(n) are theFourier-coefficients

Ψj(x + iy) =∞∑

n=−∞

ρj(n)e(nx)Kiκj(|2πny|)

of an orthonormalized basis Ψj∞j=1 of Maass wave forms, and

αj =|ρj(1)|2

cosh πκj

and tj(n) =ρj(n)

ρj(1),

are the eigenvalues for the Hecke-operators Tn. Similarly Ψj,kθ(k)j=1 is an orthonormal basis

of holomorphic cusp forms of weight k,

Ψj,k(z) =∞∑

n=1

ρj,k(n)e(nz), tj,k(n) =ρj,k(n)

ρj,k(1),

and

a(k) = 21−4kπ−2k−1Γ(2k), αj,k = a(k)|ρj,k(1)|2.

Proof. For proof, see also Bykovski-Kuznetsov-Vinogradov [7]. The notation αj, αj,k, tj(n),tj,k(n) is taken from Motohashi [9]. When D = 1 the formula reduces after the change ofvariables

Φ(t) =1

tφ

(4π

√|mnD|

t

)

135

to the usual formula (See Motohashi [9], Theorem 2.3 and Theorem 2.5.). For D 6= 1 wereduce it to D = 1 by using the identity

S(D, m, n; c) =∑

d|(D,m,c)

dS

(mD

d2, n;

c

d

), (10)

(see Theorem 1 of Andersson [1] for a proof ) and

tj(m)tj(n) =∑

d|(m,n)

tj

(mn

d2

),

together with the corresponding identities

σ2ir(m)

mir

σ2ir(n)

nir=∑

d|(m,n)

σ2ir

(mn

d2

)(mn

d2

)−ir

, tj,k(m)tj,k(n) =∑

d|(m,n)

tj,k

(mn

d2

),

for the divisor functions and the Fourier coefficients for holomorphic modular forms (seeMotohashi[9], formula (3.1.4), (3.1.14)).

2 Main result

We will now state the main result

Theorem 1. Let f be a function on A ∈ GL(2, R) : det A = D, where D is a nonzerointeger, fulfilling the conditions∣∣∣∣ ∂n

∂xn

∂m

∂ym

∂v

∂tvf

(xt ∗t yt

)∣∣∣∣ t−v

(t2+δ + t−1−δ)(x2 + y2 + 1)1+δ,

n, m, v = 0, 1, 2, 3,t > 0,

(11)

for some δ > 0, where ∗ denotes the number making the determinant of the matrix equalto D. One then has the identity

∑ad−bc=D

c>0

f

(a bc d

)= −

∞∑c=1

∑d|(D,c)

φ( c

d

)∫ ∞

−∞

∫ ∞

−∞f

(cx ∗c cy

)dxdy

+∞∑

n=−∞

∞∑c=1

∑d|(D,c)

cc/d(n)

∫ ∞

−∞

(f

(cn ∗c cx

)+ f

(cx ∗c cn

))dx

+∞∑

j=1

∑m,n6=0

αjtj(D)tj(m)tj(n)F (κj; D, m, n)

+∑

m,n6=0

1

π

∫ ∞

−∞

σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)F (r; D, m, n)dr

|Dmn|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

∑m,n6=0

αj,ktj,k(D)tj,k(m)tj,k(n)F((

12− k)i; D, m, n

),

136

where

F (r; D, m, n) =

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(mx + ny)Mr

(Dmn

t2

)f

(tx ∗t ty

)dxdydt, (12)

Mr(x) is defined by (9), and αj, αj,k, tj(n), tj,k(m) are defined as in Lemma 1, φ(c) denotesthe Euler phi function and cc(n) denotes the Ramanujan sums (7).

Proof. The proof will closely follow our proof from [4]. We will now use the Bruhatdecomposition, a simple change of summation trick, and the Kuznetsov summation formulato prove our theorem. By letting a = mc+h, and d = nc+k, with 0 ≤ h, k < c, we obtain

∑ad−bc=D

c>0

f

(a bc d

)=

∞∑c=1

∞∑m,n=−∞

∑hk≡D (mod c)

0≤h,k<c

f

(mc + h ∗

c nc + k

)

=∞∑

c=1

∑hk≡D (mod c)

0≤h,k<c

∞∑m,n=−∞

fc

(h

c+ m,

k

c+ n

),

where

ft(x, y) = f

(tx ∗t ty

). (13)

Condition (11) for x and y ensures that we can use the shifted double Poisson summationformula (see e.g Zygmund [10] (13.3)), and we get

∞∑c=1

∑hk≡D (mod c)

0≤h,k<c

∞∑m,n=−∞

e

(mh + nk

c

)fc(m, n),

where

ft(m, n) =

∫ ∞

−∞

∫ ∞

−∞e(mx + ny)ft(x, y)dxdy (14)

is the inverse two dimensional Fourier transform of ft(x, y). Lemma 5 guarantees that thissum is absolutely convergent and we can change the summation order

∞∑m,n=−∞

∞∑c=1

∑hk≡D (mod c)

0≤h,k<c

e

(mh + nk

c

)fc(m, n) =∞∑

m,n=−∞

∞∑c=1

S(D, m, n; c)fc(m, n),

(15)

and the Kloosterman sums appear. Since f fulfills (11), we have by Lemma 4 in theAppendix that the function Φ(t) = ft(m, n) is of the type (8). Hence we can apply our

137

version of the Kuznetsov summation formula, Lemma 1, and we find that the contributioncoming from mn 6= 0 is equal to

∑m,n6=0

(∞∑

j=1

αjtj(D)tj(m)tj(n)F (κj; D, m, n)

+1

π

∫ ∞

−∞

σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)F (r; D, m, n)dr

|Dmn|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

αj,ktj,k(D)tj,k(m)tj,k(n)F((

12− k)i; D, m, n

) . (16)

From the estimate for the Fourier coefficients of cusp forms∑K/2<κj≤K

αj|tj(m)|2 K2 + d3(m) log(2m)√

m, (17)

θ(k)∑j=1

αj,k|tj,k(m)|2 kd3(m)√

m log(2m), (18)

valid uniformly for k,K, m ≥ 1 (see Motohashi [9] Lemma 2.4, and equation (2.2.10)),together with the fact that

|F (r; D, m, n)| |mn|−94 |D|

34 (1 + |r|)−

52 ,∣∣F((1

2+ k)i; D, m, n

)∣∣ |mn|−2|D|k−73 ,

when f satisfies (11) (see the Appendix, Lemma 7 and Lemma 8), we see that the summa-tion and summation-integration in (16) is absolutely convergent, and hence we can changethe summation order, and we get the spectral part of Theorem 1.

We now consider the contribution coming from m = 0 or n = 0. We get from (15) thecontribution ∑

m=0 or n=0

∞∑c=1

S(D, m, n; c)fc(m, n), (19)

By the symmetry (4) of S(D, m, n; c), this equals

∞∑n=−∞

∞∑c=1

S(D, 0, n; c)(fc(0, n) + fc(0, n)

)−

∞∑c=1

S(D, 0, 0)fc(0, 0). (20)

By the Bykovsky-Kuznetsov-Vinogradov identity (3) we have that

S(D, 0, n; c) =∑d|D

S(0, n;

c

d

)which by the definition of the Ramanujan sums (7) equals∑

d|D

cc/d(n).

138

By applying the Poisson summation formula on the first term in (20) again, and usingcl(0) = φ(l) we find that equation (20) equals

∞∑n=−∞

∞∑c=1

∑d|(D,c)

cc/d(n)

∫ ∞

−∞

(f

(cn ∗c cx

)+ f

(cx ∗c cn

))dx

−∞∑

c=1

∑d|(D,c)

φ( c

d

)∫ ∞

−∞

∫ ∞

−∞f

(cx ∗c cy

)dxdy

which accounts for the second and first term in the theorem.

3 Some remarks

We believe that our main summation formula will have plenty of applications. We shallpresent some of them in forthcoming papers, e.g. [3] and [2]. An obvious application is thefollowing: Since

σ2ir(n) = 1, tj(n) = 1, and tj,k(n) = 1

the summation formula over the full modular group as proved in [4] will be the specialcase when D = 1 of this formula. About the proof: This was not our first proof. Our firstsketch of a proof used Motohashi’s work [9] on the theory of the Riemann zeta-function.For a sketch of that proof idea, see the introduction of our thesis [6]. Another proof,that we presented in Turku in the fall of 1999, used the Kuznetsov summation formula ina different way. We did not need to use the Kloosterman sum identity (3). Instead weused a formula relating the Kloosterman sums and Ramanujan sums. We will present thisproof in [5]. Another proof that we would like to develop and hope to return to later isto prove the summation formula over integer matrices of determinant D directly from thesummation formula over the full modular group from [4], by means of Hecke operators. Inthe current proof we use Hecke operators indirectly as they are used to prove the identity(3) in [1]. However a more direct proof would be nice as well.

4 Appendix - Integral transforms

In this appendix we prove estimates of integral transforms that are needed in the rest ofthe paper.

Lemma 2. One has the following Mellin-transform and Mellin-Barnes representations ofthe Kernel function Mr

(i)

∫ ∞

0

Mr(ε · x)xs−1dx = (2π)−2sΓ(s + ri)Γ(s− ri)

cos πs ε = 1cosh πr ε = −1

,

(|Im (r)| < Re(s) < 3/4)

(ii) Mr(x) =1

2πi

∫ c+∞

c−∞(2π)−2sΓ(s + ri)Γ(s− ri)

cos πs x > 0cosh πr x < 0

|x|−sds

(|Im(r)| < c < 1/2)

139

Proof. For proof, see Lemma 3 in Andersson [4].

Lemma 3. One has that when k ≥ 2 is an integer, then∣∣Mi(1/2−k)(x)∣∣ x2

|k|7/3.

Proof. This is Lemma 9 in Andersson [4].

Lemma 4. Suppose that f fulfills condition (11). Then the function Φ(t) = ft(m, n),where ft(m, n) is defined by (13), fulfills condition (8).

Proof. For a proof, see Andersson [4], Lemma 5.

Lemma 5. Suppose that f fulfills condition (11). Then

∞∑c=1

∑hh≡D (mod c)

∞∑m,n=−∞

∣∣fc(m, n)∣∣ < ∞

Proof. The proof of this result is identical to the one in Andersson [4], Lemma 10.

Lemma 6. One has the estimates∣∣∣∣∫ t

0

yMr

(nmD

y2

)dy

∣∣∣∣ t1/2(|mnD|3/4 + t3/2

)(1 + |r|)−3/2,∣∣∣∣∫ x

0

∫ t

0

yMr

(nmD

y2

)dydx

∣∣∣∣ x3/2(|mnD|3/4 + x3/2

)(1 + |r|)−5/2.

Proof. We will use the Mellin-Barnes representation, Lemma 2 with c = 1/4. We get

yMr

(nmD

y2

)=

1

2πi

∫ c+∞i

c−∞i

Γ(s + ri)Γ(s− ri)

cos πs, nmD > 0cosh πr, nmD < 0

(2π)−2s|nmD|−sy2s+1ds.

Integrating this with respect to y gives us∫ t

0

yMr

(nmD

y2

)dy =

1

2πi

∫ c+∞i

c−∞i

Γ(s + ri)Γ(s− ri)

2s + 2

cos πs, nmD > 0cosh πr, nmD < 0

(2π)−2s|nmD|−st2s+2ds. (21)

Integrating this again with respect to t we get∫ x

0

∫ t

0

yMr

(nmD

y2

)dydx =

1

2πi

∫ c+∞i

c−∞i

Γ(s + ri)Γ(s− ri)

(2s + 2)(2s + 3)

cos πs, nmD > 0cosh πr, nmD < 0

(2π)−2s|nmD|−sx2s+3ds. (22)

140

By moving the integration line to c = −3/4 we will pick up the residues coming froms = ±ir. In order to estimate these residue terms we will use the following variant of theStirling formula (see e.g. Ivic [8], A.34)

|Γ(σ + it)| |t|σ−1/2e−π2|t|, σ0 ≤ σ ≤ σ1, 0 < t0 ≤ |t|. (23)

For (21), the residue part coming from s = ±ir can be estimated by

O(t2(1 + |r|)−3/2

).

Using (23) again, the remaining integral can be estimated by

O(t1/2(1 + |r|)−3/2|mnD|3/4

).

Similarly for (22) the residue part coming from s = ±ir can be estimated by

O(x3(1 + |r|)−5/2

),

and the remaining integral can be estimated by

O(x3/2(1 + |r|)−5/2|mnD|3/4

).

Lemma 7. If f fulfills condition (11) then the integral transformF (r; D, m, n) defined by (12) satisfies

|F (r; D, m, n)| |mn|−9/4|D|3/4(1 + |r|)−5/2.

Proof. We will use partial integration with respect to x, y, t and differentiate

t−1f

(tx ∗t ty

)three times with respect to x, y and two times with respect to t, and perform the corre-sponding integration of the function

e(nx + my)tMr

(nmD

t2

).

Condition (11) enables us to do this and ensures that the “limit terms” will disappear. Weget

F (r; D, m, n) =1

(nm)3

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(nx + my)H(t)G(x, y, t)dxdydt, (24)

where

G(x, y, t) =∂8

∂x3∂y3∂t2t−1f

(tx ∗t ty

),

141

and

H(t) =

∫ t

0

∫ x

0

yMr

(nmD

y2

)dydx.

By Lemma 6 we get

|H(t)| t3/2(|mnD|3/4 + t3/2

)(1 + |r|)−5/2. (25)

By the fact that f fulfills condition (11), we have that

|G(x, y, t)| t−3

(t2+δ + t−1−δ)

1

(x2 + y2 + 1)1+δ. (26)

By combining (24), (25) and (26), the proof is completed.

Lemma 8. If f fulfills condition (11) then the integral transform satisfies F (r; D, m, n)defined by (12) fulfills ∣∣F((1

2− k)i; D, m, n

)∣∣ |mn|−2|D|k−7/3. (27)

Proof. We will use partial integration with respect to x, y and differentiate

f

(tx ∗t ty

)(28)

two times with respect to each variable and do the corresponding integration of

e(nx + my)Mi(1/2−k)

(nmD

t2

). (29)

Condition (11) ensures that we can do this and that the “limit terms” disappear. Byapplying Lemma 3 the result then follows.

References

[1] J. Andersson. A note on some Kloosterman sum identities, 2006.

[2] On the additive circle problem, forthcoming.

[3] The Selberg and Eichler-Selberg trace formulae, 2006.

[4] A summation formula on the full modular group, 2006.

[5] A summation formula over integer matrices II, 2006.

[6] Summation formulae and zeta-functions, Thesis.

[7] V. Bykovsky, N. Kuznetsov, and A. Vinogradov. Generalized summation formulafor inhomogeneous convolution. In Automorphic functions and their applications(Khabarovsk, 1988), pages 18–63. Acad. Sci. USSR Inst. Appl. Math., Khabarovsk,1990.

142

[8] A. Ivic. The Riemann zeta-function. John Wiley & Sons Inc., New York, 1985. Thetheory of the Riemann zeta-function with applications.

[9] Y. Motohashi. Spectral theory of the Riemann zeta-function. Cambridge UniversityPress, Cambridge, 1997.

[10] A. Zygmund. Trigonometric series: Vols. I, II. Cambridge University Press, London,1968.

143

A summation formula over integer matrices II

Johan Andersson∗

Abstract

In a previous paper [3] we proved a summation formula over integer matrices withfixed determinant D, that is we obtained a formula for∑

ad−bc=D

f

(a bc d

).

In doing so we used an identity by Bykovsky-Kuznetsov-Vinogradov (see [1])

S(D,m, n; c) =∑

d|(D,m,c)

dS

(mD

d2, n;

c

d

), (1)

where

S(D,m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

)(2)

are generalized Kloosterman sums. In this paper we will give an alternative proofwithout using this identity.

Contents

1 Introduction 146

2 Kloosterman sums and Ramanujan sums 147

3 Spectral theory 152

4 Some concluding remarks 156

∗Department of Mathematics, Stockholm University, johana@math.su.se

145

1 Introduction

In [2] we proved a new type of summation formula over the full modular group, and in [3]we proved the corresponding identity for sums over integer matrices of fixed determinantD. We first thought of ways to obtain an identity like this in the spring of 1999 whenwe were visiting Minneapolis. It occurred to us when we were studying a certain sixthpower moment of the Lerch zeta-function1, that an identity in the paper of Motohashi [7]on the fourth power moment of the Riemann zeta-function, could be seen as a “functionalequation” corresponding to a yet to be discovered “summation formula”2. In the summerof 1999 we presented our idea at a symposium in Turku. The method of proof was nota very practical one. Even if it was simple in principle, the integrals would turn outto be cumbersome, and we never proved an explicit version of our summation formulawith this method. Starting with that idea we searched for a simpler method of proof,and we soon found a nice proof, using just the classical Poisson summation formula, theKuznetsov summation formula and some simple ideas concerning the Ramanujan sums andthe Kloosterman sums. We presented our results at a seminar in Turku, in October 1999and in this paper we will present the proof as given at that seminar. Even if it is a bitmore involved than the proof from [2] and [3], which we found later and presented at theMillennial conference, May 2000, Urbana-Champaign, we still believe it has some interest,and is worth publishing. We do not have to define the generalized Kloosterman sums

S(D, m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

)(3)

and use the identity3

S(D, m, n; c) =∑

d|(D,m,c)

dS

(mD

d2, n;

c

d

)(4)

as we did in [3]. Instead we will use the identity

cl(nm−D) =1

l

l∑p,q=1

S(pq, D; l)e

(−mp− nq

l

), (5)

that relates the classical Kloosterman sums with the Ramanujan sums cl(m). Althoughprobably less important than the identity (4), we have not (yet) found the identity (5)anywhere in the literature, and it might be of some independent interest.

1At the time we were hoping that the main summation formula, Theorem 1 in this paper would haveapplications on estimating certain exponential sums arising from this problem. We have not returned tothat question, but our current belief is that the summation formula is better for estimating certain otherexponential sums, not these particular ones.

2For a further discussion of this matter, see the introduction in our Thesis [4].3We originally believed that we might have found a new identity in (4). However it turned out that it

had previously been stated by Bykovsky-Kuznetsov-Vinogradov [5], and in a less refined form by Heath-Brown [6]. For the full story see our paper [1].

146

2 Kloosterman sums and Ramanujan sums

Kloosterman sums

S(m, n; c) =∑

hh≡1 (mod c)0≤h,h<c

e

(mh + nh

c

)(6)

have many applications in analytic number theory. From their definition we immediatelydeduce some of their elementary properties, such as the symmetries

S(m, n; c) = S(n, m; c), (7)

S(n, m; c) = S(−n,−m, c), (8)

as well as the facts that

S(m, 0; l) = cl(m), (9)

and

S(0, 0; c) = φ(c), (10)

where

cl(m) =∑

(h,l)=10≤h<l

e

(mh

l

), (11)

denote the Ramanujan sums, and φ(c) denote Euler’s phi function. Ramanujan sums ofcourse inherits the symmetry (8), since by (9) they can be seen as special cases of theKloosterman sum. Thus we have

cl(m) = cl(−m). (12)

We will first show some results about the Ramanujan sums.

Lemma 1. Suppose that f is a function on R such that

∞∑k=1

∣∣∣f(x

k, k)∣∣∣ < ∞. (13)

One then has the identity

∑ad=ma>0

f(a, d) =∞∑

k,l=1

cl(m)

klf(kl,

m

kl

), (14)

147

Proof. The condition (13), the estimate |cl(m)| ≤ φ(l), which follows immediately fromthe definition of the Ramanujan sums, and the identity n =

∑d|n φ(d) ensures that the

sum on the right hand side in the identity (14) is absolutely convergent. Hence we canchange the summation order and Lemma (1) follows from∑

d|k

cd(m) =∑d|k

∑(h,d)=10≤h<d

e

(hm

d

),

=k∑

q=1

e(qm

k

),

=

k, if k|m,

0, if k 6 |m.

(15)

Kloosterman sums and Ramanujan sums are closely related, as is seen by the ele-mentary identity (9). We will now prove a somewhat more involved identity involvingthe Kloosterman sums as well as the Ramanujan sums, that we will need for our mainargument.

Lemma 2. One has the following identity

cl(nm−D) =1

l

l∑p,q=1

S(pq, D; l)e

(−mp− nq

l

).

Proof. We have by the definition (11) of the Ramanujan sums, and the symmetry equation(12), that

cl(nm−D) =∑

(h,l)=10≤h<l

e

(−(nm−D)h

l

),

=∑

(h,l)=10≤h<l

e

(−nmh

l

)e

(Dh

l

).

(16)

We will now express e(−nmh/l) through the identity

e

(−nmh

l

)=

1

l

l∑q=1

e

(−mq

l

) l∑p=1

e

(−p(−qh + n)

l

)(17)

This identity can be proved as follows. By starting with the right hand side of equation

148

(17) we obtain

1

l

l∑q=1

e

(−mq

l

) l∑p=1

e

(−p(−qh + n)

l

)=

l∑p=1

1

l

l∑q=1

e

(−mq + pqh− np

l

),

=l∑

p=1

e

(−np

l

)1

l

l∑q=1

e

((−m + ph)q

l

),

=l∑

p=1

e

(−np

l

)1, −m + ph ≡ 0 (mod l),

0, otherwise.

Since the only contribution will come when −m + ph ≡ 0 (mod l), or in other wordsp ≡ mh (mod l), this equals

e

(−nmh

l

).

By equations (16) and (17), and changing the summation order, we will now get that

cl(nm−D) =∑

(h,l)=10≤h<l

1

l

l∑q=1

e

(−mq

l

) l∑p=1

e

(−p(−qh + n)

l

)e

(Dh

l

),

=1

l

l∑p,q=1

∑(h,l)=10≤h<l

e

(pqh + Dh

l

)e

(−mp− nq

l

),

which by the definition of the Kloosterman sum (6) equals

=1

l

l∑p,q=1

S(pq, D; l)e

(−mp− nq

l

).

Lemma 3. Let l > 0 be an integer and f a continuously differentiable function on R2,such that

|f(x, y)| ≤ (x2 + y2)−1−ε. (ε > 0)

One then has∞∑

m,n=−∞

f(m, n)cl(mn−D) =1

l

∞∑m,n=−∞

f(m

l,n

l

)S(mn, D; l), (18)

where

f(t, τ) =

∫ ∞

−∞

∫ ∞

−∞e(−τx− ty)f(x, y)dxdy (19)

is the two dimensional Fourier transform of f , the Ramanujan sums cl(n) are defined by(11) and the Kloosterman sums S(m, n; c) by (6).

149

Proof. We will start with the right hand side of equation (18). We have with m = al+p, n =bl + q that

1

l

∞∑m,n=−∞

f(m

l,n

l

)S(mn, D; l) =

l∑p,q=1

∞∑a,d=−∞

S((al + p)(bl + q), d; l)f(a +

p

l, d +

q

l

).

Since by their definition (6), the Kloosterman sums S(m, n; l) only depend on the residueclass of m, n (mod l), we see that this equals

l∑p,q=1

Sl(pq, D; l)∞∑

a,d=−∞

f(a +

p

l, d +

q

l

). (20)

By using the fact that if

g(x, y) = f(x +

p

l, y +

q

l

)then

g(ξ, η) = e

(−ξp− ηq

l

)f(ξ, η),

we can sum over p and q (mod l) and the conditions on f ensure that we can use the twodimensional version of the Poisson summation formula (see e.g Zygmund [9] (13.3)) on theinner sum in (20). We obtain

l∑p,q=1

S(pq, D; l)∞∑

m,n=−∞

e

(−pm− qn

l

)f(m, n). (21)

By changing the summation order this equals

∞∑m,n=−∞

(l∑

p,q=1

S(pq, D; l)e

(−pm− qn

l

))f(m, n). (22)

By Lemma 2

cl(nm−D) =∑

(h,l)=10≤h<l

1

l

l∑p,q=1

S(pq, D, l)e

(−mp− nq

l

),

we obtain

∞∑m,n=−∞

f(m, n)cl(nm−D).

150

Lemma 4. Let f be a continuously differentiable test function on all real 2 × 2 matriceswith determinant D, such that∣∣∣∣f (x ∗

t y

)∣∣∣∣ ≤ C(ε)(|t|(x2 + y2)

)−1−ε. (ε > 0) (23)

One then has the following identity

∑ad−bc=D

c>0

f

(a bc d

)=

∞∑m,n=−∞

∞∑k,l=1

kS(mn, D, l)fkl(m, n),

where

ft(a, b) =

∫ ∞

−∞

∫ ∞

−∞e(−ax− by)f

(xt ∗t yt

)dxdy.

Proof. We have that

∑ad−bc=D

c>0

f

(a bc d

)=

∞∑b,c=−∞

∑ad=bc+D

f

(a bc d

),

by expressing the inner sum as a sum of Ramanujan sums with Lemma 1, this equals

∞∑a,d=−∞

∞∑k,l=1

cl(ad−D)

klf

(a ad−D

kl

kl d

),

By changing summation order and using the notation

gt(ξ, ν) = f

(ξ ∗t ν

), (24)

we see that this equals

∞∑k,l=1

∞∑a,d=−∞

cl(ad−D)

klgkl(a, d).

We now use Lemma 3 on the inner sum and we obtain

∞∑k,l=1

1

kl2

∞∑m,n=−∞

gt

(m

kl,

n

kl

)S(mn, D, l), (25)

where

gt(µ, τ) =

∫ ∞

−∞

∫ ∞

−∞e(−µξ − τν)f

(ξ ∗t ν

)dξdν

151

With the substitutions

ξ = tx, and ν = ty (26)

we obtain that

gt

(a

t,b

t

)= t2ft(a, b),

where

ft(a, b) = t2∫ ∞

−∞

∫ ∞

−∞e(−ax− by)f

(xt ∗t yt

)dxdy

is the Fourier transform of

ft(x, y) = f

(xt ∗t xy

).

Hence equation (25) equals

∞∑k,l=1

1

kl2

∞∑m,n=−∞

(kl)2ft(m, n)S(mn, D, l)

and the lemma follows from 1/(kl2) · (kl)2 = k and by changing the summation order.

3 Spectral theory

We have now expressed the sum in terms of Kloosterman sums and we are ready to proveour main summation formula by means of the Kuznetsov summation formula:

Lemma 5. (Kuznetsov summation formula) Suppose that Φ ∈ C3(0,∞) is a function onR+, and that for some δ > 0 and v = 0, 1, 2, 3 it satisfies∣∣Φ(v)(t)

∣∣ t−v

t32+δ + t−δ

. (27)

Then

∞∑c=1

S(m, n; c)Φ(c) =∞∑

j=1

αjtj(m)tj(n)Φm,n(κj) +1

π

∫ ∞

−∞

Φm,n(r)σ2ir(|m|)σ2ir(|n|)dr

|mn|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

tj,k(m)αj,ktj,k(n)Φm,n

(12− ki

),

where

Φm,n(r) =

∫ ∞

0

Mr

(mn

t2

)Φ(t)dt,

152

and Mr is defined by

Mr(x) =

iπ

2 sinh πr(J2ri(4π

√x)− J−2ri(4π

√x)) , x ≥ 0,

2 cosh πrK2ri(4π√−x), x < 0.

(28)

K2ri(x) and J2ri(x) are the K-Bessel and J-Bessel-functions respectively, ρj(n) are theFourier-coefficients

Ψj(x + iy) =∞∑

n=−∞

ρj(n)e(nx)Kiκj(|2πny|) (29)

of an orthonormalized basis Ψj∞j=1 of Maass wave forms, and

αj =|ρj(1)|2

cosh πκj

and tj(n) =ρj(n)

ρj(1), (30)

are the eigenvalues for the Hecke-operators Tn. Similarly Ψj,kθ(k)j=1 is an orthonormal basis

of holomorphic cusp forms of weight k,

Ψj,k(z) =∞∑

n=1

ρj,k(n)e(nz), tj,k(n) =ρj,k(n)

ρj,k(1), (31)

and

a(k) = 21−4kπ−2k−1Γ(2k), αj,k = a(k)|ρj,k(1)|2. (32)

Here the automorphic forms Ψj(z) and Ψj,k(z) are also chosen to be eigenfunctions of theHecke operators. (For more about the notation see Motohashi [8]).

Proof. After the change of variables

Φ(t) =1

tφ

(4π

√|mn|t

)(33)

it reduces to the classical Kuznetsov summation formula.(This version of the formula istaken from Motohashi [8], Theorem 2.3 and Theorem 2.5). The same formula appears in[2], except that we now choose to use the notation αj, αj,k, tj(n), tj,k(n).

We are now ready to prove our main theorem

Theorem 1. Let f be an even function on A ∈ GL(2, R) : det A = D, where D is anonzero integer, fulfilling the conditions∣∣∣∣ ∂n

∂xn

∂m

∂ym

∂v

∂tvf

(xt ∗t yt

)∣∣∣∣ t−v

(t2+δ + t−1−δ)(x2 + y2 + 1)1+δ,

n, m, v = 0, 1, 2, 3,t > 0,

(34)

153

for some δ > 0, where ∗ denote the number so that the determinant of the matrix is D.One then has

∑ad−bc=D

c>0

f

(a bc d

)=

∞∑n=−∞

∞∑c=1

∑d|(D,c)

cc/d(n)

∫ ∞

−∞

(f

(cn ∗c cx

)+ f

(cx ∗c cn

))dx

−∞∑

c=1

∑d|(D,c)

φ( c

d

)∫ ∞

−∞

∫ ∞

−∞f

(cx ∗c cy

)dxdy

+∞∑

j=1

∑m,n6=0

αjtj(D)tj(m)tj(n)F (κj; D, m, n)

+∑

m,n6=0

1

π

∫ ∞

−∞

σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)F (r; D, m, n)dr

|Dmn|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

∑m,n6=0

αj,ktj,k(D)tj,k(m)tj,k(n)F((

12− k)i; D, m, n

),

where

F (r; D, m, n) =

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(mx + ny)Mr

(Dmn

t2

)f

(tx ∗t ty

)dxdydt, (35)

Mr(x) is defined by (28), and αj, αj,k, tj(n), tj,k(m) are defined as in Lemma 1, φ(c) denotethe Euler phi function and cc(n) denote the Ramanujan sums(see (11)).

Proof. By Lemma 4 we have that

∑ad−bc=D

c>0

f

(a bc d

)=

∞∑m,n=−∞

∞∑k=1

∞∑l=1

S(mn, D, l)kfkl(m, n), (36)

where

ft(ξ, ν) =

∫ ∞

−∞

∫ ∞

−∞e(−ξx− νy)f

(xt ∗t yt

)dxdy.

By Lemma 7 in [3] we have that Φ(t) = ft(n, m) fulfills the test function condition (27).Hence we can apply the Kuznetsov summation formula, Lemma 5 on the inner sum withΦ(l) = kfkl(m, n). We obtain

Φm,n(r) = k

∫ ∞

−∞

∫ ∞

−∞

∫ ∞

0

Mr

(mn

t2

)e(−mnx−Dy)f

(ktx ∗kt kty

)dtdxdy,

which with the substitution τ = tk equals

=

∫ ∞

−∞

∫ ∞

−∞

∫ ∞

0

Mr

(mnk2

τ 2

)e(−mnx−Dy)f

(τx ∗τ τy

)dτdxdy,

154

which by definition of F (r; m, n,D) equals

= F (r; mn, D, k2).

The contribution for fixed m, n, k 6= 0 coming from the inner sum in (36) thus equals

∞∑j=1

tj(D)tj(mn)F (κj; D, mn, k2)

+1

π

∫ ∞

−∞

σ2ir(|D|)σ2ir(|mn|)F (r; D, mn, k2)dr

|Dmn|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

αj,ktj,k(D)tj,k(mn)F((

12− k)i; D, mn, k2

). (37)

From the estimate for the Fourier coefficients of cusp forms∑K/2<κj≤K

αj|tj(m)|2 K2 + d3(m) log(2m)√

m, (38)

θ(k)∑j=1

αj,k|tj,k(m)|2 kd3(m)√

m log(2m), (39)

valid uniformly for k,K, m ≥ 1 (see Motohashi [8] Lemma 2.4, and equation (2.2.10)), andthe fact that

|F (r; D, m, n)| |mn|−94 |D|

34 (1 + |r|)−

52 ,∣∣F((1

2+ k)i; D, m, n

)∣∣ |mn|−2|D|k−73 ,

when f satisfies the test function condition of the theorem, equation(34) (see Andersson [3],Lemma 7 and Lemma 8), we see that when we sum over m, n, k in (37) the summation andsummation-integration is absolutely convergent, and hence we can change the summationorder. To obtain the spectral part of the theorem it is now sufficient to prove the identities

∞∑k=1

∞∑m,n=−∞

m,n6=0

tj(mn)F (κj; D, mn, k2) =∞∑

m,n=−∞m,n6=0

tj(m)tj(n)F (κj; D, m, n)

and the corresponding identities for the holomorphic cusp forms and Eisenstein series.They follow from the identities

tj(m)tj(n) =∑

d|(m,n)

tj

(mn

d2

), (40)

as well as the corresponding identities

σ2ir(m)

mir

σ2ir(n)

nir=∑

d|(m,n)

σ2ir

(mn

d2

)(mn

d2

)−ir

, tj,k(m)tj,k(n) =∑

d|(m,n)

tj,k

(mn

d2

),

(41)

155

for the divisor functions and the Fourier coefficients for holomorphic modular forms (seeMotohashi [8], formula (3.1.4), (3.1.14)). It remains to consider the case when at least oneof m, n are zero. We then get the contribution

∞∑n=−∞

∞∑k=1

k∞∑l=1

S(0, D, l) + fkl(0, n) + fkl(n, 0)−∞∑

k=1

k∞∑l=1

S(0, D, l)fkl(0, 0).

Using the one dimensional version of the Poisson summation formula again as we did in[3] on the first term, and using equation (9) to express the Kloosterman sums in terms ofthe Ramanujan sums we obtain the first two terms in the theorem.

4 Some concluding remarks

The Bykovsky-Kuznetsov-Vinogradov identity, equation (4) can in fact be used to provethat the identity

∑ad−bc=D

c>0

f

(a bc d

)=

∞∑m,n=−∞

∞∑k,l=1

kS(mn, D, l)fkl(m, n),

of Lemma 4, in fact equals

=∞∑

m,n=−∞

∞∑c=1

S(D, m, n; c)fc(m, n).

This appears as equation (15) in [3]. Hence the two methods of proofs are clearly related.It is interesting4 to see that whatever method we use in the proof, the Hecke operators areinvolved in one way or another. In [1] we used the Hecke operators, and the identities (40)and (41) are clearly a direct consequence of

T (m)T (n) =∑

d|(m,n)

T(mn

d2

), (42)

(see Motohashi [8] equation (3.1.4)). As we wrote in [3] a more direct proof of the summa-tion formula over integer matrices of determinant D using the corresponding formula forthe full modular group (from [2]) and Hecke operators would be interesting.

References

[1] J. Andersson. A note on some Kloosterman sum identities, 2006.

[2] A summation formula on the full modular group, 2006.

[3] A summation formula over integer matrices, 2006.

4By the nature of the problem it is not surprising though.

156

[4] Summation formulae and zeta-functions, Thesis.

[5] V. Bykovsky, N. Kuznetsov, and A. Vinogradov. Generalized summation formulafor inhomogeneous convolution. In Automorphic functions and their applications(Khabarovsk, 1988), pages 18–63. Acad. Sci. USSR Inst. Appl. Math., Khabarovsk,1990.

[6] D. R. Heath-Brown. The fourth power moment of the Riemann zeta function. Proc.London Math. Soc. (3), 38(3):385–422, 1979.

[7] Y. Motohashi. An explicit formula for the fourth power mean of the Riemann zeta-function. Acta Math., 170(2):181–220, 1993.

[8] Spectral theory of the Riemann zeta-function. Cambridge University Press,Cambridge, 1997.

[9] A. Zygmund. Trigonometric series: Vols. I, II. Cambridge University Press, London,1968.

157

The summation formula on the modular group impliesthe Kuznetsov summation formula

Johan Andersson∗

Abstract

In a previous paper [1] we showed that the Kuznetsov summation formula impliesa new type of summation formula on the full modular group PSL(2, Z). For testfunctions on PSL(2, R) we obtained an expansion for

∑“

a bc d

”∈PSL(2,Z)

f

(a bc d

)

in terms of spectral objects such as Fourier coefficients for Maass wave forms, holo-morphic cusp forms and Eisenstein series. In this paper we shall show the converse.Our general summation formula with the choice of a suitable test function impliesthe Kuznetsov summation formula

Contents

1 Introduction 159

2 Our main summation formula 160

3 A smooth definition of the Kloosterman sum 161

4 The Kuznetsov summation formula 163

5 Some remarks 166

1 Introduction

The Kloosterman sum

S(m, n; c) =∑

hh≡1 (mod c)0≤h,h<c

e

(mh + nh

c

)(1)

∗Department of Mathematics, Stockholm University, johana@math.su.se

159

is an important exponential sum, with wide ranging applications in number theory. Specialcases of the Kloosterman sums include

S(m, 0; c) = cc(m), (2)

S(0, 0; c) = φ(c), (3)

where

cc(m) =∑

(h,l)=c0≤h<c

e

(mh

c

)(4)

denotes the Ramanujan sums, and φ(c) denotes Euler’s phi function. The sums have analgebraic-geometric interpretation (see Weil [6]). The most important tool in the theory ofKloosterman sums and the analytic theory of numbers is however the formula of Kuznetsov[5], that allows us to express a sum over the Kloosterman sums

∞∑c=1

S(m, n; c)Φ(c) =∞∑

j=1

ρj(m)ρj(n)Φm,n(κj) +1

π

∫ ∞

−∞

Φm,n(r)σ2ir(|m|)σ2ir(|n|)dr

|nm|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

ρj,k(m)ρj,k(n)Φm,n

(12− ki

),

in terms of spectral objects coming from the full modular group. Here we have

Ψj(x + iy) =∞∑

n=−∞

ρj(n)e(nx)Kiκj(|2πny|), and Ψj,k(z) =

∞∑n=1

ρj,k(n)zn, (5)

where Ψj∞j=1 is an orthonormalized basis of Maass wave forms, and Ψj,kθ(k)j=1 is an

orthonormal basis of holomorphic cusp forms of weight k (see Andersson [1]). In [1] weused the Kuznetsov summation formula as the principal tool to prove a a new type ofgeneral summation formula on the full modular group. In this paper we will show theconverse. Our summation formula on the full modular group will in a natural way implythe Kuznetsov summation formula. It is sorts of a circular proof, since we originally usedthe Kuznetsov summation formula to prove our main summation formula. However in ourforthcoming paper [2] we will present an independent proof of our summation formula,and then the approach in this paper will become a natural way to prove the Kuznetsovsummation formula.

2 Our main summation formula

We start by stating our summation formula which is Theorem 1 in [1]:

Theorem 1. Let f be a function on PSL(2, R), fulfilling the condition∣∣∣∣ ∂n

∂ξn

∂m

∂ηm

∂v

∂tvf

(ξt ∗t ηt

)∣∣∣∣ t−v

(t2+δ + t−1−δ)(ξ2 + η2 + 1)1+δ,

n, m, v = 0, 1, 2, 3,t > 0.

(6)

160

Let ∗ denote the number making the determinant of the matrix to be 1. We have

∑ad−bc=1

c>0

f

(a bc d

)= −

∞∑c=1

φ(c)

∫ ∞

−∞

∫ ∞

−∞f

(cξ ∗c cη

)dξdη

+∞∑

n=−∞

∞∑c=1

cc(n)

∫ ∞

−∞

(f

(cn ∗c cξ

)+ f

(cξ ∗c cn

))dξ

+∞∑

j=1

∑m,n6=0

ρj(m)ρj(n)F (κj; m, n) +∑

m,n6=0

1

π

∫ ∞

−∞

σ2ir(|m|)σ2ir(|n|)F (r; m, n)dr

|nm|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

∑m,n6=0

ρj,k(m)ρj,k(n)F((

12− k)i; m, n

)where

F (r; m, n) =

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(mξ + nη)Mr

(mn

t2

)f

(tξ ∗t tη

)dξdηdt, (7)

Mr is defined by

Mr(x) =

iπ

2 sinh πr(J2ri(4π

√x)− J−2ri(4π

√x)) , x ≥ 0,

2 cosh πrK2ri(4π√−x), x < 0.

(8)

K2ri(x) and J2ri(x) are the K-Bessel and J-Bessel-functions respectively, ρj(n) and ρj,k(n),defined by (5) are Fourier coefficients of Maass wave forms and holomorphic modular formsrespectively, cc(n) are the Ramanujan sums defined by (4) and φ is the Euler phi function.

3 A smooth definition of the Kloosterman sum

The Kloosterman sums are defined by

S(m, n; c) =∑

hh≡1 (mod c)0≤h,h<c

e

(mh + nh

c

). (9)

Let m, n be fixed non zero integers. It is clear that if we define

f

(a bc d

)=

Φ(c)e

(ma+nd

c

), 0 ≤ a < c, 0 ≤ d < c,

0, otherwise.(10)

then we have that ∑ad−bc=1

c>0

f

(a bc d

)=

∞∑c=1

S(m, n; c)Φ(c). (11)

161

We can not apply our main summation formula though, since the function f does notfulfill the conditions of the test function class (6). In fact f is neither differentiable norcontinuous. We will now show how we can solve this problem and see how we can use apartition of unity to express the Kloosterman sums in a smooth way.

Lemma 1. Suppose that Ψ ∈ C∞(R) is a smooth function defined so that

Ψ(x) =

1, x ≥ C,

0, x ≤ −C,(12)

for some constant C > 0. Then we have that

S(m, n; c) =∑

ad≡1 (mod c)a,d∈Z

(Ψ(a + c)−Ψ(a))(Ψ(d + c)−Ψ(d))e

(ma + nd

c

).

Proof. We first notice that the condition (12) implies

∞∑k=−∞

(Ψ(a + kc + c)−Ψ(a + kc)) = 1. (13)

By defining h, h, k, l so that

a = kc + h and d = lc + h, with 0 ≤ h, h < c,

we obtain∑ad≡1 (mod c)

a,d∈Z

(Ψ(a + c)−Ψ(a))(Ψ(d + c)−Ψ(d))e

(ma + nd

c

)=

∑hh≡1 (mod c)

0≤h,h<c

∞∑k,l=−∞

(Ψ(kc+h+c)−Ψ(kc+h))(Ψ(lc+h+c)−Ψ(lc+h))e

(m(ck + h) + n(cl + h)

c

).

(14)

Since e(x) has period 1 we have

e

(m(ck + h) + n(cl + h)

c

)= e

(mh + nh

c

),

and by changing the summation order, equation (14) equals(∞∑

k=−∞

(Ψ(kc + h + c)−Ψ(kc + h))

)(∞∑

l=−∞

(Ψ(lc + c + h)−Ψ(lc + h))

)e

(mh + nh

c

).

162

By equation (13) this equals ∑hh≡1 (mod c)

0≤h,h<c

e

(mh + nh

c

),

which by the definition (9) equals

S(m, n; c)

the Kloosterman sum.

4 The Kuznetsov summation formula

By the following Lemma we will be able to use our summation formula:

Lemma 2. Suppose that Ψ fulfills the conditions of Lemma 1 and that that Φ ∈ C(0,∞) isa continuous function on R+ such that for some δ > 0, one has for positive integers c > 0that

|Φ(c)| c−2−δ. (15)

Also suppose that

f

(a bc d

)= (Ψ(a)−Ψ(a− c))(Ψ(d)−Ψ(c− d))Φ(c)e

(ma + nd

c

).

Then ∑ad−bc=D

c>0

f

(a bc d

)=

∞∑c=1

Φ(c)S(m, n; c)

Proof. The condition (15) ensures absolute convergence, and by changing the summationorder this follows from Lemma 1.

Before stating the Kuznetsov summation formula we will prove yet another lemma thatwe will need.

Lemma 3. Suppose that Ψ ∈ C∞(R) is a function such that

Ψ(x) =

1, x ≥ C,

0, x ≤ −C,(16)

for some constant C > 0. Then for n being an integer one has that∫ ∞

−∞e(nx)(Ψ(xt)−Ψ(t(x− 1)))dx =

1, n = 0,

0, otherwise.

163

Proof. We get by using the substitution ξ = xt that∫ ∞

−∞e(nx)(Ψ(xt)−Ψ((x− 1)t))dx =

∫ ∞

−∞e(nξ/t)(Ψ(ξ)−Ψ(ξ − t))dξ,

which if n 6= 0 equals

Ψ(n/t)− e(nt/t)Ψ(n/t) = (1− e(nt/t))Ψ(n/t),

= 0,

since n is an integer. In case n = 0 we get∫ ∞

−∞e(nx)(Ψ(xt)−Ψ((x− 1)t))dx = lim

M→∞

∫ M

−M

e(0 · x)(Ψ(xt)−Ψ((x− 1)t))dx

= limM→∞

(∫ M

M−1

Ψ(xt)dx−∫ −M

−M−1

Ψ(xt)dx

),

= 1− 0,

= 0,

by equation (16).

We are now ready to prove the Kuznetsov summation formula

Theorem 2. (Kuznetsov summation formula) Suppose that Φ ∈ C3(0,∞) is a functionon R+, which for some δ > 0 and v = 0, 1, 2, 3 fulfills∣∣Φ(v)(t)

∣∣ t−v

t2+δ + t−1−δ. (17)

Then

∞∑c=1

S(m, n; c)Φ(c) =∞∑

j=1

ρj(m)ρj(n)Φm,n(κj) +1

π

∫ ∞

−∞

Φm,n(r)σ2ir(|m|)σ2ir(|n|)dr

|nm|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

ρj,k(m)ρj,k(n)Φm,n

(12− ki

),

where

Φm,n(r) =

∫ ∞

0

Mr

(mn

t2

)Φ(t)dt,

ρj(n), ρj,k(n) are defined by (5) and Mr is defined by

Proof. By Lemma 2 we have that

∞∑c=1

Φ(c)S(m, n; c) =∑

ad−bc=Dc>0

f

(a bc d

),

164

where f is defined by

f

(a bc d

)= (Ψ(a)−Ψ(a− c))(Ψ(d)−Ψ(c− d))Φ(c)e

(ma + nd

c

).

We have defined Ψ, Φ so that the test function f satisfies condition (6), which means thatwe can use Theorem 1. We have that∫ ∞

−∞

∫ ∞

−∞f

(cx ∗c cy

)dxdy

=

∫ ∞

−∞

∫ ∞

−∞e(mx + ny)Φ(c)(Ψ(cy)−Ψ(c(y − 1)))(Ψ(cx)−Ψ(c(x− 1)))dxdy,

=Φ(c)

∫ ∞

−∞e(mx)(Ψ(cx)−Ψ(c(x− 1)))dx

∫ ∞

−∞e(ny)(Ψ(cy)−Ψ(c(y − 1)))dy.

Since m, n 6= 0 we have by Lemma 3 that this vanishes, and it follows that the first term

∞∑c=1

φ(c)

∫ ∞

−∞

∫ ∞

−∞f

(cx ∗c cy

)dxdy = 0

also vanishes. Similarly by Lemma 3 the integral in the second and third terms can bewritten as∫ ∞

−∞

(f

(cn ∗c cx

)+ f

(cx ∗c cn

))dx

= Φ(c)

(∫ ∞

−∞e(mx)(Ψ(cx)−Ψ(c(x− 1)))dx +

∫ ∞

−∞e(nx)(Ψ(cx)−Ψ(c(x− 1)))dx

),

= Φ(c) · (0 + 0),

= 0,

and hence the second and third terms

∞∑n=−∞

∞∑c=1

cc(n)

∫ ∞

−∞

(f

(cn ∗c cx

)+ f

(cx ∗c cn

))dx = 0,

also vanish. We now consider the integral transform and we have that

F (r; µ, ν) =

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(µx + νy)Mr

(µν

t2

)f

(tx ∗t ty

)dxdydt,

=

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(µx + νy)Mr

(µν

t2

)Φ(t)(Ψ(xt)−Ψ((x− 1)t))×

× (Ψ(yt)−Ψ(t(y − 1)))e(mx + ny)dxdydt.

By changing the order of integration, we see that this equals

=

∫ ∞

0

Mr

(µν

t2

)Φ(t)

(∫ ∞

−∞e((m + µ)x)(Ψ(xt)−Ψ((x− 1)t))dx

)×

×(∫ ∞

−∞e((n + ν)y)(Ψ(yt)−Ψ((y − 1)t))dy

)dt.

165

The integrals with respect to x and y can be treated with Lemma 3 and we get∫ ∞

−∞e((m + µ)x)(Ψ(xt)−Ψ((x− 1)t))dx = δm,−µ,

and ∫ ∞

−∞e((n + ν)y)(Ψ(yt)−Ψ((y − 1)t))dy = δn,−ν .

We thus find that

F (r; µ, ν; r) = δµ,−mδν,−n

∫ ∞

0

Φ(t)Mr

(µν

t2

)dt.

The formula in Lemma 2 now follows from the fact that

ρj(−µ)ρj(−ν) = ρj(µ)ρj(ν),

together with the corresponding identities for Fourier coefficients of holomorphic cuspforms, and divisor functions.

5 Some remarks

Remark 1. The generalized Kuznetsov summation formula of Bykovsky-Kuznetsov- Vino-gradov [4]

∞∑c=1

S(D, m, n; c)Φ(c) =∞∑

j=1

αjtj(D)tj(m)tj(n)ΦD,m,n(κj)

+1

π

∫ ∞

−∞

ΦD,m,n(r)σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)dr

|nmD|ir|ζ(1 + 2ir)|2

+∞∑

k=1

θ(k)∑j=1

αj,ktj,k(D)tj,k(m)tj,k(n)ΦD,m,n

(12− ki

),

where S(D, m, n; c) are the generalized Kloosterman sums

S(D, m, n; c) =∑

hk≡D (mod c)0≤h,k<c

e

(mh + nk

c

).

can be proved in the same way from our summation formula for integer matrices from [3].

Remark 2. We have that the test function condition∣∣Φ(v)(t)∣∣ t−v

t32+δ + t−δ

. (v = 0, 1, 2, 3 and some δ > 0)

166

for the Kuznetsov summation formula when we use it as the starting point in our proof ofthe main summation formula in [1]. When we use the main summation formula to provethe Kuznetsov summation formula we get the condition

∣∣Φ(v)(t)∣∣ t−v

t2+δ + t−1−δ. (v = 0, 1, 2, 3 and some δ > 0)

for the functions we can use in the test function class. Hence we lose some information. Itwould be interesting to see if we can sharpen the proof of the Main summation formula sowe can get back the same class as we started with.

References

[1] J. Andersson. A summation formula on the full modular group, 2006.

[2] A summation formula on the full modular group II, forthcoming.

[3] A summation formula over integer matrices, 2006.

[4] V. Bykovsky, N. Kuznetsov, and A. Vinogradov. Generalized summation formulafor inhomogeneous convolution. In Automorphic functions and their applications(Khabarovsk, 1988), pages 18–63. Acad. Sci. USSR Inst. Appl. Math., Khabarovsk,1990.

[5] N. V. Kuznecov. The Petersson conjecture for cusp forms of weight zero and theLinnik conjecture. Sums of Kloosterman sums. Mat. Sb. (N.S.), 111(153)(3):334–383,479, 1980.

[6] A. Weil. On some exponential sums. Proc. Nat. Acad. Sci. U. S. A., 34:204–207, 1948.

167

The Selberg and Eichler-Selberg trace formulae

Johan Andersson∗

Abstract

In this paper we will show how we can use our summation formula from [2] toprove the Selberg and Eichler-Selberg trace formula in a unified way. Our versionof the Selberg trace formula is for the full modular group and with respect to Heckeoperators and it also distinguishes between odd and even forms. Related formulaehave previously been proved by Selberg [21] (without Hecke operators), Hejhal [13](with Hecke operators of prime orders), Strombergsson [23] (arbitrary Hecke opera-tors - all forms), Bogomolny-Georgeot-Giannoni-Schmit [3] (for Hecke operators andodd forms), but with other methods.

Contents

1 Introduction 169

2 Preliminaries 1702.1 The summation formula on integer matrices . . . . . . . . . . . . . . . . . 1702.2 A class number formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1702.3 Some smoothing lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1732.4 The Rankin-Selberg zeta function . . . . . . . . . . . . . . . . . . . . . . . 175

3 The trace formulae 177

4 Appendix 189

1 Introduction

The Selberg Trace Formula [21] is an important result that gives a connection between thespectral theory of the group Γ and the geometry of the Riemann surface Γ\H. In this paperwe will show how our summation formula on the full modular group implies the Selbergtrace formula for the full modular group as well as the Eichler-Selberg trace formula. Wewill give an arithmetic form of the formula that involves class numbers instead of lengthsof closed geodesics. However, as shown by Sarnak [20] it is possible to translate betweenthe arithmetic and geometric form of the trace formula.

∗Department of Mathematics, Stockholm University, johana@math.su.se

169

2 Preliminaries

2.1 The summation formula on integer matrices

We will start by stating our summation formula which is Theorem 1 in [2]:

Theorem 1. Let f be a function on A ∈ GL(2, R) : det A = D, where D is a nonzerointeger, fulfilling the conditions∣∣∣∣ ∂n

∂xn

∂m

∂ym

∂v

∂tvf

(xt ∗t yt

)∣∣∣∣ t−v

(t2+δ + t−1−δ)(x2 + y2 + 1)1+δ,

n, m, v = 0, 1, 2, 3,t > 0,

(1)

for some δ > 0, where ∗ denotes the number which makes the determinant equal to D. Onethen has the identity

∑ad−bc=D

c>0

f

(a bc d

)= −

∞∑c=1

∑d|(D,c)

φ( c

d

)∫ ∞

−∞

∫ ∞

−∞f

(cx ∗c cy

)dxdy

+∞∑

n=−∞

∞∑c=1

∑d|(D,c)

cc/d(n)

∫ ∞

−∞

(f

(cn ∗c cx

)+ f

(cx ∗c cn

))dx

+∞∑

j=1

∑m,n6=0

αjtj(D)tj(m)tj(n)F (κj; D, m, n)

+∑

m,n6=0

1

π

∫ ∞

−∞

σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)F (r; D, m, n)

|Dmn|ir|ζ(1 + 2ir)|2dr

+∞∑

k=1

θ(k)∑j=1

∑m,n6=0

αj,ktj,k(D)tj,k(m)tj,k(n)F((

12− k)i; D, m, n

),

where

F (r; D, m, n) =

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(mx + ny)Mr

(Dmn

t2

)f

(tx ∗t ty

)dxdydt, (2)

the function Mr(x) is defined by

Mr(x) =

iπ

2 sinh πr(J2ri(4π

√x)− J−2ri(4π

√x)) , x ≥ 0,

2 cosh πrK2ri(4π√−x), x < 0,

(3)

and αj, αj,k, tj(n), tj,k(m) are defined as in Andersson [2] or Motohashi [18], φ(c) denotesthe Euler phi function and the cc(n) denote the Ramanujan sums.

2.2 A class number formula

We will first prove an asymptotic formula involving the Hurwitz class number H(∆), thatcounts the number of reduced positive definite quadratic forms (a, b, c) of discriminant∆ = 4ac− b2.

170

Lemma 1. Let ∆ be a non square integer. One has then that

limN→∞

1

N

∑b2−4ac=∆

0<b≤c

e−cN =

6

π2× H(−∆)√

|∆|×

2 log εD, ∆ > 0,

π, ∆ < 0,

where

H(∆) =h(−D)

w(−D)f−1

∑cd|f

(D

d

)µ(d)c, (∆ = Df 2, D square free)

h(∆) is the narrow class number (see Gauss [10]), and

εD =xd +

√Dyd

2

the fundamental unit in the number field O(√

D).

Remark 1. The number H(∆) as defined in Lemma 1 coincides with the Hurwitz classnumber for ∆ > 0 (see Zagier [25] page 132) and is a natural generalization for ∆ < 0,where the Hurwitz class number is usually defined to be zero.

Proof. Following Zagier [25], equation (6), we introduce the zeta function

ζ(s, ∆) =∑

φ mod Γ|φ|=∆

∑(m,n)∈Z2/ Aut(Φ)

φ(m,n)>0

1

φ(m, n)s. (Re(s) > 1)

where the summation is over quadratic forms φ of discriminant ∆. When ∆ = Df 2, andD 6= 1 is square free we have ([25] Proposition 3.1 (iii))

ζ(s, ∆) =

0, ∆ ≡ 2, 3 (mod 4),

ζ(s)ζ(2s− 1), ∆ = 0,

ζ(s)LD(s)∑

d|f µ(d)(

Dd

)d−sσ1−2s

(fd

), ∆ ≡ 0, 1 (mod 4), ∆ 6= 0,

where

LD(s) =∞∑

n=1

(D

n

)n−s (Re(s) > 1)

is the Dirichlet L-series and(

Dn

)denotes the Kronecker symbol. We have that the product

ζ(s)LD(s) = ζQ(√

D)

171

is also the Dedekind zeta-function of the quadratic number field Q(√

D) (see e.g the dis-cussion in Cohn [4] Chapter X, section 7). The residue of the Dedekind zeta-function ats = 1 can be expressed (see e.g. Frohlich-Taylor [9], Theorem 61) by

lims→1

ζK(s)(s− 1) =2s+tπtRKhK

wK

√|dK |

,

where hK is the class number of K, RK is the regulator of K, s (resp 2t) is the number ofreal (resp. imaginary) embeddings of K, wK is the number of roots of unity in K, and dK

denotes the discriminant of K. In the case of a real quadratic number field K = Q(√

D)we have that

s = 1, t = 0, RK = log(εD), hK = h(−D), wK = 1,

where εD is the fundamental unit in Q(√

D). In the case of an imaginary quadratic numberfield Q =

√−D we have that

s = 0, t = 2, RK = 1, hK = h(D), wK = w(D),

and we get when ∆ ≡ 0, 1 (mod D) that

lims→1

ζ(s, ∆)(1− s) =h(−D)√

|d|

∑d|f

µ(d)

(D

d

)d−1σ−1

(f

d

)×

2π

w(D), D > 0,

log εD, D < 0.(4)

The factor ∑d|f

µ(d)

(D

d

)d−1σ−1

(f

d

)

can be written as

f−1∑cd|f

(D

d

)µ(d)c.

With the notation

a∆(n) = #x : x2 ≡ ∆ (mod n), (5)

we have that ([25] Proposition 3.1 (i))

ζ(s, ∆) = ζ(2s)∞∑

n=1

a∆(n)n−s. (Re(s) > 1) (6)

172

From equations (4), (6), the fact that ζ(2) = π2/6, and the general fact that

lims→1+

∞∑n=1

ann−s =

C

s− 1=⇒ lim

N→∞

1

N

∞∑n=1

ane− n

N = C,

we get that

limN→∞

1

N

∞∑n=1

a∆(n)e−nN =

6

π2

H(−∆)√|∆|

×

2 log εD, ∆ > 0,

π, ∆ < 0.

The lemma now follows from noticing that by the definition (5), and identifying n = c andb = x, we obtain

∞∑n=1

a∆(n)e−nN =

∑b2−4ac=∆

0<b≤c

e−cN .

2.3 Some smoothing lemmas

In this section we will prove some auxiliary results that we will need later.

Lemma 2. Suppose that Ψ ∈ C∞(R) is a function such that

Ψ(x) =

1, x ≥ C,

0, x ≤ −C,(7)

for some constant C > 0. Let n be an integer and take t > 0. Then∫ ∞

−∞e(nx)(Ψ(xt)−Ψ(t(x− 1)))dx =

1, n = 0,

0, otherwise.

Proof. We get by using the substitution ξ = xt that∫ ∞

−∞e(nx)(Ψ(xt)−Ψ((x− 1)t))dx =

∫ ∞

−∞e(nξ/t)(Ψ(ξ)−Ψ(ξ − t))dξ,

which, if n 6= 0, equals

Ψ(n/t)− e(nt/t)Ψ(n/t) = (1− e(nt/t))Ψ(n/t),

and since n is an integer this is equal to

(1− 1)Ψ(n/t) = 0.

173

In case n = 0 we get∫ ∞

−∞e(nx)(Ψ(xt)−Ψ((x− 1)t))dx = lim

M→∞

∫ M

−M

e(0 · x)(Ψ(xt)−Ψ((x− 1)t))dx

= limM→∞

(∫ M

M−1

Ψ(xt)dx−∫ −M

−M−1

Ψ(xt)dx

),

which by equation (7) equals

= 1− 0 = 1.

Lemma 1 is not quite in a form that we can apply later, so we will now prove a variationof it.

Lemma 3. Let ∆ be a non square integer. Suppose that θN ∈ C∞0 (R) is a sequence of test

functions converging to the unit step function, with the further properties

0 ≤θN(x) ≤ 1, (x ∈ R) (8)

θN(x) =

1, x ≥ 0,

0, x ≤ −1/N2.(9)

Then that

limN→∞

1

N

∑b2−4ac=∆

(θN

(b

c

)− θN

(b

c− 1

))e−

cN =

6

π2

H(−∆)√|∆|

×

2 log εD ∆ > 0

π ∆ < 0

where H(∆) is defined as in Lemma 1.

Proof. This follows from Lemma 1 and∑b2−4ac=∆

0<b≤c

e−cN =

∑b2−4ac=∆

c>0

(θN

(b

c

)− θN

(b

c− 1

))e−

cN + O(1). (10)

This can be seen as follows: If we choose not to exclude the case when b = c and c = 0from the sum, the difference between the different sides of the equality in (10) will be oforder

∞∑c=N2

ce−c/N = O(N2e−N

).

For the remaining case b = 0 or b = c there will be at most 2d(∆) different choices of cwhich is bounded.

174

2.4 The Rankin-Selberg zeta function

Before we proceed with our main theorems we will state some facts about the Rankin-Selberg zeta functions. We will consider the following versions.

1. The Rankin-Selberg zeta function

Zj(s) = αj

∞∑n=1

|tj(n)|2n−s (Re(s) > 1)

associated with Maass wave forms;

2. The Rankin-Selberg zeta function

Zj,k(s) = αj,k

∞∑n=1

|tj,k(n)|2n−s (Re(s) > 1)

associated with holomorphic cusp forms;

3. The Rankin-Selberg zeta function

Z(r; s) =∞∑

n=1

|σ2ir(n)|2n−s (Re(s) > 1)

associated with the Eisenstein series.

The Rankin-Selberg zeta function associated with the Eisenstein series is easiest to treatsince the Ramanujan identity (see [11], theorem 305)

∞∑n=1

σα(n)σβ(n)n−s =ζ(s)ζ(s− α)ζ(s− β)ζ(s− α− β)

ζ(2s− α− β), Re(s−α),Re(s−β),

Re(s−α−β),Re(s)>1,

implies that

Z(s; r) =1

|ζ(1 + 2ir)|2ζ(2ir + s)ζ(s− 2ir)ζ2(s)

ζ(2s). (11)

After logarithmic differentiation this implies the following result.

Lemma 4. The function Z(r; s) has the following Laurent expansion at s = 1:

Z(s; r) =c−2

(s− 1)2+

c−1(r)

s− 1+ O(1),

where

c−2 =6

π2, and c−1(r

′) = Re

(ζ ′(1 + 2ir)

ζ(1 + 2ir)

)− ζ ′(2)

ζ(2).

175

The corresponding result for the Rankin-Selberg zeta functions of Maass wave formscan be found in Motohashi [18] Lemma 3.5, and for the Rankin-Selberg zeta functions ofholomorphic cusp form in Rankin [19] and Selberg [22]:

Lemma 5. The functions Zj(s) and Zj,k(s) are analytic for Re(s) > 0 with the exceptionof a simple pole at s = 1 with residue 12/π2.

Equation (11) together with the functional equation for the Riemann zeta function alsoimplies that

Z∗(r; s) = Z∗(r; 1− s), (12)

where

Z∗(r; s) = (2π2)−sζ(2s)Γ(s)Γ(s/2)2Γ(12(s + ri))Γ(1

2(s− ri))Z(r; s),

The corresponding functional equation can also be proved for the Rankin-Selberg zetafunction

Z∗j (s) = Z∗

j (1− s), (13)

where

Z∗j (s) = (2π2)−sζ(2s)Γ(s)Γ(s/2)2Γ(1

2(s + κji))Γ(1

2(s− κji))Zj(s),

corresponding to Maass wave forms (Lemma 3.5 in Motohashi [18]), as well as the Rankin-Selberg zeta function

Z∗j,k(s) = Z∗

j,k(1− s), (14)

where

Z∗j,k(s) = (2π2)−sζ(2s)Γ(s)Γ(s/2)2Γ(1

2(s + 1

2− k))Γ(1

2(s− 1

2+ k))Zj,k(s),

associated with holomorphic cusp forms (Rankin [19] or Selberg [22]).

Lemma 6. Let ε > 0 be given. There exist a constant 1/2 < c < 1 such that

|Zj(c + it)| ≤ (κj(1 + |t|))ε,

|Zj,k(c + it)| ≤ (k(1 + |t|))ε,

|Z(r; c + it)| ≤ ((1 + |r|)(1 + |t|))ε.

Proof. This follows from the functional equations (12), (13) and (14), the Phragmen-Lindelof principle and the fact that

x∑n=1

|tj(n)|2 ≤ xκεj,

(see Iwaniec [15] Theorem 8.3) together with The Ramanujan-Petersson conjecture

|tj,k(n)| ≤ d(n),

which was proved for holomorphic cusp forms by Deligne [5].1

1To refer to Deligne is really overkill and not necessary since a weaker version is sufficient.

176

Remark 2. Motohashi and Jutila [16] have recently proved sharp estimates on the corre-sponding Hecke L-functions Hj(1/2+ it) and Hj,k(1/2+ it) in both κj and t simultaneously(similarly to Lemma 6). Even though it is clearly an interesting problem to find sharp es-timates (in particular sub convexity estimates) for the Rankin-Selberg zeta-functions inboth κj and t, for our purposes the weak estimates of Lemma 6 are sufficient.

3 The trace formulae

We are now ready to prove the trace formula

Theorem 2. Suppose that g ∈ C∞0 (R) is a smooth function with compact support, such

that

g

(d2

4|D|

)= 0 whenever d2 − 4D is a square. (15)

One then has the identity

∞∑d=1

H(4D − d2)√|4D − d2|

×

2 log εd2−4D 4D − d2 < 0

π 4D − d2 > 0

× g

(d2

4|D|

)=

π

6

∫ ∞

−∞σ2ir(|D|)|D|−ir Re

(ζ ′(1 + 2ir)

ζ(1 + 2ir)

)g(r)dr +

∞∑k=1

θ(k)∑j=1

tj,k(D)g((1

2− k)i

)+

∞∑j=1

tj(D)g(κj) + σ−2(|D|)√|D|∫ ∞

0

g(y)dy√

y− 1

2

∑d|D

∫ ∞

−∞

g(x2) dx∣∣∣2x− (√|D|/d + d/

√|D|)

∣∣∣ ,where ε∆, H(∆), are defined as in Lemma 1,

g(r) =

∫∞

1Re((

x +√

x2 − 1)2ri)(x2 − 1)−1/2g(x2)dx, D > 0,∫∞

0Re((

1 +√

1 + x2)2ri)(1 + x2)−1/2g(x2)dx, D < 0,

and

g((

12− k)i)

=

(−1)k

∫ 1

0cos((2k − 1) arcsin x)(1− x2)−1/2g(x2)dx, D > 0,

0, D < 0.

Proof. We have that

det

(b2− d

2a

c b2

+ d2

)=

b2 − 4ac

4− d2

4.

Now −∆ = 4ac− b2 is the discriminant of the quadratic form. We see that if

gN

(x yz w

)=

π2

6×

1N

g(

(x−w)2

4|D|

)e−

zN , 0 ≤ x + w ≤ z,

0, otherwise,

177

then we get with b/2− d/2 = e and b/2 + d/2 = f that (Since the function g has compactsupport there are only finitely many non zero elements in d. Hence the limit and summationcan be interchanged.):

limN→∞

∑ef−ac=−D

c>0

gN

(e ac f

)=

∞∑d=−∞

limN→∞

1

N

π2

6g

(d2

4|D|

) ∑b,a,c

b2−4ac−d2=−4D0≤b≤c

e−cN . (16)

Notice that by condition (15), g(d2/4|D|) vanishes when −4D + d2 is square, and by usingLemma 1 on the inner sum, we find that this equals

∞∑d=−∞

H(4D − d2)√|4D − d2|

×

2 log εd2−4D 4D − d2 < 0

π 4D − d2 > 0

× g

(d2

4|D|

).

The function gN tends to 0 nicely as the argument tends to infinity. However it is notsmooth at x + w = z. That is the reason why we proved the smoothed class numberformula Lemma 3. By choosing instead

fN

(x yz w

)=

π2

6

1

Ng

((x− w)2

4|D|

)e−

zN (θN(x + w)− θN(x + w − z)),

where θN is a function converging to the unit step function that satisfies equations (8) and(9), we have that fN fulfill equation (1), the condition of Theorem 1. By Lemma 3 wehave that (Again since function g has compact support, the summation and limit can beinterchanged):

∞∑d=−∞

H(4D − d2)√|4D − d2|

×

2 log εd2−4D, 4D − d2 < 0

π, 4D − d2 > 0

× g

(d2

4|D|

)= lim

N→∞

∑ad−bc=−D

c>0

fN

(a bc d

),

(17)

and we can apply Theorem 1. We get that the first term is

−∞∑

c=1

∑d|(D,c)

φ( c

d

)∫ ∞

−∞

∫ ∞

−∞fN

(cx ∗c cy

)dxdy =

−π2

6

∞∑c=1

∑d|(D,c)

φ( c

d

)e−c/N

N

∫ ∞

−∞

∫ ∞

−∞g

(c2(x− y)2

4|D|

)(θN(c(x + y))− θN(c(x + y − 1)))dxdy.

With the substitution

ν =cx− cy

2√|D|

, µ = y + x, we have dxdy =

√|D|dµdν

c,

178

the integral splits and we obtain

−π2√|D|

6

∞∑c=1

∑d|(D,c)

φ( c

d

)e−c/N

cN

∫ ∞

−∞g(ν2)dν

∫ ∞

−∞(θN(µ)− θN(µ− 1))dµ.

By Lemma 2 we get that the first term equals

−π2√|D|

6

∞∑c=1

∑d|(D,c)

φ( c

d

)e−c/N

cN

∫ ∞

−∞g(ν2)dν. (18)

The second term is

∞∑c=1

∞∑n=−∞

∑d|(c,D)

cc/d(n)

∫ ∞

−∞

(fN

(cn ∗c cx

)+ fN

(cx ∗c cn

))dx =

π2

3

∞∑c=1

∞∑n=−∞

∑d|(c,D)

cc/d(n)

∫ ∞

−∞

1

Ne−

cN g

(c2(x− n)2

4|D|

)(θN(c(x + n))− θN(c(x + n− 1)))dx.

Since g has finite support this can be estimated by

π2

3

∞∑c=1

∞∑n=−∞

∑d|(c,D)

cc/d(n)

∫ ∞

−∞

1

Ne−

cN g

(c2x2

4|D|

)(θN(cx)− θN(c(x− 1)))dx

+ O

√|D|N

∞∑c=1

c−1

∑1≤|n|≤M

∑d|(c,D)

∣∣cc/d(n)∣∣e−

cN

∫ ∞

−∞

1

Ne−

cN

∣∣∣∣g(c2(x− n)2

4|D|

)∣∣∣∣dx

.

With the substitution ν = cx/2√|D| the error term can be estimated by O

(√|D|/N

),

and by first using (2) and then the same substitution ν = cx/2√|D| on the main term we

obtain

2π2√|D|

3

∞∑c=1

∞∑n=−∞

∑d|(c,D)

φ( c

d

)e−cN

cN

∫ c

0

g(ν2)dν + O

(√|D|N

).

Since g has compact support we see that if we integrate to infinity instead of c we introduce

an error O(√

|D|/N). Also since g(x2) is even we can write the second term as

π2√|D|

3

∞∑c=1

∞∑n=−∞

∑d|(c,D)

φ( c

d

)e−cN

cN

∫ ∞

−∞g(ν2)dν + O

(√|D|N

). (19)

We see that for fixed D, the sum of the first two terms equations (18) and (19) will be

π2√|D|

6

∞∑c=1

∞∑n=−∞

∑d|(c,D)

φ( c

d

)e−cN

cN

∫ ∞

−∞g(ν2)dν + O

(√|D|N

).

179

By letting k = c/d this equals

π2√|D|

6

∑d|D

∞∑k=1

φ(k)e−dk/N

dkN

∫ ∞

−∞g(ν2)dν. (20)

By the fact that ζ(s)/ζ(2s) =∑∞

n=1 φ(n)n−s we can prove that

∞∑n=1

φ(n)e−n/M =6

π2M + O

(√M),

and by this result we find that eq. (20) equals

σ−2(|D|)√|D|∫ ∞

−∞g(ν2)dν + O

(√|D|N

).

After the substitution y = ν2 this equals

σ−2(|D|)√|D|∫ ∞

0

g(y)dy√

y+ O

(√|D|N

). (21)

We will now consider the integral transform FN(r;−D, m, n). We get

FN(r;−D, m, n) =π2

6

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(mx + ny)Mr

(−Dmn

t2

)fN

(tx ∗t ty

)dxdydt

=π2

6

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(mx + ny)Mr

(−Dmn

t2

)1

Ne−

tN g

(t2(x− y)2

4|D|

)×

× (θN(t(x + y))− θN(t(x + y − 1)))dxdydt.

By using the substitution

µ = x + y, ν = x− y,

we have

dxdy = 12dµdν, x = 1

2(µ + ν), y = 1

2(µ− ν),

and we get that

FN(r;−D, m, n) =π2

12

∫ ∞

0

∫ ∞

−∞

∫ ∞

−∞e(

12(m + n)µ + 1

2(m− n)ν

)Mr

(−Dmn

t2

)×

× g

(t2ν2

4|D|

)1

Ne−

tN (θN(µt)− θN(µ(t− 1)))dµdνdt.

We first consider the integral with respect to ν. We have

π2

12

∫ ∞

−∞e(

12(m− n)ν

)g

(t2ν2

4|D|

)dν,

180

which with the change of variables

ξ =tν√|D|

, dν =

√|D|t

dξ,

becomes

π2

12

√|D|t

∫ ∞

−∞e

((m− n)

√|D|ξ

2t

)g

(ξ2

4

)dξ =

π2

12

√|D|t

G

((m− n)

√|D|

2t

),

where

G(t) =

∫ ∞

−∞e(xt)g

(x2

4

)dx, (22)

is the Fourier transform of g(ξ2/4). In particular we have that G belongs to the Schwartzclass since g(x) and hence also g(ξ2/4) belongs to the Schwartz class, and the Fouriertransform maps the Schwartz class onto the Schwartz class2. By Lemma 2 we get that∫ ∞

−∞(θN(µt)− θN(µ(t− 1)))e

(12(m + n)µ

)dµ = δm,−n.

We thus obtain

FN(r;−D, m, n) =π2

12δm,−n

√|D|∫ ∞

0

G

((m− n)

√|D|

2t

)Mr

(−Dmn

t2

)1

Ne−

tN

dt

t.

With the substitution

x =

√|Dn2|t

,dt

t=

dx

x,

we then get

FN(r;−D, m, n) =π2

12δm,−n

√|D|∫ ∞

0

G(x)Mr

(sign(D) · x2

)e−√|Dn2|xN

N

dx

x.

We will first consider the contribution coming from a single Maass wave form. The factor1/2 disappears when we consider the pair (m, n) = (n,−n), and (m, n) = (−n, n) jointly.By further using the identity

tj(−D)tj(n)tj(−n) = tj(D)t2j(n),

we see that the contribution from a single Maass wave form will be

π2

6tj(D)

∞∑n=1

αjt2j(n)

∫ ∞

0

G(x)Mκj

(sign(D) · x2

)e−√|D|nxN

N

√|D|dx

x.

2See any text book on Fourier analysis.

181

By using the Rankin-Selberg zeta function

Zj(s) = αj

∞∑n=1

t2j(n)n−s,

together with Mellin transforms and complex integration, we find that this equals

tj(D)π2

6

1

2πi

∫ 2+∞i

2−∞i

Zj(s)

∫ ∞

0

ts−1

∫ ∞

0

G(x)Miκj

(sign(D) · x2

)e−√|D|txN

N

√|D|dx

xdtds.

By changing the integration in the Mellin transform the inner integral becomes

∫ ∞

0

tse−

√|D|t

xN

N

dt

t.

With the substitution

y =

√|D|t

xN,

dt

t=

dy

y,

this equals

x√|D|

(xN√|D|

)s−1 ∫ ∞

0

ys−1e−ydt =x√|D|

(N√|D|

)s−1

Γ(s)xs−1,

and the contribution from a cusp form can be expressed as a Mellin-Barnes integral in-volving the Rankin-Selberg zeta-function

π2

6αjtj(D)

∞∑n=1

t2j(n)

∫ ∞

0

G(x)Mκj

(sign(D) · x2

)e−√|D|nxN

Ndx =

π2

6tj(D)

1

2πi

∫ 2+∞i

2−∞i

Zj(s)Γ(s)

(N√|D|

)s−1 ∫ ∞

0

xs−1G(x)Mκj

(sign(D) · x2

)dxds. (23)

Since by Lemma 5 the Rankin-Selberg zeta-function is holomorphic for Re(s) > 0 with theexception of a simple pole at s = 1 with residue 12/π2 we can move the integration line toc as given by Lemma 7, and we see that equation (23) equals

2tj(D)

∫ ∞

0

G(x)Mr

(sign(D) · x2

)dx

+π2

6tj(D) · 1

2πi

∫ c+∞i

c−∞i

Γ(s)

(N√|D|

)s−1 ∫ ∞

0

xs−1G(x)Mκj

(sign(D) · x2

)dxds.

182

Since G belongs to the Schwartz class it follows from Lemma 17∣∣∣∣∫ ∞

0

xc−1+itG(x)Mr

(sign(D) · x2

)dx

∣∣∣∣ (|t|+ 1)2

(|r|+ 1)5/2,

the Stirling formula (see e.g. Ivic [14], A.34)

|Γ(c + it)| |t|c−1/2e−π2|t|,

the growth estimate for the Rankin-Selberg zeta function given by Lemma 7

|Zj(c + it)| ((|t|+ 1)(|κj|+ 1))ε,

and the estimate ∣∣∣∣∣∣(

N√|D|

)c+it−1∣∣∣∣∣∣

(N√|D|

)c−1

,

that

π2

6tj(D)

1

2πi

∫ c+∞i

c−∞i(1/2<c<1)

Zj(s)Γ(s)

(N√|D|

)s−1 ∫ ∞

0

xs−1G(x)Mκj

(sign(D) · x2

)dxds

|tj(D)|

(N√|D|

)c−1

(|κj|+ 1)−5/2,

and the contribution from a single Maass wave form will be

2tj(D)

∫ ∞

0

G(x)Mκj

(sign(D) · x2

)dx + O

tj(D)(|κj|+ 1)ε−5/2

(N√|D|

)c−1. (24)

Similarly by the corresponding identities for the Rankin-Selberg zeta-function of holomor-phic cusp forms from Lemma 5 and Lemma 6

Zj,k(s) = αj,k

∞∑n=1

t2j,k(n)n−s, lims→1

(s− 1)Zj,k(s) =12

π2, Zj(c + it) |(1 + |t|)k|ε,

Lemma 18 ∣∣∣∣∫ ∞

0

xc+it−1G(x)Mi(k−1/2)

(sign(D) · x2

)dx

∣∣∣∣ 1

|k|7/3,

and the Stirling formula, we get that the contribution from a holomorphic cusp form willbe

2tj,k(D)

∫ ∞

0

G(x)Mi(k−1/2)

(sign(D) · x2

)dx + O

tj,k(D)

(N√|D|

)c−1

kε−7/3

. (25)

183

The contribution coming from the Eisenstein series∑m,n6=0

1

π

∫ ∞

−∞

σ2ir(|D|)σ2ir(|m|)σ2ir(|n|)F (r; D, m, n)dr

|Dmn|ir|ζ(1 + 2ir)|2

can in a similar way be written as

π

6

∫ ∞

−∞

σ2ir(|D|)|D|ir

1

2πi

∫ 2+∞i

2−∞i

Z(r; s)Γ(s)

(N√|D|

)s−1

×

×∫ ∞

0

xs−1G(x)Mr

(sign(D) · x2

)dxdsdr, (26)

where

Z(r; s) =1

|ζ(1 + 2ir)|2∞∑

n=1

σ22ir(|n|)n−2ir−s.

In the same way as in the Maass wave form case we can now move the integration path in(26) to c given by Lemma 7. We will pick up the residue coming from s = 1, and we seethat (26) equals

π

6

∫ ∞

−∞

[σ2ir(|D|)|D|ir

× Ress=1

[Z(r; s)Γ(s)

(N√|D|

)s−1

×

×∫ ∞

0

xs−1G(x)Mr

(sign(D) · x2

)dx

]+ O

σ2ir(|D|)|D|ir

(N√|D|

)c−1

(|r|+ 1)ε−5/2

]dr.

(27)

By Lemma 5

Z(s; r) =c−2

(s− 1)2+

c−1(r)

s− 1+ O(1) (s → 1)

and the fact that

d

dsΓ(s)

(Nx√|D|

)s−1

=

(Γ′(s)

Γ(s)+ log

(Nx√|D|

))(Nx√|D|

)s−1

,

we obtain that

Ress=1

Z(r; s)Γ(s)

(N√|D|

)s−1 ∫ ∞

0

xs−1G(x)Mr

(sign(D) · x2

)dx

=

∫ ∞

0

G(x)Mr

(sign(D) · x2

)(Re

(ζ ′(1 + 2ir)

ζ(1 + 2ir)

)+ c0 + c1 log x

)dx

184

where the constants

c0 = −ζ ′(2)

ζ(2)+

6

π2

(γ + log

(N√|D|

)), and c1 =

6

π2

come from the definition of c−2 and c−1(r) in Lemma 4. By integrating the error term in(27) we see that the contribution from the Eisenstein series equals

π

6

∫ ∞

−∞

σ2ir(|D|)|D|ir

Re

(ζ ′(1 + 2ir)

ζ(1 + 2ir)

)∫ ∞

0

G(x)Mr

(sign(D) · x2

)dxdr

+π

6

∫ ∞

−∞

σ2ir(|D|)|D|ir

∫ ∞

0

(c0 + c1 log x)G(x)Mr

(sign(D) · x2

)dxdr

+ O

( N√|D|

)c−1. (28)

We will thus treat the integral

π

6

∫ ∞

−∞

∫ ∞

0

σ2ir(|D|)|D|ir

(c0 + c1 log x)G(x)Mr

(sign(D) · x2

)dxdr. (29)

By changing the integration order we find that it equals

π

6

∫ ∞

0

G(x)(c0 + c1 log x)

∫ ∞

−∞

σ2ir(|D|)|D|ir

Mr

(sign(D) · x2

)drdx.

By using Lemma 14

∫ ∞

−∞

σ2ir(|D|)|D|ir

Mr(ε · x2)dr = π∑d|D

cos

(2πx

(√|D|d

+ εd√|D|

)),

we see that equation (29) equals

π2

6

∫ ∞

0

G(x)(c0 + c1 log x)∑d|D

cos

(2πx

(√|D|d

+ sign(D)d√|D|

))dx,

which with the definition of G, equation (22), equals

π2

6c0 log

(√|D|N

)∑d|D

g

1

4·

(√|D|d

+ sign(D)d√|D|

)2

+

∫ ∞

0

G(x)∑d|D

cos

(2πx

(√|D|d

+ sign(D)d√|D|

))log x dx. (30)

185

But if

k =D

d+ d, then

k

2√|D|

=

(√|D|d

+ sign(D)d√|D|

), (31)

and we have that

g

1

4·

(√|D|d

+ sign(D)d√|D|

)2 = g

(k2

4|D|

).

We also have that

k2 − 4D = D2/d2 + 2D + d2 − 4D

=

(D

d− d

)2

is a square, and by the assumption (15) we find that

g

(k2

4|D|

)= 0.

Hence the first term in equation (30) vanishes, and eq. (29) equals

∫ ∞

0

G(x)∑d|D

cos

(2πx

(√|D|d

+ sign(D)d√|D|

))log x dx.

According to Lemma 7 in the Appendix and equation (31) this equals

−1

2

∑d|D

∫ ∞

−∞

g(x2) dx∣∣∣2x− k/√|D|∣∣∣ ,

with k defined by (31), and we get the contribution

−1

2

∑d|D

∫ ∞

−∞

g(x2) dx∣∣∣2x− (√|D|/d + d/

√|D|)

∣∣∣ .We thus find that the contribution coming from the Eisenstein series equals

π

6

∫ ∞

−∞

σ2ir(|D|)|D|ir

Re

(ζ ′(1 + 2ir)

ζ(1 + 2ir)

)∫ ∞

0

G(x)Mr

(sign(D) · x2

)dxdr

− 1

2

∑d|D

∫ ∞

−∞

g(x2)dx∣∣∣2x− (√|D|/d + d/

√|D|)

∣∣∣ + O

( N√|D|

)c−1. (32)

186

The main term in Theorem 2 now comes from summing over the first two terms in oursummation formula (21), the Maass wave form parts (24), the holomorphic cusp form parts(25), and the Eisenstein series part (32). The error terms coming from the first two termsin our main summation formula (21), Maass wave form parts (24), the holomorphic cuspform parts (25), and the Eisenstein series part (32), now sum up to

O

(√|D|N

)+

∞∑j=1

O

tj(D)(|κj|+ 1)−5/2

(N√|D|

)c−1

+∞∑

k=1

θ(k)∑j=1

O

tj,k(D)

(N√|D|

)c−1

k−7/3

+ O

( N√|D|

)c−1.

From the estimate for the Fourier coefficients of cusp forms∑K/2<κj≤K

αj|tj(m)|2 K2 + d3(m) log(2m)√

m,

θ(k)∑j=1

αj,k|tj,k(m)|2 kd3(m)√

m log(2m),

valid uniformly for k,K, m ≥ 1 (see Motohashi [18] Lemma 2.4, and equation (2.2.10)), weget that the error terms can be estimated by

O

(√|D|N

)+ O

( N√|D|

)c−1.

Since c− 1 < 0 and D is a constant with respect to N we have that the error term tendsto zero when N tends to infinity. By the definition of G, equation (22), we have

2

∫ ∞

0

G(x)Mr

(sign(D) · x2

)dx = 2

∫ ∞

0

∫ ∞

−∞e(xt)g

(t2

4

)dtMr

(sign(D) · x2

)dx,

= 2

∫ ∞

−∞

∫ ∞

0

e(xt)Mr

(sign(D) · x2

)dx g

(t2

4

)dt,

which under the substitution

2x = ξ, 2dx = dξ,

transforms to ∫ ∞

−∞

∫ ∞

0

e

(ξt

2

)Mr

(sign(D) · ξ2

4

)dξ g

(t2

4

)dt.

By Lemma 11 this equals

∫ ∞

−∞

0 |t/2| ≤ 1, ε = 1

Re

((|t/2|+

√(t/2)2 − 1

)2ri)

((t/2)2 − 1)−1/2 |t/2| > 1, ε = 1

Re

((1 +

√1 + (t/2)2

)2ri)

(1 + (t/2)2)−1/2 ε = −1

g

(t2

4

)dt,

187

for r real, and∫ ∞

−∞

(−1)k cos((2k − 1) arcsin |t/2|)(1− (t/2)2)−1/2, |t/2| < 1

0, |t/2| > 1

g

(t2

4

)dt,

for r = i(k − 1/2). With the further substitution x = t/2, we finish our proof.

We will see that both the Eichler-Selberg trace formula (see Eichler [6]), as well as theclassical Selberg trace formula (see e.g. Selberg [21] or Hejhal ([12] and [13])) both withHecke characters, and the factor εj that determines whether the form is odd or even willfollow. This means that our proof gives a unified way of proving both formulae. Note thatthe Eichler-Selberg trace formula can be obtained in an elementary way (see e.g. Zagier[26]), so this proof is certainly not the simplest. We should also remark that Kuznetsov’spaper [17] might be relevant, but we have not yet gotten hold on it to verify this.

Corollary 1. (Eichler-Selberg) Let D > 0 be a positive integer. Suppose that g ∈C∞([0, 1]) and g(1) = 0. One then has the identity

∑d2≤4D

H(4D − d2

)g

(d2

4D

)=√

D∞∑

k=1

θ(k)∑j=1

tj,k(D)g(k)

+2σ−2(D)D

π

∫ 1

0

√1− y

yg(y)dy −

√D

π

∑d|D

∫ 1

−1

g(x2)√

1− x2 dx√D/d + d/

√D − 2x

,

where H(∆) denotes the Hurwitz class number, and

g(k) =(−1)k

π

∫ 1

0

cos((2k − 1) arcsin x)g(x2)dx.

Proof. This follows from Theorem 2 by specializing to D > 0, and to a function g withsupport on [−1, 1].

Corollary 2. (Selberg trace formula with Hecke operators) Let D > 0 be a positive integer.Suppose that g ∈ C∞

0 (R) has support on [1,∞[ and is such that

g

(d2

4|D|

)= 0 whenever d2 − 4D is a square.

One then has the identity

∞∑d=1

H(4D − d2

)log εd2−4Dg

(d2

4D

)=

π

6

∫ ∞

−∞σ2ir(D)D1/2−ir Re

(ζ ′(1 + 2ir)

ζ(1 + 2ir)

)g(r)dr +

∞∑j=1

tj(D)√

Dg(κj)

−√

D

2

∑d|D

∫ ∞

−∞

g(x2)√

x2 − 1 dx∣∣∣2x− 2(√

D/d + d/√

D)∣∣∣ + σ−2(D)D

∫ ∞

1

√y − 1

yg(y)dy,

188

where ε∆, H(∆), are defined as in Lemma 1, and

g(r) =

∫ ∞

1

g(x2) Re

((x +

√x2 − 1

)2ri)

dx.

Proof. This follows from Theorem 2 by specializing to D > 0, and to a function g withsupport on [1,∞[.

Corollary 3. (Selberg trace formula with Hecke operators, and εj) Let D > 0 be a positiveinteger. Suppose that g ∈ S(R) belongs to the Schwartz class, such that

g

(d2

4|D|

)= 0 whenever d2 + 4D is a square.

We have

∞∑d=1

H(−4D − d2

)log εd2+4Dg

(d2

4D

)=

π

6

∫ ∞

−∞σ2ir(D)D1/2−ir Re

(ζ ′(1 + 2ir)

ζ(1 + 2ir)

)g(r)dr +

∞∑j=1

εjtj(D)√

Dg(κj)

−√

D

2

∑d|D

∫ ∞

−∞

g(x2)√

1 + x2 dx∣∣∣2x− 2(√

D/d− d/√

D)∣∣∣ + σ−2(D)D

∫ ∞

0

√1 + y

yg(y)dy,

where ε∆3, H(∆), are defined as in Lemma 1, and

g(r) =

∫ ∞

0

Re

((1 +

√1 + x2

)2ri)

g(x2)dx.

Proof. This follows from Theorem 2 by specializing to D < 0 in that result and by lettingD → −D.

Remark 3. The classical Selberg trace formula is usually given in a more geometric form,rather than arithmetic form, using length of closed geodesics instead of class numbers. Itis well known that these approaches are equivalent, see e.g. Sarnak [20].

Remark 4. In the sense of the common distinction between the Kuznetsov summationformula, and the Kuznetsov trace formula, our results should maybe have been calledthe Selberg and Eichler-Selberg summation formulae instead of trace formulae (the traceformula being the inversion of the summation formula).

4 Appendix

In the Appendix we will prove some results on integral transforms and special functionsneeded in the main text.

3Notice that εd2+4D is a unit in a quadratic number field, and εj determines weather the j’th Maasswave form is odd or even. Hence despite using the same notation, they are not the same.

189

Lemma 7. Suppose that g ∈ C∞0 (R), the function G is defined as

G(t) =

∫ ∞

−∞e(xt)g

(x2

4

)dx, and g

(α2

4

)= 0. (33)

One then has the identity∫ ∞

0

G(t) cos(αt) log t dt = −1

2

∫ ∞

−∞

g(x2)dx

|2x− α|.

Proof. Without loss of generality we will assume that g and hence also G are real functions.We will consider the Laplace transform

LG(s) =

∫ ∞

0

e−stG(t) log t dt. (Re(s) > 0)

With the definition of G this equals∫ ∞

0

∫ ∞

−∞e(yt)g

(y2

4

)dy e−st log t dt

This integral is absolutely convergent and hence we can change the integration order. Weobtain ∫ ∞

−∞

∫ ∞

0

e(2πiy−s)t log t dt g

(y2

4

)dy.

By the the Laplace transform (see e.g. Erdelyi-Magnus-Oberhettinger-Tricomi [7] 4.6 (1))∫ ∞

0

e−zt log t dt = − log(γz)

z(Re(z) > 0)

we obtain with z = s− 2πiy that

LG(s) =

∫ ∞

−∞

(− log(γ(s− 2πiy))

s− 2πiy

)g

(y2

4

)dy. (Re(s) > 0) (34)

Since by the assumption g and G are real we have for a real ε > 0 that

Re(LG(ε− 2πiα)) =

∫ ∞

0

e−εt cos(2πα)G(t) log t dt. (35)

By the assumption that g(α2/4) = 0 this will converge when ε → 0+ and by the equations(34) and (35) we obtain∫ ∞

0

cos(2πα)G(t) log t dt = Re

(∫ ∞

−∞

(− log(γ(−2πi(α− y)))

−2πi(α− y)

)g

(y2

4

)dy

)which simplifies to

−1

4

∫ ∞

−∞g

(y2

4

)dy

|α− y|

The substitution x = y/2 finishes our proof.

190

Lemma 8. The Kernel function Mr has the following Mellin-transform and Mellin-Barnesrepresentations:

(i)

∫ ∞

0

Mr(ε · x)xs−1dx = (2π)−2sΓ(s + ri)Γ(s− ri)

cos πs ε = 1cosh πr ε = −1

,

(|Im (r)| < Re(s) < 3/4)

(ii) Mr(x) =1

2πi

∫ c+∞

c−∞(2π)−2sΓ(s + ri)Γ(s− ri)

cos πs x > 0cosh πr x < 0

|x|−sds

(|Im(r)| < c < 1/2)

Proof. For a proof, see Lemma 3 in Andersson [1].

Lemma 9. Suppose that k is an integer. Then

Mi(1/2−k)(x) = π(−1)kJ2k−1

(4π√

x). (x > 0)

Proof. See Andersson [1] Lemma 8.

Lemma 10. Suppose that n ≥ 1, and x is real. Then

|Jn(x)| ≤ 18

πn1/3.

Proof. See Andersson [1] Lemma 7.

Lemma 11. For real r the Fourier transform of Mr(ε · x2/4)/x equals

∫ ∞

−∞e(itx)Mr

(ε · x2

4

)dx =

0, |t| ≤ 1, ε = 1,

Re((|t|+

√t2 − 1)2ri

)(t2 − 1)−1/2, |t| > 1, ε = 1,

Re((1 +

√1 + t2)2ri

)(1 + t2)−1/2, ε = −1.

Proof. By using the substitution

ξ = 2πx,dξ

2π= dx,

we obtain ∫ ∞

−∞e(itx)Mr

(ε · x2

4

)dx =

1

π

∫ ∞

0

cos(ξt)Mr

(ε ·(

x

4πξ

)2)

dξ.

With the definition (3) this equals

1

π

∫ ∞

0

cos(ξt)×

iπ

2 sinh πr(J2ri(ξ)− J−2ri(ξ)) , ε = 1

2 cosh πrK2ri(ξ), ε = −1

dξ.

For ε = −1 the result follows from Erdelyi-Magnus-Oberhettinger-Tricomi [7], 2.12 (40)∫ ∞

0

cos(xt)Kν(x)dx =π((1 +

√1 + t2)ν + (1 +

√1 + t2)−ν)

4 cos(νπ/2)√

t2 + 1

191

when we put ν = 2ri. For ε = 1 it follows from the discontinuous integral (see Watson[24], 13.42 (5))

∫ ∞

0

cos(xt)Jν(x)dx =

cos (ν arcsin |t|)√

1−t2, |t| < 1,

sin(πν/2)√t2−1

(|t|+

√t2 − 1

)ν, t > 1.

(Re(ν) > −1) (36)

By putting ν = 2ri and noticing that by (36) the integral∫ ∞

0

cos(xt)J2ir(x)dx (37)

is odd in r for |t| ≤ 1 and thus we see from the definition of Mr that it vanishes when|t| < 1. For |t| > 1 we obtain

(|t|+√

t2 − 1)2ri + (|t|+√

t2 − 1)−2ri

2√

t2 − 1=

Re((|t|+

√t2 − 1)2ri

)√

t2 − 1.

Lemma 12. For integer k the Fourier transform of M(1/2−k)i(±x2/4) is equal to∫ ∞

−∞e(itx)M(1/2−k)i

(x2

4

)dx =

(−1)k cos((2k − 1) arcsin |t|)(1− t2)−1/2, |t| < 1,

0, |t| > 1.

Proof. By using Lemma 9 we have that∫ ∞

−∞e(itx)M(1/2−k)i

(x2

4

)dx = (−1)k2π

∫ ∞

0

cos(2πtx)J2k−1(2πx)dx.

With the substitution

ξ = 2πx, dξ = 2πdx,

we find that this equals

(−1)k

∫ ∞

0

cos(tξ)J2k−1(ξ)dξ,

and the result follows from (36).

Lemma 13. Suppose that r and x > 0 are real numbers. One then has that

Mr(x) = |x|−ir

∫ ∞

0

t2ir−1 cos(2π(t +

x

t

))dt. (38)

Proof. We will consider the Mellin transform (in z)∫ ∞

0

xz−1−ir

∫ ∞

0

t2ir−1 cos(2π(t± x

t

))dtdx,

(x > 0

0 < Re(z) < 1/2

)

192

We change the integration order (That we can do this has to be justified. See Remark 5)∫ ∞

0

t2ir−1

∫ ∞

0

xz−1−ir cos(2π(t± x

t

))dxdt. (39)

By the summation formula for the cosine function we have

cos(2π(t± x

t

))= cos 2πt cos

(2πx

t

)∓ sin 2πt sin

(2πx

t

), (40)

and by the following expressions for the gamma function∫ ∞

0

sin(xt)xs−1dx = Γ(s) sin(πs

2

)t−s,

∫ ∞

0

cos(xt)xs−1dx = Γ(s) cos(πs

2

)t−s, (41)

we see that the integral given by eq. (39) equals

(2π)ir−zΓ(z − ir)

∫ ∞

0

tir+z−1

(cos 2πt cos

(π(z + ir)

2

)∓ sin 2πt sin

(π(z + ir)

2

))dt.

By using (40) and (41) again we see that it equals

(2π)ir−zΓ(z − ir)(2π)−z−irΓ(z + ir)×

×(

cos

(π(z − ir)

2

)cos

(π(z + ir)

2

)∓ sin

(π(z + ir)

2

)sin

(π(z − ir)

2

)),

which according to (40) equals

(2π)−2zΓ(z − ir)Γ(z + ir) cos(π

2((z − ir)± (z + ir))

). (42)

The result finally follows from the fact that by Lemma 8 this is the Mellin transform ofMr(x).

Remark 5. In the proof of Lemma 13 we have to justify that we can change the integrationorder, since the integral is not absolutely convergent. Doing a limit argument/analyticcontinuation in r and z also present some difficulties since if the situation improves at ∞it will get worse at 0. A way it can be done is as follows: Divide the integral∫ ∞

0

∫ ∞

0

t2ir−1xz−1−ir cos(2π(t± x

t

))dxdt =(∫ 1

0

+

∫ ∞

1

)(∫ 1

0

+

∫ ∞

1

)t2ir−1xz−1−ir cos

(2π(t± x

t

))dxdt,

and do a limit argument/analytic continuation in r and z separately for all four integrals.An alternative way to prove Lemma 13 is to use to the following results (see Erdelyi-Magnus-Oberhettinger-Tricomi [7], 6.5 (35) and (36))∫ ∞

0

cos[a(x + b2x−1)]xs−1dx = −πbs[Js(2ab) sin(πs/2) + Ys(2ab) cos(πs/2)], (|Re(s)| < 1)∫ ∞

0

cos[a(x− b2x−1)]xs−1dx = 2bsKs(2ab) cos(πs/2), (|Re(s)| < 1)

193

and (see Erdelyi-Magnus-Oberhettinger-Tricomi [8], 7.2.1 (4))

Yν(z) =1

sin νπ(Jν(z) cos(νπ)− J−ν(z)),

together with the definition of Mr(x), equation (3).

Lemma 14. Suppose that x 6= 0 is a real number, and D > 0. Then∫ ∞

−∞σ2ir(D)D−irMr(ε · x2)dr = π

∑d|D

cos

(2πx

(√D

d+ ε

d√D

)).

Proof. We use the expression from Lemma 13 to get∫ ∞

−∞σ2ir(D)D−irMr(ε · x2)dr =

∫ ∞

−∞σ2ir(D)D−ir|x|−ir

∫ ∞

0

t2ir−1 cos

(2π

(t + ε · x2

t

))dtdr,

With the substitution s = 2ri this equals

π · 1

2πi

∫ +∞i

−∞i

σs(D)(|x|√

D)−s

∫ ∞

0

ts−1 cos

(2π

(t + ε · x2

t

))dtds.

We recognize this as a Mellin Barnes integral and by Mellin inversion this equals

π∑d|D

cos

(2π|x|

(√D

d+ ε

d√D

)).

Lemma 15. When k ≥ 2 is an integer one has the estimate∣∣Mi(1/2−k)(x)∣∣ x2

|k|7/3.

Proof. This is Lemma 9 in Andersson [1].

Lemma 16. We have that∣∣∣∣∫ t

0

yMr

(nmD

y2

)dy

∣∣∣∣ t1/2(|mnD|3/4 + t3/2

)(1 + |r|)−3/2,∣∣∣∣∫ x

0

∫ t

0

yMr

(nmD

y2

)dydx

∣∣∣∣ x3/2(|mnD|3/4 + x3/2

)(1 + |r|)−5/2.

Proof. This is Lemma 6 in Andersson [2].

Lemma 17. If f ∈ S(R) belongs to the Schwartz class, then for r real, and for each0 < c1 < c2 one has the estimate∫ ∞

0

f(x)Mr(ε · x2)xs−1dx (|s|+ 1)2

(1 + |r|)5/2. (c1 ≤ Re(s) ≤ c2)

194

Proof. With the substitution

t =1

x,

dx

x=

dt

t,

we get ∫ ∞

0

f(x)Mr(ε · x2)xs−1dx =

∫ ∞

0

f

(1

t

)t−s−2 ·Mr

( ε

t2

)dt.

Since f belongs to the Schwartz class, we get by doing partial integration twice that thisequals∫ ∞

0

(f ′′(

1

t

)+ f ′

(1

t

)t2(3 + s) + f

(1

t

)t2(2 + s)(3 + s)

)t−6−s

∫ t

0

∫ x

0

yMr

(ε

y2

)dydxdt.

(43)

By Lemma 16 we have with D = ε, n = 1, m = 1∣∣∣∣∫ t

0

∫ x

0

yMr

(ε

y2

)dydx

∣∣∣∣ t3/2(1 + t3/2

)(1 + |r|)−5/2,

and it is clear that equation (43) can be estimated by

(|s|+ 1)2

(1 + |r|)5/2

∫ ∞

0

2∑v=0

∣∣∣∣f (v)

(1

t

)∣∣∣∣(t2 + 1)t−4−Re(s)t3/2(1 + t3/2

)dt. (c1 ≤ Re(s) ≤ c2)

(44)

Since f belongs to the Schwartz class it is of the order

(|s|+ 1)2

∫ ∞

1

t−1−Re(s)dt (1 + |r|)−5/2 (|s|+ 1)2

(1 + |r|)5/2. (c > 0) (45)

Lemma 18. If f ∈ S(R) belongs to the Schwartz class, then for k integer, and for each0 < c1 < c2 one has the estimate∫ ∞

0

f(x)Mi(1/2−k)(ε · x2)xs−1dx k−7/3, (c1 ≤ Re(s) ≤ c2)

Proof. By Lemma 15, we have for k ≥ 2

Mi(1/2−k)(x) = O( x

k7/3

),

and we get that∣∣∣∣∫ ∞

0

f(x)Mi(1/2−k)(ε · x2)xs−1dx

∣∣∣∣ ∫ ∞

0

xRe(s)+1|f(x)|dx k−7/3, (c1 ≤ Re(s) ≤ c2)

195

whenever f belongs to the Schwartz class. The case k = 1 follows from Lemma 9

Mi(1/2−k)(x) = −πJ1

(4π√

x),

and Lemma 10

|J1(x)| ≤ 18

π.

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