Reciprocity Theorems for Bettin–Conrey Sums

# Reciprocity Theorems for Bettin–Conrey Sums

Juan S. Auli Department of Mathematics
Dartmouth College
Hanover, NH 03755
U.S.A.
Université d’Évry Val d’Essonne
Laboratoire de Mathématiques et Modélisation d’Évry (CNRS-UMR 8071), I.B.G.B.I., 23 Bd. de France, 91037 Évry Cedex, France
and  Matthias Beck Department of Mathematics
San Francisco State University
San Francisco, CA 94132
U.S.A.
Dedicated to the memory of Tom M. Apostol
20 January 2017
###### Abstract.

Recent work of Bettin and Conrey on the period functions of Eisenstein series naturally gave rise to the Dedekind-like sum

 ca(hk) = kak−1∑m=1cot(πmhk)ζ(−a,mk),

where , and are positive coprime integers, and denotes the Hurwitz zeta function. We derive a new reciprocity theorem for these Bettin–Conrey sums, which in the case of an odd negative integer can be explicitly given in terms of Bernoulli numbers. This, in turn, implies explicit formulas for the period functions appearing in Bettin–Conrey’s work. We study generalizations of Bettin–Conrey sums involving zeta derivatives and multiple cotangent factors and relate these to special values of the Estermann zeta function.

###### Key words and phrases:
Dedekind sum, cotangent sum, Bettin–Conrey sum, reciprocity theorem, Hurwitz zeta function, period function, quantum modular form, Estermann zeta function.
###### 2010 Mathematics Subject Classification:
Primary 11F20; Secondary 11L03, 11M35.

## 1. Introduction and Statement of Results

Our point of departure is recent work of Bettin and Conrey [9, 8] on the period functions of Eisenstein series. Their initial motivation was the derivation of an exact formula for the second moments of the Riemann zeta function, but their work naturally gave rise to a family of finite arithmetic sums of the form

 ca(hk) = kak−1∑m=1cot(πmhk)ζ(−a,mk),

where , and are positive coprime integers, and denotes the Hurwitz zeta function

 ζ(a,x) = ∞∑n=01(n+x)a,

initially defined for and meromorphically continued to the -plane. We call and its natural generalizations appearing below Bettin–Conrey sums.

There are two major motivations to study these sums. The first is that is essentially a Vasyunin sum, which in turn makes a critical appearance in the Nyman–Beurling–Báez-Duarte approach to the Riemann hypothesis through the twisted mean-square of the Riemann zeta function on the critical line (see, e.g., [6, 17]). Bettin–Conrey’s work, for , implies that there is a hidden symmetry of this mean-square.

The second motivation, and the central theme of our paper, is that the Bettin–Conrey sums satisfy a reciprocity theorem:

 ca(hk)−(kh)1+aca(−kh)+kaaζ(1−a)πh

extends from its initiation domain to an (explicit) analytic function on , making nearly an example of a quantum modular form in the sense of Zagier [19]. In fact, Zagier’s “Example 0” is the Dedekind sum

 s(h,k) = 14kk−1∑m=1cot(πmhk)cot(πmk),

which is, up to a trivial factor, . Dedekind sums first appeared in the transformation properties of the Dedekind eta function and satisfy the reciprocity theorem [11, 12]

 s(h,k)+s(k,h) = −14+112(hk+1hk+kh).

We now recall the precise form of Bettin–Conrey’s reciprocity theorem.

###### Theorem 1.1 (Bettin–Conrey [8]).

If and are positive coprime integers then

where

 ψa(z) = iπzζ(1−a)ζ(−a)−iz1+acotπa2+iga(z)ζ(−a)

and

 ga(z) = −2∑1≤n≤M(−1)nB2n(2n)!ζ(1−2n−a)(2πz)2n−1 +1πi∫(−12−2M)ζ(s)ζ(s−a)Γ(s)cosπa2sinπs−a2(2πz)−sds.

Here denotes the Bernoulli number, is any integer , and the integral notation indicates that our integration path is over the vertical line .

We note that Bettin and Conrey initially defined through

 ψa(z) = Ea+1(z)−1za+1Ea+1(−1z),

in other words, is the period function of the Eisenstein series of weight ,

 Ea+1(z) = 1+2ζ(−a)∑n≥1σa(n)e2πinz,

where , and then showed that satisfies the properties of Theorem 1.1.

We have several goals. We start by showing that the right-hand side of Theorem 1.1 can be simplified by employing an integration technique for Dedekind-like sums that goes back to Rademacher [12]. This yields our first main result:

###### Theorem 1.2.

Let and suppose and are positive coprime integers. Then for any ,

Theorem 1.2 implies that the function

has a holomorphic continuation to the whole complex plane. In particular, in this sense Theorem 1.2 can be extended to all complex .

Second, we employ Theorem 1.2 to show that in the case that is an odd negative integer, the right-hand side of the reciprocity theorem can be explicitly given in terms of Bernoulli numbers.

###### Theorem 1.3.

Let be an odd integer and suppose and are positive coprime integers. Then

 h1−nc−n(hk)+k1−nc−n(kh) = (2πihk)n1i(n+1)!(nBn+1+n+1∑m=0(n+1m)BmBn+1−mhmkn+1−m).

Our third main result is, in turn, a consequence of Theorem 1.3: in conjunction with Theorem 1.1, it implies the following explicit formulas for and when is an odd negative integer.

###### Theorem 1.4.

If is an odd integer then for all

 ψ−n(z) = (2πi)nζ(n)(n+1)!n+1∑m=0(n+1m)BmBn+1−mzm−1

and

 g−n(z) = (2πi)ni(n+1)!n∑m=0(n+1m+1)Bm+1Bn−mzm.

In [8, Theorem 2], Bettin and Conrey computed the Taylor series of and remarked that, if is a negative integer, is a rational polynomial in . Theorem 1.4 generalizes this remark. We will prove Theorems 1.21.4 in Section 2. We note that both Theorem 1.3 and 1.4 can also be derived directly from Theorem 1.1.

Our next goal is to study natural generalizations of . Taking a leaf from Zagier’s generalization of to higher-dimensional Dedekind sums [18] and its variation involving cotangent derivatives [10], let be positive integers such that for , let be nonnegative integers, a complex number, and define the generalized Bettin–Conrey sum

 ca(k0k1⋯knm0m1⋯mn) = ka0k0−1∑l=1ζ(m0)(−a,lk0)n∏j=1cot(mj)(πkjlk0).

Here denotes the derivative of the Hurwitz zeta function with respect to .

This notation mimics that of Dedekind cotangent sums; note that

 cs(hk) = cs(kh00).

In Section 3, we will prove reciprocity theorems for generalized Bettin–Conrey sums, paralleling Theorems 1.2 and 1.3, as well as more special cases that give, we think, interesting identities.

Our final goal is to relate the particular generalized Bettin–Conrey sum

 q−1∑m=1cot(k)(πmx)ζ(−a,mq)

with evaluations of the Estermann zeta function at integers ; see Section 4.

## 2. Proofs of Main Results

In order to prove Theorem 1.2, we need two lemmas.

###### Lemma 2.1.

Let be a nonnegative integer. Then

 limy→∞cot(m)π(x±iy) = {∓iif m=0,0if m>0.

Furthermore, this convergence is uniform with respect to in a fixed bounded interval.

###### Proof.

Since , we may estimate

Given that the rightmost term in this inequality vanishes as , we see that

 limy→∞cotπ(x+iy) = −i.

Similarly, the inequality

 |−i+cotπ(x−iy)| = 2∣∣ei(2πx)e2πy−1∣∣ ≤ 2∣∣∣∣ei(2πx)e2πy∣∣−1∣∣ = 2|e2πy−1|

implies that . Since

it follows that . Similarly,

 |cscπ(x−iy)| = 2eπy|eiπxe2πy−e−iπx| ≤ 2eπy||eiπxe2πy|−|e−iπx|| = 2eπy|e2πy−1|

implies that . We remark that and

 ddz(cscz) = −csczcotz,

so all the derivatives of have a factor, and therefore,

 limy→∞cot(m)π(x±iy) = {∓i%if$m=0,$0if m>0.

Since the convergence above is independent of , the limit is uniform with respect to in a fixed bounded interval. ∎

Lemma 2.1 implies that

 limy→∞cot(m)πh(x±iy) = limy→∞cot(m)πk(x±iy) = {∓iif m=0,0if m>0,

uniformly with respect to in a fixed bounded interval.

The proof of the following lemma is hinted at by Apostol [4].

###### Lemma 2.2.

If and , then vanishes uniformly with respect to as .

###### Proof.

We begin by showing that vanishes as if . Since and , we have the integral representation [15, eq. 25.11.25]

 ζ(a,z) = 1Γ(a)∫∞0ta−1e−zt1−e−tdt,

which may be written as

 (2.1) ζ(a,z) = 1Γ(a)∫∞0ta−1e−tR(z)1−e−te−itI(z)dt.

Note that for fixed ,

 ∫∞0ta−1e−tR(z)1−e−tdt = ζ(a,R(z))Γ(a)

and

 ∫∞0∣∣∣ta−1e−tR(z)1−e−t∣∣∣dt = ∫∞0tR(a)−1e−tR(z)1−e−tdt = ζ(R(a),R(z))Γ(R(a)),

so the Riemann–Lebesgue lemma (see, for example, [14, Theorem  16]) implies that

 ∫∞0ta−1e−tR(z)1−e−te−itI(z)dt

vanishes as . By (2.1), this means that for fixed, vanishes as . In other words, pointwise with respect to as .

Moreover, the vanishing of as is uniform with respect to . Indeed, denote , then (2.1) implies that

 ∫∞0g(t)dt = Γ(a)ζ(a,R)

and

 ∫∞0|g(t)|dt = Γ(R(a))ζ(R(a),R).

It then follows from the Riemann–Lebesgue lemma that . If , we may write

 Γ(a)ζ(a,x±iy) = ∫∞0ta−1e−tx1−e−te∓itydt = ∫∞0g(t)e−it(±y−i(x−R))dt.

Since does not depend on , the speed at which vanishes depends on and . However, we know that , so the speed of the vanishing depends only on .

Finally, note that

 ζ(a,iy) = ∞∑n=01(iy+n)a = ∞∑n=01(1+iy+n)a+1(iy)a = ζ(a,1+iy)+1(iy)a,

so as , and the speed at which vanishes depends on that of . Thus, uniformly as , as long as . ∎

###### Proof of Theorem 1.2.

The idea is to use Cauchy’s residue theorem to integrate the function

 f(z) = cot(πhz)cot(πkz)ζ(a,z)

along as , where denotes the positively oriented rectangle with vertices , , and , for and (see Figure 1).

Henceforth, is such that , is a pair of coprime positive intergers, and and are as above, unless otherwise stated. Since is analytic inside , the only poles of are those of the cotangent factors. Thus, the fact that and are coprime implies that a complete list of the possible poles of inside is

 E={1h,…,h−1h,1k,…,k−1k,1}

and each of these poles is (at most) simple, with the exception of 1, which is (at most) double. For ,

 Resz=mhf(z) = cot(πkmh)cos(πm)ζ(a,mh)Resz=mh1sin(πhz) = 1πhcot(πkmh)ζ(a,mh).

Of course, an analogous result is true for for all , and therefore

 ∑z0∈EResz=z0f(z) = Resz=1f(z)+1πhh−1∑m=1cot(πkmh)ζ(a,mh)+1πkk−1∑m=1cot(πhmk)ζ(a,mk)

or, equivalently,

 (2.2) h1−ac−a(hk)+k1−ac−a(kh) = π(hk)1−a⎛⎝⎛⎝∑z0∈EResz=z0f(z)⎞⎠−Resz=1f(z)⎞⎠.

We now determine . The Laurent series of the cotangent function about 0 is given by

 cotz = 1z−13z−145z3−2945z5+⋯,

so, by the periodicity of , for in a small neighborhood of ,

 cot(πkz) = (1πk)1z−1−πk3(z−1)−(πk)345(z−1)3−2(πk)5945(z−1)5+⋯

and, similarly,

 cot(πhz) = (1πh)1z−1−πh3(z−1)−(πh)345(z−1)3−2(πh)5945(z−1)5+⋯.

Since is analytic in a small neighborhood of , Taylor’s theorem implies that

 ζ(a,z)=∞∑n=0bn(z−1)n,

where for (derivatives relative to ). Thus, the expansion of about 1 is of the form

 b0π2hk(1z−1)2+(b1π2hk)1z−1+(analytic part).

Given that , we know that [15, eq. 25.11.17], so . We conclude that and it then follows from (2.2) that

 (2.3) h1−ac−a(hk)+k1−ac−a(kh) = aζ(a+1)π(hk)a+π(hk)a−1∑z0∈EResz=z0f(z).

We now turn to the computation of via Cauchy’s residue theorem, which together with (2.3) will provide the reciprocity we are after. Note that the function is analytic on any two closed contours and and since the poles inside these two contours are the same, we may apply Cauchy’s residue theorem to both contours and deduce that

 ∫C(M1,ϵ)f(z)dz = ∫C(M2,ϵ)f(z)dz.

In particular, this implies that

 (2.4) limM→∞∫C(M,ϵ)f(z)dz = 2πi∑z0∈EResz=z0f(z).

Let be the path along from to . Similarly, define from to , from to , and from to (see Figure 1). Since ,

 ζ(a,z+1) = ∞∑n=01(n+z+1)a = ∞∑n=11(n+z)a = ζ(a,z)−1za,

and so the periodicity of implies that

Lemmas 2.1 and 2.2 imply that vanishes uniformly as (uniformity with respect to ) so

 limM→∞∫γ1f(z)dz = 0 = limM→∞∫γ2f(z)dz.

This means that

 limM→∞∫C(M,ϵ)f(z)dz = limM→∞(∫γ3f(z)dz+∫γ4f(z)dz)

and it follows from (2.3) and (2.4) that

This completes the proof of Theorem 1.2. ∎

To prove Theorem 1.3, we now turn to the particular case in which is an odd integer and study Bettin–Conrey sums of the form .

Let denote the -th polygamma function (see, for example, [15, Sec. 5.15]). It is well known that for a positive integer,

 ζ(n+1,z) = (−1)n+1Ψ(n)(z)n!

whenever (see, for instance, [15, eq. 25.11.12]), so for , we may write

 c−n(hk) = (−1)nkn(n−1)!k−1∑m=1cot(πmhk)Ψ(n−1)(mk).

By the reflection formula for the polygamma functions [15, eq. 5.15.6],

 Ψ(n)(1−z)+(−1)n+1Ψ(n)(z) = (−1)nπcot(n)(πz),

we know that if is odd, then

 Ψ(n−1)(1−mk)−Ψ(n−1)(mk) = πcot(n−1)(πmk)

for each . Therefore,

 2k−1∑m=1cot(πmhk)Ψ(n−1)(mk) = k−1∑m=1cot(πmhk)Ψ(n−1)(mk) +k−1∑m=1cot(π(k−m)hk)Ψ(n−1)(1−mk),

which implies that

 2k−1∑m=1cot(πmhk)Ψ(n−1)(mk) = k−1∑m=1cot(πmhk)(Ψ(n−1)(mk)−Ψ(n−1)(1−mk)) = −πk−1∑m=1cot(πmhk)cot(n−1)(πmk).

This means that for odd, is essentially a Dedekind cotangent sum. Indeed, using the notation in [10],

 c−n(hk) = π2kn(n−1)!k−1∑m=1cot(πmhk)cot(n−1)(πmk) = π2(n−1)!c⎛⎜ ⎜⎝kh1n−10n−1000⎞⎟ ⎟⎠.

Thus Theorem 1.3 is an instance of Theorem 1.2. Its significance is a reciprocity instance for Bettin–Conrey sums of the form in terms of Bernoulli numbers. For this reason we give the details of its proof.

###### Proof of Theorem 1.3.

We consider the closed contour defined as the positively oriented rectangle with vertices , , and , with indentations (to the right) of radius around 0 and 1 (see Figure 2).

Since contains the same poles of as the closed contour in Figure 1 used to prove Theorem 1.2, we may apply Cauchy’s residue theorem, letting , and we only need to determine in order to deduce a reciprocity law for the sums .

As in the case of , the integrals along the horizontal paths vanish, so using the periodicity of the cotangent to add integrals along parallel paths, as we did when considering , we obtain

 (2.5) limM→∞∫˜C(M,ϵ)f(z)dz = limM→∞(∫iϵiMg(z)dz+∫−iM−iϵg(z)dz)+∫γ3g(z)dz,

where denotes the indented path around 0 and

 g(z) = cot(πkz)cot(πhz)zn.

Given that is an odd function, the vertical integrals cancel and we may apply Cauchy’s residue theorem to integrate along the positively oriented circle of radius and centered at 0, to deduce that

 limM→∞∫˜C(M,ϵ)f(z)dz = −πiResz=0g(z).

This is the main reason to use the contour instead of . Indeed, integration along exploits the parity of the function , allowing us to cancel the vertical integrals in (2.5).

The expansion of the cotangent function is

 πzcot(πz) = ∞∑m=0(2πi)mBmm!zm,

with the convention that must be redefined to be zero. Thus, we have the expansion

 cot(πkz) = ∞∑m=−1(2i)(2πik)mBm+1(m+1)!zm

and of course, an analogous result holds for . Hence,

 (2.6) Resz=0g(z) = (2i)(2πi)nπhk(n+1)!n+1∑m=0(n+1m)BmBn+1−mhmkn+1−m,

and given that [15, eq. 25.6.2], the Cauchy residue theorem and (2.3) yield

 (2.7) (2πihk)n1i(n+1)!(nBn+1+n+1∑m=0(n+1m)BmBn+1−mhmkn+1−m)

Finally, note that the convention is irrelevant in (2.7), since in this sum is always multiplied by a Bernoulli number with odd index larger than 1. ∎

Note that Theorem 1.3 is essentially the same as the reciprocity deduced by Apostol for Dedekind–Apostol sums [3]. This is a consequence of the fact that for an odd integer, is a multiple of the Dedekind–Apostol sum . Indeed, for such [4, Theorem  1]

 sn(h,k) = in!(2πi)−nc−n(hk).

It is worth mentioning that although the Dedekind–Apostol sum is trivial for even [3, eq. (4.13)], in the sense that is independent of , the Bettin–Conrey sum is not.

The following corollary is an immediate consequence of Theorems 1.2 and 1.3.

###### Corollary 2.3.

Let be an odd integer and suppose and are positive coprime integers, then for any ,

 ∫ϵ−i∞ϵ+i∞cot(πhz)cot(πkz)zndz = 2(2πi)nhk(n+1)!n+1∑m=0(n+1m)BmBn+1−mhmkn+1−m.
###### Proof of Theorem 1.4.

Given that , it follows from Theorems 1.1 and 1.3 that

 (2.8) ψ−n(hk) = (2πi)nζ(n)(n+1)!n+1∑m=0(n+1m)BmBn+1−m(hk)m−1.

The function

 ϕ−n(z) = (2πi)nζ(n)(n+1)!n+1∑m=0(n+1m)BmBn+1−mz