# A Generalization of the Secant Zeta Function as a Lambert Series

Authors: H.-Y. Li; B. Maji; T. Kuzumaki
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

Recently, Lalín, Rodrigue, and Rogers have studied the secant zeta function and its convergence. They found many interesting values of the secant zeta function at some particular quadratic irrational numbers. They also gave modular transformation properties of the secant zeta function. In this paper, we generalized secant zeta function as a Lambert series and proved a result for the Lambert series, from which the main result of Lalín et al. follows as a corollary, using the theory of generalized Dedekind eta-function, developed by Lewittes, Berndt, and Arakawa.


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## Body

## 1. Introduction

The Dedekind eta-function and its limiting values have been considered by several authors starting from Riemann’s posthumous fragment [1] and Wintner [2] and later by Reyna [3] and Wang [4]. There are many generalizations of the Dedekind eta-function as a Lambert series including those of Lewittes [5], Berndt [6], and Arakawa [7, 8]. In particular cases, they reduce to the cotangent or the cosecant zeta function. Lerch [9] in 1904 introduced the cotangent zeta function for an algebraic irrational number z and an odd positive integer s as(1)ξz,s:=∑n=1∞cotnπzns.He stated the following functional equation for the cotangent zeta function, but without proof.Theorem 1 (see [9]).
For any algebraic irrational numberz and sufficiently large positive integer k=kz, we have(2)ξz,2k+1+z2kξ1z,2k+1=2π2k+1ϕz,2k+1,where(3)ϕz,n:=∑i=0n+1BiBn+1−jj!n+1−j!zj−1,where Bi is the i-th Bernoulli number.Berndt [10], in 1973, focused on the cotangent zeta function for general s∈ℂ and proved Lerch’s functional equation for cotangent zeta function. He found many interesting explicit formulae for ξz,s when z is a quadratic irrational and s≥3 is an odd integer. One such pleasing formula is(4)ξ1+52,3=−π3455.In fact, Berndt’s work implies thatjξj,sπ−s∈ℚ, where j is any positive integer and s≥3 is an odd integer.


## 2. Secant Zeta Function

Recently, Lalín et al. [11] considered the secant zeta function(5)ψz,s:=∑n=1∞secnπznsand found its special values at some particular quadratic irrational arguments. They proved the following results.Theorem 2 (see [11], Theorem 1).
The series (5) is absolutely convergent in the following cases:(1)
Whenz=p/q is a rational number with q odd and s>1.(2)
Whenz is an algebraic irrational number and s≥2.To prove this theorem, they have used the celebrated Thue–Siegel–Roth theorem.Theorem 3 (see [11], Theorem 3).
LetEm denote the Euler numbers and let Bm denote the Bernoulli numbers. Suppose that l is an even positive integer. Then, for appropriate values of α,(6)α+1l−1ψαα+1,l−−α+1l−1ψα−α+1,l=πill!∑n=0l2n−1−1BnEl−nln1+αn−1−1−αn−1.They found the values of the secant zeta function at some quadratic irrational numbers. Forj∈ℤ,(7)ψ2j2j+1,2=3j+1π26,ψ8j2j+1,2=π26,ψ2j2j+1,4=75j2+46j+68j+3π4180.After observing these values, they conjectured the following.Conjecture 1 (see [11], Conjecture 1).
Ifj is any positive integer and s is an even positive integer, then(8)ψj,sπ−s∈ℚ.By a clever use of residue theorem, Berndt and Straub [12] proved the above functional equation (6), and from it they derived(9)ψr,sπ−s∈ℚ,r∈ℚ+,s∈2ℕ.Furthermore, they connected the secant Dirichlet series with Eichler integrals of Eisenstein series and checked unimodularity of period polynomials. On the contrary, Charollais and Greenberg [13] related the secant Dirichlet series ψα,s to the generalized eta-function which was studied by Arakawa [7]. They proved that for s∈2ℕ,(10)ψα,sπ−s∈ℚα,for all real quadratic irrationals α. They used Arakawa’s result to give an explicit formula for ψα,s for real quadratic irrational numbers α.We will introduce a generalization of the secant zeta function as a Lambert series. Using the theory of generalized Dedekind eta-function due to Lewittes [5], Berndt [6], and Arakawa [7], we shall give a generalization of Theorem 3.We begin by briefly describing the theory of generalized Dedekind eta-function, developed by Lewittes [5], Berndt [6], and Arakawa [7], which is a main tool in our study.


## 3. Work of Lewittes and Berndt

Lewittes and Berndt treat the case of the upper half-planeℍ while Arakawa treats the case of upper half plane limiting to an algebraic irrational number. Hereafter, we use the following notations:(11)ew:=exp2πiw,w∈ℂ,x∈ℝ,0<x≦1,x−x∈ℤ,x∈ℝ,0≦x<1,x−x∈ℤ.Lewittes [5] defined the generalization of the Dedekind eta-function as a Lambert series. For a pair r1,r2 of real numbers, z∈ℍ and arbitrary s∈ℂ, he considered the series(12)Az,s,r1,r2:=∑m>−r1∑k=1∞ks−1ekr2+km+r1z,where the first summation is over all integers m with m>−r1. He also introduced its associate as(13)Hz,s,r1,r2:=Az,s,r1,r2+es2Az,s,−r1,−r2.Lets=r1=r2=0. Put Az,0,0,0=Az, then Hz,0,0,0=2Az. Using the product definition of Dedekind eta-function ηz, it is easy to show that(14)logηz=πi12−Az.Let us see a couple of examples.Example 1.
For special choices of parametersr1 and r2, the A- and H-functions reduce to the cosecant and cotangent zeta functions:(15)11+es/2Hz,s,12,0=Az,s,12,0=∑m>−1/2∑k=1∞ks−1ekm+12z=∑k=1∞ks−1e1/2kz1−ekz=i2∑k=1∞cosecπkzk1−s.
Also,(16)11+es/2Hz,s,1,0=Az,s,1,0=∑m>−1∑k=1∞ks−1ekm+1z=∑k=1∞ks−1ekz1−ekz=12∑k=1∞ks−11+ekz1−ekz−1=i2∑k=1∞cotπkzk1−s−12ζ1−s.
Some more definitions will be required.Definition 1 (Hurwitz zeta function).
For a positive number a, the Hurwitz zeta function(17)ζs,a:=∑n=0∞n+a−s,ℜs>1.Definition 2.
LetΩ denote the characteristic function of integers, i.e.,(18)Ωa:=1,a∈ℤ,0,a∉ℤ.
For any positive numberλ, let Iλ,∞ denote the integration path consisting of the oriented line segment +∞,λ, the positively oriented circle of radius λ with center at the origin, and the oriented line segment λ,+∞.
Let(19)G2z,ω1,ω2;t:=exp−zt1−exp−ω1t1−exp−ω2t,for any pair ω1,ω2 of positive numbers and for z,t∈ℂ.
Berndt [6] proved the following transformation formula.Theorem 4 (see [6], Theorem 2).
LetV=abcd∈SL2,ℤ with c>0. For any pair r1,r2 of real numbers, set R1=r1a+r2c,R2=r1b+r2d,ρ=R2c−R1d. For z∈ℍ with cℜz+d>0, let β=cz+d. Then, for arbitrary s∈ℂ, we have(20)β−sHVz,s,r1,r2−Hz,s,R1,R2=−Ωr12π−ses4β−sΓsζs,r2+es2ζs,−r2+ΩR12π−se−s4Γsζs,−R2+es2ζs,R2+2π−se−s4Lz,s,R1,R2,c,d,where(21)Lz,s,R1,R2,c,d=−∑j=1c∫Iλ,∞ts−1exp−1−jd+ϱ/c+cz+dj−R1/ct1−exp−t1−exp−cz+dtdt,0<λ<2π,2πβ.Here,logt is understood to be real-valued on the upper segment +∞,λ of Iλ,∞.


## 4. Work of Arakawa

Arakawa studied certain Lambert series associated to a complex variables and an irrational real algebraic number α. Those Lambert series are defined as limiting (boundary) values of the generalized Dedekind eta-functions studied by Berndt [6]. Arakawa obtained transformation formulae under the action of SL2,ℤ on those α.For an irrational real algebraic numberα and a pair p,q of real numbers, Arakawa [7] introduced a generalized eta-function defined as(22)ηα,s,p,q:=∑n=1∞ns−1enpα+q1−enα,s∈ℂ,and its associate by(23)Hα,s,p,q:=ηα,s,p,q+es2ηα,s,−p,−q.Example 2.
Again, if we considerp,q=1/2,0 and p,q=1,0, then also we will get the cosecant and cotangent zeta function:(24)11+es/2Hα,s,12,0=ηα,s,12,0=∑k=1∞ks−1e1/2kα1−ekα=i2∑k=1∞cosecπkαk1−s,(25)11+es/2Hα,s,1,0=ηα,s,1,0=∑k=1∞ks−1ekz1−ekz=i2∑k=1∞cotπkαk1−s−12ζ1−s,where s∈ℂ with ℜs<0.Theorem 5 (see [7], Lemma 1 and Theorem 2).
Supposeα∈ℝ∩ℚ¯ and α∉ℚ. Then, the infinite series ηα,s,p,q is absolutely convergent if ℜs<0. If, in addition, ℚα:ℚ=2 and p,q∈ℚ2, then Hα,s,p,q has analytic continuation to ℂ−0, and the singularity at s=0 is at worst a simple pole.Arakawa proved the absolute convergence ofηα,s,p,q for ℜs<0, by using the Thue–Siegel–Roth theorem.Consider the generalized eta-function(26)ηz,s,p,q=∑n=1∞ns−1enpz+q1−enz,s∈ℂcorresponding to (22), for z∈ℍ and a pair p,q∈ℝ2 with p>0. Then, one can see that this series is absolutely convergent for arbitrary s∈ℂ. It can be easily checked that there is a link between the infinite series Az,s,r1,r2 and ηz,s,r1,r2.Lemma 1.
For any pairr1,r2∈ℝ2 and z∈ℍ, we have(27)Az,s,r1,r2=ηz,s,r1,r2,s∈ℂ.Now, from the definition ofH-function (13), we have(28)Hz,s,r1,r2=Az,s,r1,r2+es2Az,s,−r1,−r2.Hence, using Lemma1, we get(29)Hz,s,r1,r2=ηz,s,r1,r2+es2ηz,s,−r1,−r2.Similarly, we haveLemma 2.
For any algebraic irrational numberα and a pair p,q∈ℝ2,(30)Aα,s,p,q=ηα,s,p,q,ℜs<0.Again by the definition ofH-function (23)(due to Arakawa), we have(31)Hα,s,p,q=ηα,s,p,q+es2ηα,s,−p,q.Therefore, by Lemma2, we get(32)Hα,s,p,q=Aα,s,p,q+es2Aα,s,−p,q.Proposition 1 (see [7], Proposition 1).
LetV=abcd∈SL2,ℤ, α be an irrational real algebraic number, and p,q∈ℝ2 with p>0. Let z=α+iy with y>0. Set z∗=Vz and β=Vα=aα+bcα+d−1. If ℜs<−3, then(33)limy⟶0+ηz∗,s,p,q=ηβ,s,p,q.Arakawa obtained the following transformation formulae forHα,s,p,q, by virtue of Theorem 4 of Berndt and Proposition 1.Theorem 6 (see [7], Theorem 1).
Letα be any real algebraic irrational, and let V=abcd∈SL2,ℤ with c>0 such that β:=cα+d>0. For any pair p,q of real numbers, set p′=pa+qc,q′=pb+qd, and ρ=q′c−p′d. Then, for ℜs<0,(34)D1V,α,s,p,q:=β−sHVα,s,p,q−Hα,s,p,qV=β−sHVα,s,p,q−Hα,s,p′,q′=−Ωp2π−ses4β−sΓsζs,q+es2ζs,−q+Ωp′2π−se−s4Γsζs,−q′+es2ζs,q′+2π−se−s4Lα,s,p′,q′,c,d,where(35)Lα,s,p′,q′,c,d=−∑j=1c∫Iλ,∞ts−1G21−jd+ρc+j−p′βc,1,β;tdt,0<λ<2π,2πβ.Berndt [6] (p. 499) found the special values of Lα,s,p′,q′,c,d at nonnegative integral arguments s=−m:(36)Lα,−m,p′,q′,c,d=2πim+2!∑j=1c∑k=0m+2m+2kBkj−p′cB¯m+2−kjd+ρc−βk−1,where Bnx denotes the nth Bernoulli polynomial and B¯nx=Bnx.Lemma 3 (see [7], Lemma 4).
Letα be an irrational number in a real quadratic field ℚΔ and let p,q be a pair of rational numbers. Then, there exist a totally positive unit β of ℚΔ and an element V=abcd of SL2,ℤwhich satisfy the conditions:(i)
c>0(ii)
p,qV≡p,qmod1(iii)
βα1=Vα1
We choose suchβ∈ℚΔ and V∈SL2,ℤ, i.e., which satisfy the conditions of Lemma 3. Then, using condition (ii), we have(37)Hα,s,p,q=Hα,s,p,qV.
SinceVα=α and c>0, we can see easily from Theorem 6 that(38)Hα,s,p,q=−Ωp2π−ses4Γsζs,q+Ωp2π−se−s4Γsζs,q1−esβ−sβ−s−1+2π−se−s/4β−s−1Lα,s,p,q,c,d.Example 3.
Letα,β, and V as in Lemma 3 and with p,q=1,0 and p,q=1/2,0. Then,(39)Hα,s,1,0=2π−s−es4+e−s41−esβ−sβ−s−1Γsζs+2π−se−s/4β−s−1Lα,s,1,0,c,d,Hα,s,12,0=2π−se−s/4β−s−1Lα,s,12,0,c,d.
Values at some particular matrices. Let(40)V0=0−110,V1=1011,V2=V02V1−1=−101−1.Example 4.
Theorem6 gives the following:(41)D1V0,α,s,p,q=α−sH−1α,s,p,q−Hα,s,q,−p=−Ωp2π−ses4α−sΓsζs,q+es2ζs,−q+Ωq2π−se−s4Γsζs,p+es2ζs,−p+2π−se−s4Lα,s,q,−p,1,0,(42)D1V1,α,s,p,q=α+1−sHαα+1,s,p,q−Hα,s,p+q,q=−Ωp2π−ses4α+1−sΓsζs,q+es2ζs,−q+Ωp+q2π−se−s4Γsζs,−q+es2ζs,q+2π−se−s4Lα,s,p+q,q,1,1,(43)D1V2,α,s,p,q=α−1−sH−αα−1,s,p,q−Hα,s,−p+q,−q=−Ωp2π−ses4α−1−sΓsζs,q+es2ζs,−q+Ω−p+q2π−se−s4Γsζs,q+es2ζs,−q+2π−se−s4Lα,s,−p+q,−q,1,−1.
In particular, whenp,q=1,0, we have(44)D1V0,α,s,1,0=2π−ses4e−s2−α−sΓs1+es2ζs+2π−se−s4Lα,s,0,−1,1,0,D1V1,α,s,1,0=2π−ses4e−s2−α+1−sΓs1+es2ζs+2π−se−s4Lα,s,1,0,1,1,D1V2,α,s,1,0=2π−ses4e−s2−α−1−sΓs1+es2ζs+2π−se−s4Lα,s,−1,0,1,−1.
If we choosep,q=1/2,0, we get(45)D1V0,α,s,12,0=α−sH−1α,s,12,0−Hα,s,0.−12=2π−ses4+e−s4Γsζs,12+2π−se−s4Lα,s,0,−12,1,0,D1V1,α,s,12,0=α+1−sHαα+1,s,12,0−Hα,s,12,0=2π−se−s4Lα,s,12,0,1,1,D1V2,α,s,12,0=α−1−sH−αα−1,s,12,0−Hα,s,−12,0=2π−se−s4Lα,s,−12,0,1,−1.
Note that for nonnegative integersm, we have the following explicit formulae for Vj, where j=0,1,2:(46)Lα,−m,1,0Vj,c,d=2πim+2!∑k=0m+2m+2kBk1B¯m+2−k1−βk−1,Lα,−m,12,0Vj,c,d=2πim+2!∑k=0m+2m+2kBk12B¯m+2−k12−βk−1.


## 5. Generalization of the Secant Zeta Function

We introduce two Lambert series corresponding to (22) and (12). These include the generalizations of secant and tangent zeta functions as shown in Example 5. Let α be any algebraic irrational number and p,q a pair of real numbers. Then, we define the series η∗ by(47)η∗α,s,p,q:=∑n=1∞ns−1enpα+q1+enα,ℜs<0and another infinite series A∗ by(48)A∗z,s,r1,r2:=∑m>−r1−1m∑k=1∞ks−1ekr2+km+r1z,for a pair r1,r2∈ℝ2, z∈ℍ, and s∈ℂ.Example 5.
If we taker1,r2=1,0, and 1/2,0, then (48) becomes(49)A∗α,s,1,0=η∗α,s,1,0=∑k=1∞ks−1ekα1+ekα=12∑k=1∞ks−1ekα−11+ekα+1=i2∑k=1∞ks−1tanπkα+12ζ1−s,(50)A∗α,s,12,0=ηα,s,12,0=12∑k=1∞ks−11cosπkα=12ψα,1−s,respectively.
By virtue of the results of Lewittes, Berndt, and Arakawa, we have the following results.Lemma 4.
Letα be an algebraic irrational number and p,q be a pair of real numbers. The series η∗α,s,p,q is absolutely convergent, if s∈ℂ with ℜs<0.Proof.
One can prove this result applying the Thue–Siegel–Roth theorem, in a similar manner to Arakawa’s procedure for proving the absolute convergence of the seriesηα,s,p,q.Lemma 5.
Ifz∈ℍ and a pair p,q∈ℝ2 with p>0, then the series η∗z,s,p,q is absolutely convergent for any s∈ℂ.Proof.
Sincez∈ℍ, assume z=x+iy with y>0. We have(51)η∗z,s,p,q=∑n=1∞ns−1enpz+q1+enz≦∑n=1∞nσ−1exp−2πnpy1−exp−2πny,for ℜs=σ. 1−exp−2πny≧1−exp−2πy. And we can choose a large enough positive integer K such that for n>K(52)nσ−1exp−2πnpy=expσ−1logn−2πnpy≦exp−πnpy.
Thus,(53)η∗z,s,p,q≤∑n=1∞nσ−1exp−2πnpy1−exp−2πny≤∑n=1Knσ−1exp−2πnpy1−exp−2πny+∑n=K+1∞nσ−1exp−2πnpy1−exp−2πy≤∑n=1Knσ−1exp−2πnpy1−exp−2πny+∑n=K+1∞exp−πnpy1−exp−2πy≤∑n=1Knσ−1exp−2πnpy1−exp−2πny+exp−πK+1py1−exp−2πpy11−exp−2πy<∞.Lemma 6.
Letz∈ℍ and α be an irrational algebraic number. Then, for any pair of real numbers r1,r2, we have(54)A∗z,s,r1,r2=−1−r1+r1η∗z,s,r1,r2,s∈ℂ,A∗α,s,r1,r2=−1−r1+r1η∗α,s,r1,r2,ℜs<0.Proof.
Ifr1∈ℤ, then m>−r1 implies m=−r1+r for r=1,2,3,…,. By the definition of A∗z,s,r1,r2, we know(55)A∗z,s,r1,r2=∑m>−r1−1m∑k=1∞ks−1ekr2+km+r1z=∑k=1∞ks−1∑r=1∞−1−r1+rekr2+krz=−1−r1+1∑k=1∞ks−1ekr2+kz∑r=0∞−1rekrz=−1−r1+1∑k=1∞ks−1ekr2+kz1+ekz=−1−r1+r1η∗z,s,r1,r2,sincer1=1.
Again, ifr1∉ℤ, m>−r1 implies m=−⌊r1⌋+r for r=0,1,2,…,. So, we will have(56)A∗z,s,r1,r2=∑m>−r1−1m∑k=1∞ks−1ekr2+km+r1z=∑k=1∞ks−1∑r=0∞−1−r1+rekr2+kr1+rz=−1−r1∑k=1∞ks−1ekr2+kr1z∑r=0∞−1rekrz=−1−r1∑k=1∞ks−1ekr2+kr1z1+ekz=−1−r1+r1η∗z,s,r1,r2.
Similarly, we can see that(57)A∗α,s,r1,r2=−1−r1+r1η∗α,s,r1,r2,forℜs<0.Lemma 7.
Ifz∈ℍ, A∗z,s,r1,r2 is absolutely convergent for any s∈ℂ.Proof.
Using Lemmas5 and 6, we can show that A∗z,s,r1,r2 is absolutely convergent for s∈ℂ.


## 6. Main Results

Consider the difference(58)D∗V:=D∗V,α,s,12,0:=β−sA∗Vα,s,12,0−A∗α,s,12,0,for each V from (40). Now, the second term in the above expression is the secant zeta function in view of (50). This difference is quite natural in the sense that it expresses the surplus after the modular transformation is applied.We interpret the main result of Lalín et al. Theorem3 in this setting as a special case of(59)α+1−sA∗V1α,s,12,0+α−1−sA∗V2α,s,12,0,for ℜs<0, and locate it in a natural way as we will see in Corollary 1. Our main theorem is the following.Theorem 7.
For a real algebraic irrationalα and a complex variable s with ℜs<0, we have(60)D∗V0=α−sA∗−1α,s,12,0−A∗α,s,12,0=21−2sπ−se−s4Φ0+ΓsΩ0+21−sΨ0=−2π−se−s/41−es/2∫Iλ,∞ts−1exp−1/2+αt1+exp−t1−exp−αtdt+2−2sπ−se−s4Γsζs,14−ζs,34−21−s∑n=1∞ns−1enα/2+1/4enα/2+11−enα+22−s∑n=1∞2ns−1e3πα/21−e2nα,(61)D∗V1=α+1−sA∗αα+1,s,12,0−A∗α,s,12,0=21−2sπ−se−s4Φ1+21−sΨ1=−2π−se−s/41−es/2∫Iλ,∞ts−1exp−t/21+exp−texp−α+1t/21−exp−α+1tdt+2−s∑n=1∞ns−1−1n−1cosπnα/2.
Also,(62)D∗V2=α−1−sA∗−αα−1,s,12,0−A∗α,s,12,0=21−2sπ−se−s4Φ2+21−sΨ2=−2π−se−s/41−es/2∫Iλ,∞ts−1exp−t/21+exp−texp−α−1t/21−exp−α−1tdt−2−s∑n=1∞ns−11cosπnα/2,where Φk and Ψk and (k=0,1,2) are defined later. They indicate the block of L-integrals and the block of H-functions, corresponding to the matrix Vk, respectively. Also, Ω0 is defined in (90).
We recover the main result of Lalín et al. ([11], Theorem 3), i.e., Theorem 3 by adding the equations (61) and (62). We note it as a corollary.Corollary 1.
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d
t
.The genesis of the transformation formula of Lalín et al. ([11], Theorem 3) for the secant zeta function is given by the sum of D∗V1 and D∗V2, which we have seen in Corollary 1. We will see in the proof of Corollary 1 that the term 2A∗α,s,1/2,0 on the left side and the secant zeta function on the right hand side naturally cancel each other. As this occurs only in such a pairing, this elucidates the hidden structure of the paired transformation formula from a more general standpoint.Deduction of the Main Theorem of Lalín et al. Firstly, we deduce Theorem 3 from Corollary 1. To do that, let l=2k be an even positive integer and s=1−l. Then, (63) amounts to(64)α+12k−1A∗αα+1,−2k+1,12,0+α−12k−1A∗−αα−1,−2k+1,12,0=−2π2k−1e−−2k+1/41−e−2k+1/2∫Iλ,∞t−2k∑m=0∞2−m−1Emtmm!∑n=0∞21−n−1Bnα+1n−1+α−1n−1tn−1n!dt=−22k−1π2k−1k2πi∫Iλ,∞t−2k∑m=0∞∑n=0∞2−m−121−n−1EmBnα+1n−1+α−1n−1tm+n−1m!n!dt=−22k−1π2k−1k∑n=02k12k−n!n!2−2k+n−121−n−1E2k−nBnα+1n−1+α−1n−1=12π2k−1k∑n=02k12k−n!n!2n−1−1E2k−nBnα+1n−1+α−1n−1.This proves Theorem3.The following conjecture seems to be plausible.Conjecture 2.
LetW1=a1b1c1d1 and W2=a2b2c2d2 be two matrices in PSL2ℤ which are inverses to each other. Then, for a pair p,q∈ℝ2,(65)c1α+d1−sA∗W1α,s,p,q+c2α+d2−sA∗W2α,s,p,qcan be expressible in terms of special values of the zeta and L-functions as we have seen for the sum of two explicit expressions for(66)c1α+d1−sA∗Vjα,s,12,0−A∗α,s,12,0,j=1,2.


## 7.A∗ in Terms of A- and H-Functions

Before proving our main theorem we need to expressA∗ in terms of A and H. We know that given a sum S=∑nan with its even and odd parts Se and So, where the even part is over all even integer values and odd part over odd integer values, the sum 2Se−S is the alternating sum ∑n−1nan. Using this observation, we have the following result.Lemma 8.
A
∗
z
,
s
,
r
1
,
r
2
=
2
A
2
z
,
s
,
r
1
/
2
,
r
2
−
A
z
,
s
,
r
1
,
r
2
.Proof.
By the definition ofA∗z,s,r1,r2, we have(67)A∗z,s,r1,r2=∑m>−r1−1m∑k=1∞ks−1ekr2+km+r1z=2∑m>−r1m:even∑k=1∞ks−1ekr2+km+r1z−∑m>−r1∑k=1∞ks−1ekr2+km+r1z=2∑2m>−r1∑k=1∞ks−1ekr2+k2m+r1z−∑m>−r1∑k=1∞ks−1ekr2+km+r1z=2A2z,s,r12,r2−Az,s,r1,r2.There is a duplication formula forAz,s,r1,r2 which is as follows:Lemma 9.
A
z
,
s
,
r
1
,
r
2
+
A
z
,
s
,
r
1
,
r
2
+
1
/
2
=
2
s
A
2
z
,
s
,
r
1
,
2
r
2
.Proof.
From Definition1 of Az,s,r1,r2, we have(68)Az,s,r1,r2+Az,s,r1,r2+12=∑m>−r1∑k=1∞ks−1ekr2+km+r1z+∑m>−r1∑k=1∞ks−1ekr2+12+km+r1z=∑m>−r1∑k=1∞ks−1ekr2+km+r1z1+e12k=2∑m>−r1∑k=1∞2ks−1e2kr2+2km+r1z=2s∑m>−r1∑k=1∞ks−1ek2r2+km+r12z=2sA2z,s,r1,2r2.Using the duplication formula, i.e., Lemma9 in Lemma 8, we getLemma 10.
(69)
A
∗
z
,
s
,
r
1
,
r
2
=
2
1
−
s
A
z
,
s
,
r
1
2
,
r
2
2
+
2
1
−
s
A
z
,
s
,
r
1
2
,
r
2
2
+
1
2
−
A
z
,
s
,
r
1
,
r
2
.On the other hand,(70)Hz,s,r1,r2=Az,s,r1,r2+es2Az,s,−r1,−r2,Hz,s,−r1,−r2=Az,s,−r1,−r2+es2Az,s,r1,r2,Hz,s,−r1,−r2=Az,s,−r1,−r2+es2Az,s,r1,r2.Therefore,(71)Az,s,r1,r2=11−esHz,s,r1,r2−es2Hz,s,−r1,−r2.Substituting (71) in Lemma 10, we deduce the following proposition.Proposition 2.
For a real algebraic irrationalα, a pair p,q of real numbers with p>0, and a complex variable s with ℜs<0, we have(72)1−esA∗α,s,p,q=21−sHα,s,p2,q2−es2Hα,s,−p2,−q2+21−sHα,s,p2,q2+12−es2Hα,s,−p2,−q2−12−1−esAα,s,p,q,where(73)1−esAα,s,p,q=Hα,s,p,q−es2Hα,s,−p,−q,as in equation (71).Example 6.
If we considerp,q=1,0 and 1/2,0, then we get(74)A∗α,s,1,0=11+es/221−sHα,s,12,0+21−sHα,s,12,12−Hα,s,1,0,A∗α,s,12,0=21−s1−esHα,s,14,0−21−ses/21−esHα,s,−14,0+21−s1−esHα,s,14,12−21−ses/21−esHα,s,−14,−12−1−es/21−esHα,s,12,0.
For the last term, withs an even integer, we use either(75)Hα,s,12,0=2π−se−s/4β−s−1Lα,s,12,0,c,dor(76)11+es/2Hα,s,12,0=i2∑k=1∞1k1−s1sinπkα,which follows from Examples 1 and 3, respectively.


## 8. General Procedure

The general procedure is to transform(77)D∗V:=D∗V,α,s,p,q:=β−sA∗Vα,s,p,q−A∗α,s,p,q.We recall the following notations(78)D1V,α,s,p,q=β−sHVα,s,p,q−Hα,s,p,qV,(79)D0∗V,α,s,p,q=β−sAVα,s,p,q−Aα,s,p,q,(80)Az,s,r1,r2=11−esHz,s,r1,r2−es2Hz,s,−r1,−r2.Now using Proposition2, we can write(81)D∗V,α,s,p,q+D0∗V,α,s,p,q=21−s1−esβ−sHVα,s,p2,q2−Hα,s,p2,q2−21−ses/21−esβ−sHVα,s,−p2,−q2−Hα,s,−p2,−q2+21−s1−esβ−sHVα,s,p2,q2+12−Hα,s,p2,q2+12−21−ses/21−esβ−sHVα,s,−p2,−q2−12−Hα,s,−p2,−q2−12.Forp,q=1/2,0, we have(82)D0∗V,α,s,12,0=11+es/2β−sHVα,s,12,0−Hα,s,12,0.We now transform (81) by using (78):(83)D∗V,α,s,p,q+D0∗V,α,s,p,q=21−s1−esD1V,α,s,p2,q2+Hα,s,p2,q2V−Hα,s,p2,q2−21−ses/21−esD1V,α,s,−p2,−q2+Hα,s,−p2,−q2V−Hα,s,−p2,−q2+21−s1−esD1V,α,s,p2,q2+12+Hα,s,p2,q2+12V−Hα,s,p2,q2+12−21−ses/21−esD1V,α,s,−p2,−q2−12+Hα,s,−p2,−q2−12V−Hα,s,−p2,−q2−12,where(84)D0∗V,α,s,p,q=11−esD1V,α,s,p,q+Hα,s,p,qV−Hα,s,p,q−es/21−esD1V,α,s,−p,−q+Hα,s,−p,−qV−Hα,s,−p,−q,in the case of (79), while(85)D0∗V,α,s,12,0=11+es/2D1V,α,s,12,0+Hα,s,12,0V−Hα,s,12,0,in the case of (82). Hence,(86)D∗V,α,s,p,q+D0∗V,α,s,p,q=21−s1−esD1V,α,s,p2,q2−21−ses/21−esD1V,α,s,−p2,−q2+21−s1−esD1V,α,s,p2,q2+12−21−ses/21−esD1V,α,s,−p2,−q2−12+21−s1−esHα,s,p2,q2V−Hα,s,p2,q2−21−ses/21−esHα,s,−p2,−q2V−Hα,s,−p2,−q2+21−s1−esHα,s,p2,q2+12V−Hα,s,p2,q2+12−21−ses/21−esHα,s,−p2,−q2−12V−Hα,s,−p2,−q2−12,where the last term is either (79) or (85).


## 9. Proof of Theorem 7 (60)

The three identities in Theorem7 are proved on similar lines. We begin by using (83) and (85).(87)D∗V0=D∗V0,α,s,12,0=α−sA∗−1α,s,12,0−A∗α,s,12,0=21−s1−esD1V0,α,s,14,0+21−ses/21−esD1V0,α,s,−14,0+21−s1−esD1V0,α,s,14,12−21−ses/21−esD1V0,α,s,−14,−12−1−es/21−esD1V0,α,s,12,0+21−s1−esHα,s,14,0V0−Hα,s,14,0−21−s1−esHα,s,14,0V0−Hα,s,−14,0+21−s1−esHα,s,14,12V0−Hα,s,14,12−21−ses/21−esHα,s,−14,−12V0−Hα,s,−14,−12−1−es/21−esHα,s,12,0V0−Hα,s,12,0.Then, applying (41), we deduce that(88)D∗V0=21−s1−es2π−se−s4Lα,s,0,−14,1,0−es4Lα,s,0,14,1,0+21−s1−es2π−se−s4Lα,s,12,−14,1,0−es4Lα,s,−12,14,1,0−1−es/21−es2π−se−s4Lα,s,0,−12,1,0−21−ses/21−es2π−se−s4Γsζs,34+es2ζs,14+21−s1−es2π−se−s4Γsζs,14+es2ζs,34−1−es/21−es2π−se−s4Γsζs,12+es2ζs,12−1−es/21−esHα,s,0,−12−Hα,s,12,0+21−s1−esHα,s,0,−14−Hα,s,14,0−21−ss/21−esHα,s,0,14−Hα,s,−14,0+21−s1−esHα,s,12,−14−Hα,s,14,12−21−ses/21−esHα,s,−12,14−Hα,s,−14,−12.Let(89)1−esΦ0=Lα,s,0,−14,1,0+Lα,s,12,−14,1,0−es2Lα,s,0,14,1,0−es2Lα,s,−12,14,1,0−1−es22s−1Lα,s,0,−12,1,0,(90)1−esΩ0=ζs,14−esζs,14−2s−11−es2ζs,12+es2ζs,12,(91)1−esΨ0=Hα,s,0,−14−Hα,s,14,0−es2Hα,s,0,14+es2Hα,s,−14,0+Hα,s,12,−14−Hα,s,14,12−es2Hα,s,−12,14+es2Hα,s,−14,−12−2s−11−es2Hα,s,0,−12−Hα,s,12,0.Now, we can express the differenceD∗V0 as(92)D∗V0=21−2sπ−se−s4Φ0+ΓsΩ0+21−sΨ0.Using the integral representation (35) of Lα,s,p′,q′,c,d, we calculate Φ0. Therefore,(93)1−esΦ0=−∫Iλ,∞ts−1exp−1/4+αt+exp−1/4+α/2t1−exp−t1−exp−αtdt+es2∫Iλ,∞ts−1exp−3/4+αt+exp−3/4+α/2t1−exp−t1−exp−αtdt+2s−11−es2∫Iλ,∞ts−1exp−1/2+αt1−exp−t1−exp−αtdt.Combining the first two integrals, we have(94)1−esΦ0=I0+2s−11−es2∫Iλ,∞ts−1exp−1/2+αt1−exp−t1−exp−αtdt,where(95)I0=∫Iλ,∞ts−1−exp−t/41−expπis−t/2exp−αt+exp−αt/21−exp−t1−exp−αtdt.Now, making the change of variablet⟷2t, we get(96)I0=2s∫Iλ,∞ts−1−exp−t/21−expπis−texp−2αt+exp−αt1+exp−t1−exp−t1+exp−αt1−exp−αtdt.Hence, after eliminating the common factor, we arrive at(97)1−esΦ0=2s∫Iλ,∞ts−1exp−t/2−1+expπis−texp−αt1+exp−t1−exp−t1−exp−αtdt+2s−11−es2∫Iλ,∞ts−1exp−1/2+αt1−exp−t1−exp−αtdt.Therefore,(98)Φ0=−2s−11−es/2∫Iλ,∞ts−1exp−1/2+αt1+exp−t1−exp−αtdt.Our next target is to calculateΨ0. Using (23), we have(99)Ψ0=ηα,s,1,−14−ηα,s,14,0+ηα,s,12,−14−ηα,s,14,12−2sηα,s,1,12−ηα,s,12,0.Now, using the definition of theη-function, we get(100)Ψ0=∑n=1∞ns−1enα−1/41−enα+∑n=1∞ns−1enα/2−1/41−enα−∑n=1∞ns−1enα/41−enα−∑n=1∞ns−1enα/4+1/21−enα−2s∑n=1∞ns−1enα+1/21−enα+2s∑n=1∞ns−1enα/21−enα=∑n=1∞ns−1enα/2−1/4enα/2+11−enα−2∑n=1∞2ns−1e2nα/2−1/41+enα1−e2nα−∑n=1∞ns−1enα/41+−1n1−enα+2∑n=1∞2ns−1e2nα/41+enα1−e2nα=−∑n=1∞ns−1enα/2+1/4enα/2+11−enα+2∑n=1∞2ns−1e3nα/21−e2nα.To calculateΩ0, we use 2sζs,1/2=ζs,1/4+ζs,3/4, and we get(101)Ω0=ζs,14−2s−1ζs,12=2−1ζs,14−ζs,34.Finally, combining the expressions forΦ0,Ψ0, and Ω0 we deduce Theorem 7 (60).


## 10. Proof of Theorem 7(61)

By using Proposition2 and from (42), we have(102)D∗V1=D∗V1,α,s,12,0=α+1−sA∗αα+1,s,12,0−A∗α,s,12,0=21−s1−es2π−se−s4Lα,s,14,0,1,1−es4Lα,s,−14,0,1,1+21−s1−es2π−se−s4Lα,s,34,12,1,1−es4Lα,s,−34,−12,1,1+21−s1−esHα,s,34,12−Hα,s,14,12−21−ses/21−esHα,s,−34,−12−Hα,s,−14,−12−1−es/21−es2π−se−s4Lα,s,12,0,1,1.Let(103)1−esΦ1=Lα,s,14,0,1,1+Lα,s,34,12,1,1−es2Lα,s,−14,0,1,1−es2Lα,s,−34,−12,1,1−1−es22s−1Lα,s,12,0,1,1.(104)1−esΨ1=Hα,s,34,12−Hα,s,14,12−es2Hα,s,−34,−12+es2Hα,s,−14,−12.We now express (102) as(105)D∗V1=21−2sπ−se−s4Φ1+21−sΨ1.Now, utilizing the integral representation (35) of Lα,s,p′,q′,c,d, we have(106)1−esΦ1=−∫Iλ,∞ts−1exp−1/4+3α+1/4t+exp−1/4+α+1/4t1−exp−t1−exp−α+1tdt+es2∫Iλ,∞ts−1exp−3/4+α+1/4t+exp−3/4+3α+1/4t1−exp−t1−exp−α+1tdt+2s−11−es2∫Iλ,∞ts−1exp−1/2+α+1/2t1−exp−t1−exp−α+1tdt.Again, we write the left hand side of the above equation as(107)1−esΦ1=I1+2s−11−es2∫Iλ,∞ts−1exp−1/2+α+1/2t1−exp−t1−exp−α+1tdt,where(108)I1=∫Iλ,∞ts−1−exp−t/4+expπis−−3t/4exp−α+1t/4+exp−3α+1t/41−exp−t1−exp−α+1tdt.Now, by change of variablet⟷2t followed by the elimination of the common factor 1+exp−α+1t, we get(109)I1=2s∫Iλ,∞ts−1exp−t/2−1+expπis−texp−α+1t/21+exp−t1−exp−t1−exp−α+1tdt.Thus, substitutingI1 in (107), we see that(110)Φ1=−2s−11−es/2∫Iλ,∞ts−1exp−1/2+α+1/2t1+exp−t1−exp−α+1tdt.Using the definition ofH-function, from (104), we have(111)Ψ1=ηα,s,34,12−ηα,s,14,12=∑n=1∞ns−1e3nα/4+1/21−enα−∑n=1∞ns−1enα/4+1/21−enα.Thenth summand is(112)ns−1enα/4+1/2enα/2−11−enα,from which we may eliminate the common factor e1/2nα−1. Therefore,(113)Ψ1=∑n=1∞ns−1−1n−1enα/4enα/2+1=12∑n=1∞ns−1−1n−1cosπnα/2.Now, we substitute (110) and (113) in (102) and finally get(114)α+1−sA∗αα+1,s,12,0−A∗α,s,12,0=−2π−se−s/41−es/2∫Iλ,∞ts−1exp−t/21+exp−texp−α+1t/21−exp−α+1tdt+2−s∑n=1∞ns−1−1n−1cosπαn/2.This completes the proof of Theorem 7 (61).


## 11. Proof of Theorem 7 (62)

We follow the same route: first, we use Proposition2 and then using (43), we obtain(115)D∗V2=D∗V2,α,s,12,0=α−1−sA∗−αα−1,s,12,0−A∗α,s,12,0=21−s1−es2π−se−s4Lα,s,−14,0,1,−1−es4Lα,s,14,0,1,−1+21−s1−es2π−se−s4Lα,s,14,−12,1,−1−es4Lα,s,−14,12,1,−1−1−es/21−es2π−se−s4Lα,s,−12,0,1,−1+21−s1−esHα,s,−14,0−Hα,s,14,0−21−ses/21−esHα,s,14,0−Hα,s,−14,0=21−2sπ−se−s4Φ2+21−sΨ2,where(116)1−esΦ2=Lα,s,−14,0,1,−1−es2Lα,s,14,0,1,−1+Lα,s,14,−12,1,−1−es2Lα,s,−14,12,1,−1−1−es22s−1Lα,s,−12,0,1,−1,(117)1−esΨ2=Hα,s,−14,0−Hα,s,14,0−es2Hα,s,14,0+es2Hα,s,−14,0=1+es2Hα,s,−14,0−Hα,s,14,0.To simplifyΦ2, we make use of the integral representation (35) of Lα,s,p′,q′,c,d. So, we have(118)1−esΦ2=I2+2s−11−es2∫Iλ,∞ts−1exp−1/2+α−1/2t1−exp−t1−exp−α−1tdt,where(119)I2=∫Iλ,∞ts−1−exp−t/4+expπis−3t/4exp−α−1t/4+exp−3α−1t/41−exp−t1−exp−α−1tdt.As before, by eliminating the common factor1−exp−t, we obtain(120)I2=2s∫Iλ,∞ts−1exp−t/2exp−α−1t/2−1+expπis−t1+exp−t1−exp−t1−exp−α−1tdt.Whence, it follows that(121)Φ2=−2s−11−es/2∫Iλ,∞ts−1exp−1/2+α−1/2t1+exp−t1−exp−α−1tdt.While handling (117), we decompose it as(122)Ψ2=ηα,s,34,0−ηα,s,14,0.In the series expression ofΨ2, we factor out enα/4 as before and eliminate the common factor enα/2−1 to obtain(123)Ψ2=−∑n=1∞ns−1enα/4enα/2+1=−12∑n=1∞ns−11cosπnα/2.Finally, substituting the expressions forΦ2 and Ψ2, we have(124)α−1−sA∗−αα−1,s,12,0−A∗α,s,12,0=−2π−se−s/41−es/2∫Iλ,∞ts−1exp−t/21+exp−texp−α−1t/21−exp−α−1tdt−2−s∑n=1∞ns−11cosπnα/2.This finishes the proof of Theorem 7 (62).


## 12. Proof of Corollary 1

We conclude this chapter by finally proving Corollary1. We add (114) and (124) and derive that(125)α+1−sA∗αα+1,s,12,0+α−1−sA∗−αα−1,s,12,0−2A∗α,s,12,0=−2π−se−s/41−es/2∫Iλ,∞ts−1exp−t/21+exp−texp−α+1t/21−exp−α+1tdt−2π−se−s/41−e−s/2∫Iλ,∞ts−1exp−t/21+exp−texp−α−1t/21−exp−α−1tdt+2−s∑n=1∞ns−1−1n−1cosπnα/2−2−s∑n=1∞ns−11cosπnα/2=−2π−se−s/41−e−s/2∫Iλ,∞ts−1exp−t/21+exp−texp−α+1t/21−exp−α+1t+exp−α+1t/21−exp−α−1tdt−2⋅2−s∑n=1∞2ns−11cos2πnα/2.Now, in the above expression,2A∗α,s,1/2,0 on the left hand side and secant zeta function on the right hand side will cancel each other, as they are the same (from (50)). Therefore, we have(126)α+1−sA∗αα+1,s,12,0+α−1−sA∗−αα−1,s,12,0=−2π−se−s/41−es/2∫Iλ,∞ts−1exp−1/2t1+exp−texp−1/2α+1t1−exp−α+1t+exp−1/2α−1t1−exp−α−1tdt=−2π−se−s/41−es/2∫Iλ,∞ts−1∑m=0∞Em12tm2m!,and thus Corollary 1 follows.


## 13. Future Work

By the virtue of the work of Lewittes, Berndt, and Arakawa, it would be interesting to find the general modular transformation formula forA∗α,s,p,q for all p,q∈ℝ2 and from which one would like to see the truth of our Conjecture 2.



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