Four Variants of Riemann Zeta Function

Abstract

By means of the generating function method and Dougall’s formulae for bilateral hypergeometric series, we examine four classes of infinite series, which may be considered as variants of Riemann zeta function. Several summation formulae are established in closed form, which shows remarkably that the values of these series result in multiples of integer powers of $\pi$ either by rational numbers or by algebraic numbers.

1. Introduction and Motivation

Recently, Zhou [1] proposed the following problem. For $m\in {\mathbb{N}}_0$, let

$$A_m=\sum _{k=0}^{\infty }\left\{\frac{1}{{(6k+1)}^{2m+1}}-\frac{1}{{(6k+5)}^{2m+1}}\right\}.$$

(1)

Prove $A_0=\frac{\pi }{2\sqrt{3}}$ and, for $m\ge 1$,

$$2A_m+\sum _{n=1}^m\frac{{(-1)}^n{\pi }^{2n}}{\left(2n\right)!}{A}_{m-n}={(-1)}^m\frac{(1+4^m)\sqrt{3}}{2\left(2m\right)!}{\left(\frac{\pi }{3}\right)}^{2m+1}.$$

(2)

We find that this problem can be resolved by the generating function approach in conjunction with Dougall’s bilateral ${}_2{H}_2$ series identity. Define the generating function

$$F\left(x\right)=\sum _{m=0}^{\infty }{A}_m{x}^{2m}.$$

We can manipulate it by exchange of the summation order

$$\begin{array}{cc}\hfill F\left(x\right)& =\sum _{m=0}^{\infty }{x}^{2m}\sum _{k=0}^{\infty }\left\{\frac{1}{{(6k+1)}^{2m+1}}-\frac{1}{{(6k+5)}^{2m+1}}\right\}\hfill \\ & =\sum _{k=0}^{\infty }\sum _{m=0}^{\infty }{x}^{2m}\left\{\frac{1}{{(6k+1)}^{2m+1}}-\frac{1}{{(6k+5)}^{2m+1}}\right\}\hfill \\ & =\sum _{k=0}^{\infty }\left\{\frac{6k+1}{{(6k+1)}^2-x^2}-\frac{6k+5}{{(6k+5)}^2-x^2}\right\}.\hfill \end{array}$$

The next step is crucial by rewriting the summand

$$\begin{array}{cc}\hfill 2& \left\{\frac{6k+1}{{(6k+1)}^2-x^2}-\frac{6k+5}{{(6k+5)}^2-x^2}\right\}\hfill \\ \hfill =& \left\{\frac{1}{6k+1+x}-\frac{1}{6k+5-x}\right\}-\left\{\frac{1}{6k+5+x}-\frac{1}{6k+1-x}\right\}.\hfill \end{array}$$

Then, we can express $F\left(x\right)$ as a bilateral series

$$\begin{array}{cc}\hfill F\left(x\right)& =\frac{1}{2}\sum _{k=-\infty }^{\infty }\left\{\frac{1}{6k+1+x}-\frac{1}{6k+5+x}\right\}\hfill \\ & =2\sum _{k=-\infty }^{\infty }\frac{1}{(6k+1+x)(6k+5+x)}.\hfill \end{array}$$

Recall that, for an indeterminate $\lambda$ and $k\in \mathbb{Z}$, the shifted factorial is defined by the $\Gamma$-function quotient

$${\left(\lambda \right)}_k=\frac{\Gamma (\lambda +k)}{\Gamma \left(\lambda \right)}=\left\{\begin{array}{cc}\lambda (\lambda +1)\cdots (\lambda +k-1),\hfill & k>0;\hfill \\ 1,\hfill & k=0;\hfill \\ {\displaystyle \frac{1}{(\lambda -1)\cdots (\lambda +k)}},\hfill & k<0.\hfill \end{array}\right.$$

For the sake of brevity, the multiparameter $\Gamma$-function will be shortened as

$$\Gamma \left[\begin{array}{c}\alpha ,\beta ,\cdots ,\gamma \\ A,B,\cdots ,C\end{array}\right]=\frac{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)\cdots \Gamma \left(\gamma \right)}{\Gamma \left(A\right)\Gamma \left(B\right)\cdots \Gamma \left(C\right)}.$$

For the four complex numbers $a,b,c,d\notin \mathbb{Z}$ subject to condition $\Re (c+d-a-b)>1$, Dougall ([2]) discovered the following formula for the bilateral series:

$${}_2{H}_2\left[\begin{array}{c}a,b\\ c,d\end{array}|1\right]=\sum _{k=-\infty }^{+\infty }\frac{{\left(a\right)}_k{\left(b\right)}_k}{{\left(c\right)}_k{\left(d\right)}_k}=\Gamma \left[\begin{array}{c}1-a,1-b,c,d,c+d-a-b-1\\ c-a,d-a,c-b,d-b\end{array}\right].$$

(3)

Therefore, $F\left(x\right)$ can further be reformulated and evaluated as follows:

$$F\left(x\right)=\frac{2}{(1+x)(5+x)}\sum _{k=-\infty }^{\infty }\frac{{\left(\frac{1+x}{6}\right)}_k{\left(\frac{5+x}{6}\right)}_k}{{\left(\frac{7+x}{6}\right)}_k{\left(\frac{11+x}{6}\right)}_k}=\frac{\Gamma \left(\frac{1-x}{6}\right)\Gamma \left(\frac{5-x}{6}\right)\Gamma \left(\frac{1+x}{6}\right)\Gamma \left(\frac{5+x}{6}\right)}{18\Gamma \left(\frac{1}{3}\right)\Gamma \left(\frac{5}{3}\right)}.$$

By making use of the reciprocal property (cf. Rainville ([3], §17))

$$\Gamma \left(y\right)\Gamma (1-y)=\frac{\pi }{sin\left(\pi y\right)},$$

we find the simplified generating function

$$F\left(x\right)=\frac{\pi }{8\sqrt{3}sin\left(\frac{1+x}{6}\pi \right)sin\left(\frac{1-x}{6}\pi \right)}.$$

Observing further that

$$4sin\left(\frac{1+x}{6}\pi \right)sin\left(\frac{1-x}{6}\pi \right)=2cos\left(\frac{\pi x}{3}\right)-1,$$

we can express the above generating function as

$$F\left(x\right)=\frac{\pi }{2\sqrt{3}\{2cos\left(\frac{\pi x}{3}\right)-1\}}.$$

(4)

Recall Glaisher’s numbers (cf. A002114 in [4]) with the initial terms

$$\{1/2,1,11,301,15371,1261501\}$$

that have the following exponential generating function

$$\frac{1}{2(2cosx-1)}=\sum _{k=0}^{\infty }{\mathcal{G}}_k\frac{x^{2k}}{\left(2k\right)!},$$

we find from (4) the following explicit formula:

$$A_m=\frac{{\pi }^{2m+1}\sqrt{3}}{3^{2m+1}\left(2m\right)!}{\mathcal{G}}_m.$$

(5)

The values of the first few series $A_m$ are highlighted below as examples:

$$\begin{array}{cc}{A}_0=\frac{\pi }{2\sqrt{3}},\hfill & A_1=\frac{{\pi }^3}{18\sqrt{3}},\hfill \\ A_2=\frac{11{\pi }^5}{1944\sqrt{3}},\hfill & A_3=\frac{301{\pi }^7}{524880\sqrt{3}},\hfill \\ A_4=\frac{15371{\pi }^9}{264539520\sqrt{3}},\hfill & A_5=\frac{1261501{\pi }^{11}}{214277011200\sqrt{3}}.\hfill \end{array}$$

In view of the power series expansion

$$2cos\left(\frac{\pi x}{3}\right)-1=1+2\sum _{k=1}^{\infty }{\left(-\frac{{\pi }^2}{9}\right)}^k\frac{x^{2k}}{\left(2k\right)!},$$

rewriting further (4) by

$$F\left(x\right)\left\{2cos\left(\frac{\pi x}{3}\right)-1\right\}=\frac{\pi }{2\sqrt{3}}$$

and extracting the coefficient of $x^{2m}$ across the above equation, we find the following homogeneous recurrence relation

$$A_m+2\sum _{k=1}^m{\left(-\frac{{\pi }^2}{9}\right)}^k\frac{A_{m-k}}{\left(2k\right)!}=0\mathrm{for}m>0.$$

This is simpler than Zhou’s recursion (2), since the latter contains an extra constant term. In order to derive Zhou’s equation, we need the triple angle identity

$$\begin{array}{cc}\hfill 1+cos3y& =1+4{cos}^{3}y-3cosy\hfill \\ & =(2cosy-1)(cosy+2{cos}^{2}y-1)\hfill \\ & =(2cosy-1)(cosy+cos2y).\hfill \end{array}$$

Multiplying (4) by $1+cos\left(\pi x\right)$, we obtain another functional equation

$$F\left(x\right)\left\{1+cos\left(\pi x\right)\right\}=\frac{\pi }{2\sqrt{3}}\left\{cos\left(\frac{\pi x}{3}\right)+cos\left(\frac{2\pi x}{3}\right)\right\}.$$

Finally, Zhou’s Equation (2) follows by extracting the coefficient of $x^{2m}$ across.

Let p and q be two natural numbers such that $gcd(p,q)=1$. The above solution of Zhou’s problems motivates the authors to investigate further the following four classes of generalized series:

$$\begin{array}{cc}\hfill {\varphi }_m(p,q)& =\sum _{k=-\infty }^{\infty }\frac{1}{{(p+pk+qk)}^{2m}},\hfill \end{array}$$

(6)

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_m(p,q)& =\sum _{k=-\infty }^{\infty }\frac{{(-1)}^k}{{(p+pk+qk)}^{2m}},\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill {\psi }_m(p,q)& =\sum _{k=-\infty }^{\infty }\frac{1}{{(p+pk+qk)}^{2m-1}},\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}_m(p,q)& =\sum _{k=-\infty }^{\infty }\frac{{(-1)}^k}{{(p+pk+qk)}^{2m-1}}.\hfill \end{array}$$

(9)

Recall the Hurwitz zeta function and its alternating form (cf. ([5], §25.11))

$${\zeta }^{+}(m,z)=\sum _{n=0}^{\infty }\frac{1}{{(n+z)}^m}\mathrm{and}{\zeta }^{-}(m,z)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{{(n+z)}^m},$$

which play an important role in mathematics and have been investigated extensively. The afore-defined four functions can be considered as bilateral generalizations of ${\zeta }^{+}(m,z)$ and ${\zeta }^{-}(m,z)$. They will be computed, in closed form, in the remaining four separate sections, with their values being multiples of integer powers of $\pi$ either by rational numbers or by algebraic numbers. These results can be considered as counterparts of two well-known zeta function formulae discovered by Euler:

$$\begin{array}{cc}& \zeta \left(2m\right)=\sum _{n=1}^{\infty }\frac{1}{n^{2m}}={(-1)}^{m-1}\frac{B_{2m}}{2\left(2m\right)!}{\left(2\pi \right)}^{2m},\hfill \\ & \sum _{n=1}^{\infty }\frac{{(-1)}^{n-1}}{{(2n-1)}^{2m+1}}={(-1)}^m\frac{E_{2m}}{2\left(2m\right)!}{\left(\frac{\pi }{2}\right)}^{2m+1},\hfill \end{array}$$

where $B_{2m}$ and $E_{2m}$ are Bernoulli and Euler numbers with the following trigonometric generating functions:

$$\frac{y}{2}cot\frac{y}{2}=\sum _{n=0}^{\infty }{(-1)}^n\frac{B_{2n}}{\left(2n\right)!}{y}^{2n}\mathrm{and}secy=\sum _{n=0}^{\infty }{(-1)}^n\frac{E_{2n}}{\left(2n\right)!}{y}^{2n}.$$

The strategy for realizing this objective consists of the generating function method and bilateral hypergeoemtric series summation formulae (see [6] and ([7], Chapter 6) for notations), as sketched in presenting our solution to Zhou’s problem. In addition to the identity (3), we shall also utilize another formula of Dougall ([2]) for well poised ${}_5{H}_5$-series:

$$\begin{array}{cc}\hfill {}_5{H}_5& \left[\begin{array}{ccccc}\hfill 1+a/2,& b,& c,& d,& e\\ \hfill a/2,& 1+a-b,& 1+a-c,& 1+a-d,& 1+a-e\end{array}|1\right]\hfill \\ \hfill =& \sum _{k=-\infty }^{\infty }\frac{(a+2k){\left(b\right)}_k{\left(c\right)}_k{\left(d\right)}_k{\left(e\right)}_k}{a{(1+a-b)}_k{(1+a-c)}_k{(1+a-d)}_k{(1+a-e)}_k}\hfill \\ \hfill =& \Gamma \left[\begin{array}{c}1+2a-b-c-d-e,1-b,1-c,1-d,1-e,1+a-b,1+a-c,1+a-d,1+a-e\\ 1+a,1-a,1+a-b-c,1+a-b-d,1+a-b-e,1+a-c-d,1+a-c-e,1+a-d-e\end{array}\right],\hfill \end{array}$$

(10)

provided that $\Re (1+2a-b-c-d-e)>0$ and $\{a,b,c,d,e\}$ are complex numbers such that the bilateral series is well-defined. The original proofs due to Dougall for both ${}_2{H}_2$ and ${}_5{H}_5$ are standard, which consist of computing the contour integral of certain complex functions through the Cauchy residue theorem (see also Slater ([7], §6.1), and ([8], §5.3)). There are also proofs through transformation from bilateral series to unilateral series, which can be found in Chu [9,10].

In order to ensure the accuracy, we have verified numerically, throughout the paper, all the infinite series evaluations by appropriately devised Mathematica commands.

2. Evaluation of ${\varphi }_m(p,q)$-Series

Consider the generating function

$$\begin{array}{cc}\hfill \varphi \left(x\right|p,q)& =\sum _{m=1}^{\infty }{x}^{2m}{\varphi }_m(p,q)=\sum _{m=1}^{\infty }{x}^{2m}\sum _{k=-\infty }^{\infty }\frac{1}{{(p+pk+qk)}^{2m}}\hfill \\ & =\sum _{k=-\infty }^{\infty }\sum _{m=1}^{\infty }\frac{x^{2m}}{{(p+pk+qk)}^{2m}}=\sum _{k=-\infty }^{\infty }\frac{x^2}{{(p+pk+qk)}^2-x^2}\hfill \\ & =x^2\sum _{k=-\infty }^{\infty }\frac{1}{(p+pk+qk+x)(p+pk+qk-x)}\hfill \\ & =\frac{x^2}{(p+x)(p-x)}{}_2{H}_2\left[\begin{array}{c}\frac{p+x}{p+q},\frac{p-x}{p+q}\\ \\ 1+\frac{p+x}{p+q},1+\frac{p-x}{p+q}\end{array}|1\right].\hfill \end{array}$$

According to (3), we have the closed form expression in terms of $\Gamma$-function

$$\begin{array}{cc}\hfill \varphi \left(x\right|p,q)& =\frac{x^2}{(p+x)(p-x)}\Gamma \left[\begin{array}{c}1-\frac{p+x}{p+q},1-\frac{p-x}{p+q},1+\frac{p+x}{p+q},1+\frac{p-x}{p+q}\\ \\ 1+\frac{2x}{p+q},1-\frac{2x}{p+q}\end{array}\right],\hfill \end{array}$$

which can be simplified into trigonometric functions as in the theorem below.

Theorem 1.

The generating function for${\varphi }_m(p,q)$defined in (6) is given by

$$\begin{array}{c}\hfill \varphi \left(x\right|p,q)=\frac{\pi xsin\left(\frac{2\pi x}{p+q}\right)}{2(p+q)sin\left(\frac{p+x}{p+q}\pi \right)sin\left(\frac{p-x}{p+q}\pi \right)}=\frac{\pi xsin\left(\frac{2\pi x}{p+q}\right)}{(p+q)\left\{cos\left(\frac{2\pi x}{p+q}\right)-cos\left(\frac{2\pi p}{p+q}\right)\right\}}.\end{array}$$

As applications of this theorem, we exhibit five classes of infinite series evaluations by specifying concrete numbers for p and q under the condition $gcd(p,q)=1$.

2.1. $p=1\&q=2$

In this case, the generating function in Theorem 1 becomes

$$\varphi \left(x\right|1,2)=\frac{2\pi xsin\left(\frac{2\pi x}{3}\right)}{3\left\{1+2cos\left(\frac{2\pi x}{3}\right)\right\}}.$$

Taking into account that

$$\frac{ysiny}{1+2cosy}=\frac{y}{4}cot\frac{y}{2}-\frac{3y}{4}cot\frac{3y}{2}=\sum _{n=1}^{\infty }{(-1)}^n\frac{1-9^n}{2\left(2n\right)!}{B}_{2n}{y}^{2n},$$

(11)

we derive the following explicit formula:

$${\varphi }_m(1,2)={(-1)}^m{\left(\frac{2\pi }{3}\right)}^{2m}\frac{1-9^m}{2\left(2m\right)!}{B}_{2m}.$$

The first five values are displayed below as examples:

$$\begin{array}{cc}\hfill {\varphi }_1(1,2)& =\frac{4{\pi }^2}{27},\hfill \\ \hfill {\varphi }_2(1,2)& =\frac{8{\pi }^4}{729},\hfill \\ \hfill {\varphi }_3(1,2)& =\frac{104{\pi }^6}{98415},\hfill \\ \hfill {\varphi }_4(1,2)& =\frac{656{\pi }^8}{6200145},\hfill \\ \hfill {\varphi }_5(1,2)& =\frac{5368{\pi }^{10}}{502211745}.\hfill \end{array}$$

2.2. $p=1\&q=3$

In this case, the generating function becomes simpler

$$\varphi \left(x\right|1,3)=\frac{\pi x}{4}tan\left(\frac{\pi x}{2}\right).$$

The same series was previously examined by Elkies [11] in a similar manner.

Observing that

$$ytany=\frac{2y}{sin2y}-ycoty=\sum _{n=1}^{\infty }{(-4)}^n\frac{1-4^n}{\left(2n\right)!}{B}_{2n}{y}^{2n},$$

we find the explicit formula

$${\varphi }_m(1,3)={(-{\pi }^2)}^m\frac{1-4^m}{2\left(2m\right)!}{B}_{2m}$$

together with the first five values:

$$\begin{array}{cc}\hfill {\varphi }_1(1,3)& =\frac{{\pi }^2}{8},\hfill \\ \hfill {\varphi }_2(1,3)& =\frac{{\pi }^4}{96},\hfill \\ \hfill {\varphi }_3(1,3)& =\frac{{\pi }^6}{960},\hfill \\ \hfill {\varphi }_4(1,3)& =\frac{17{\pi }^8}{161280},\hfill \\ \hfill {\varphi }_5(1,3)& =\frac{31{\pi }^{10}}{2903040}.\hfill \end{array}$$

2.3. $p=1\&q=5$

In this case, the generating function is given by

$$\varphi \left(x\right|1,5)=\frac{\pi xsin\left(\frac{\pi x}{3}\right)}{3\left\{2cos\left(\frac{\pi x}{3}\right)-1\right\}}.$$

Keeping in mind that

$$\frac{ysiny}{2cosy-1}=\frac{3y}{4}tan\frac{3y}{2}-\frac{y}{4}tan\frac{y}{2}=\sum _{n=1}^{\infty }{(-1)}^n\frac{(4^n-1)(1-9^n)}{2\left(2n\right)!}{B}_{2n}{y}^{2n},$$

(12)

we derive the following explicit formula:

$${\varphi }_m(1,5)={(-{\pi }^2)}^m\frac{(4^m-1)(9^{-m}-1)}{2\left(2m\right)!}{B}_{2m}$$

as well as the evaluations of the first five series:

$$\begin{array}{cc}\hfill {\varphi }_1(1,5)& =\frac{{\pi }^2}{9},\hfill \\ \hfill {\varphi }_2(1,5)& =\frac{5{\pi }^4}{486},\hfill \\ \hfill {\varphi }_3(1,5)& =\frac{91{\pi }^6}{87480},\hfill \\ \hfill {\varphi }_4(1,5)& =\frac{697{\pi }^8}{6613488},\hfill \\ \hfill {\varphi }_5(1,5)& =\frac{228811{\pi }^{10}}{21427701120}.\hfill \end{array}$$

2.4. $p=1\&q=4$

The corresponding generating function reads as

$$\varphi \left(x\right|1,4)=\frac{\pi xsin\left(\frac{2\pi x}{5}\right)}{5\left\{cos\left(\frac{2\pi x}{5}\right)-cos\left(\frac{2\pi }{5}\right)\right\}},\mathrm{where}cos\left(\frac{2\pi }{5}\right)=\frac{\sqrt{5}-1}{4}.$$

For the series labeled by “$p=1\&q=4$” and “$p=2\&q=3$”, the resulting expressions are very complicated. Therefore, we are limited, hereafter, to highlighting the first five series by extracting the initial five coefficients from their generating functions:

$$\begin{array}{cc}\hfill {\varphi }_1(1,4)& =\frac{2(5+\sqrt{5})}{125}{\pi }^2,\hfill \\ \hfill {\varphi }_2(1,4)& =\frac{4(13+5\sqrt{5}){\pi }^4}{9375},\hfill \\ \hfill {\varphi }_3(1,4)& =\frac{4(155+67\sqrt{5}){\pi }^6}{1171875},\hfill \\ \hfill {\varphi }_4(1,4)& =\frac{8(4069+1805\sqrt{5}){\pi }^8}{615234375},\hfill \\ \hfill {\varphi }_5(1,4)& =\frac{4(924775+412751\sqrt{5}){\pi }^{10}}{692138671875}.\hfill \end{array}$$

2.5. $p=2\&q=3$

The corresponding generating function reads as

$$\varphi \left(x\right|2,3)=\frac{\pi xsin\left(\frac{2\pi x}{5}\right)}{5\left\{cos\left(\frac{2\pi x}{5}\right)-cos\left(\frac{4\pi }{5}\right)\right\}},\mathrm{where}cos\left(\frac{4\pi }{5}\right)=\frac{-1-\sqrt{5}}{4}.$$

Extracting the first five coefficients across gives the following evaluations:

$$\begin{array}{cc}\hfill {\varphi }_1(2,3)& =\frac{2(5-\sqrt{5}){\pi }^2}{125},\hfill \\ \hfill {\varphi }_2(2,3)& =\frac{4(13-5\sqrt{5}){\pi }^4}{9375},\hfill \\ \hfill {\varphi }_3(2,3)& =\frac{4(155-67\sqrt{5}){\pi }^6}{1171875},\hfill \\ \hfill {\varphi }_4(2,3)& =\frac{8(4069-1805\sqrt{5}){\pi }^8}{615234375},\hfill \\ \hfill {\varphi }_5(2,3)& =\frac{4(924775-412751\sqrt{5}){\pi }^{10}}{692138671875}.\hfill \end{array}$$

Comparing the above expressions with those displayed in the last subsection, we notice a quite curious fact that they are “algebraic conjugates”, i.e., one group of formulae becomes another group if the algebraic numbers therein are replaced by their conjugate ones. As the reader can see from the rest of the paper that, for each class of the series defined in (6)–(9), this phenomenon occurs among all the series paired by labels “$p=1\&q=4$” and “$p=2\&q=3$”.

3. Evaluation of ${\mathsf{\Phi }}_m(p,q)$-Series

Similarly, we can manipulate the generating function

$$\begin{array}{cc}\hfill \mathsf{\Phi }\left(x\right|p,q)& =\sum _{m=1}^{\infty }{x}^{2m}{\mathsf{\Phi }}_m(p,q)=\sum _{m=1}^{\infty }{x}^{2m}\sum _{k=-\infty }^{\infty }\frac{{(-1)}^k}{{(p+pk+qk)}^{2m}}\hfill \\ & =\sum _{k=-\infty }^{\infty }{(-1)}^k\sum _{m=1}^{\infty }\frac{x^{2m}}{{(p+pk+qk)}^{2m}}=\sum _{k=-\infty }^{\infty }\frac{x^2{(-1)}^k}{{(p+pk+qk)}^2-x^2}\hfill \\ & =x^2\sum _{k=-\infty }^{\infty }\frac{{(-1)}^k}{(p+pk+qk+x)(p+pk+qk-x)}\hfill \\ & =\frac{x^2}{(p+x)(p-x)}{}_2{H}_2\left[\begin{array}{c}\frac{p+x}{p+q},\frac{p-x}{p+q}\\ \\ 1+\frac{p+x}{p+q},1+\frac{p-x}{p+q}\end{array}|-1\right].\hfill \end{array}$$

Now letting $b=\frac{a}{2}$ and $e\to -\infty$ in (10), we obtain the formula

$${}_2{H}_2\left[\begin{array}{c}c,d\\ \\ 1+a-c,1+a-d\end{array}|-1\right]=\Gamma \left[\begin{array}{c}1+\frac{a}{2},1-\frac{a}{2},1-c,1-d,1+a-c,1+a-d\\ \\ 1+a,1-a,1+\frac{a}{2}-c,1+\frac{a}{2}-d,1+a-c-d\end{array}\right].$$

This can be employed to evaluating the generating function $\mathsf{\Phi }\left(x\right|p,q)$ as follows:

$$\begin{array}{cc}\hfill \mathsf{\Phi }\left(x\right|p,q)& =\frac{x^2}{(p+x)(p-x)}\Gamma \left[\begin{array}{c}1+\frac{p}{p+q},1-\frac{p}{p+q},1+\frac{p+x}{p+q},1-\frac{p+x}{p+q},1+\frac{p-x}{p+q},1-\frac{p-x}{p+q}\\ \\ 1+\frac{2p}{p+q},1-\frac{2p}{p+q},1+\frac{x}{p+q},1-\frac{x}{p+q}\end{array}\right].\hfill \end{array}$$

By means of the $\Gamma$-function reciprocity, we obtain the following simplified expression:

Theorem 2.

The generating function for${\mathsf{\Phi }}_m(p,q)$defined in (7) is given by

$$\begin{array}{c}\hfill \mathsf{\Phi }\left(x\right|p,q)=\frac{\pi x}{(p+q)}\frac{sin\left(\frac{\pi x}{p+q}\right)cos\left(\frac{\pi p}{p+q}\right)}{sin\left(\frac{p+x}{p+q}\pi \right)sin\left(\frac{p-x}{p+q}\pi \right)}=\frac{2\pi x}{(p+q)}\frac{cos\left(\frac{\pi p}{p+q}\right)sin\left(\frac{\pi x}{p+q}\right)}{cos\left(\frac{2\pi x}{p+q}\right)-cos\left(\frac{2\pi p}{p+q}\right)}.\end{array}$$

By assigning small integer values for p and q subject to $gcd(p,q)=1$, we can further derive from this theorem concrete infinite series evaluations.

3.1. $p=1\&q=2$

In this case, the generating function in Theorem 2 becomes

$$\mathsf{\Phi }\left(x\right|1,2)=\frac{2\pi xsin\left(\frac{\pi x}{3}\right)}{3\left\{2cos\left(\frac{2\pi x}{3}\right)+1\right\}}.$$

Taking into account that

$$\frac{ysiny}{2cos2y+1}=\frac{3y}{4sin3y}-\frac{y}{4siny}=\sum _{n=1}^{\infty }\frac{{(-1)}^n(1-2^{2n-1})(9^n-1)}{2\left(2n\right)!}{B}_{2n}{y}^{2n},$$

(13)

we obtain the following explicit formula:

$${\mathsf{\Phi }}_m(1,2)=\frac{{(-{\pi }^2)}^m(1-2^{2m-1})(1-9^{-m})}{\left(2m\right)!}{B}_{2m}$$

as well as the evaluations of the first five series:

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_1(1,2)& =\frac{2{\pi }^2}{27},\hfill \\ \hfill {\mathsf{\Phi }}_2(1,2)& =\frac{7{\pi }^4}{729},\hfill \\ \hfill {\mathsf{\Phi }}_3(1,2)& =\frac{403{\pi }^6}{393660},\hfill \\ \hfill {\mathsf{\Phi }}_4(1,2)& =\frac{5207{\pi }^8}{49601160},\hfill \\ \hfill {\mathsf{\Phi }}_5(1,2)& =\frac{48983{\pi }^{10}}{4591650240}.\hfill \end{array}$$

3.2. $p=1\&q=3$

We have the corresponding generating function

$$\mathsf{\Phi }\left(x\right|1,3)=\frac{\sqrt{2}\pi x}{4}sin\left(\frac{\pi x}{4}\right)sec\left(\frac{\pi x}{2}\right).$$

In view of the power series expansion

$$\begin{array}{cc}\hfill sinysec\left(2y\right)& =\sum _{i=1}^{\infty }{(-1)}^{i-1}\frac{y^{2i-1}}{(2i-1)!}\sum _{k=0}^{\infty }{(-4)}^k\frac{E_{2k}}{\left(2k\right)!}{y}^{2k}\hfill \\ & =\sum _{n=1}^{\infty }{(-1)}^{n-1}{y}^{2n-1}\sum _{k=0}^n\frac{4^k{E}_{2k}}{\left(2k\right)!(2n-2k-1)!},\hfill \end{array}$$

we derive the following analytical formula

$${\mathsf{\Phi }}_m(1,3)={(-1)}^{m-1}\frac{{\pi }^{2m}\sqrt{2}}{4^{2m}}\sum _{k=0}^m\frac{4^k{E}_{2k}}{\left(2k\right)!(2m-2k-1)!}.$$

Its first five values are given explicitly as follows:

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_1(1,3)& =\frac{{\pi }^2}{8\sqrt{2}},\hfill \\ \hfill {\mathsf{\Phi }}_2(1,3)& =\frac{11{\pi }^4}{768\sqrt{2}},\hfill \\ \hfill {\mathsf{\Phi }}_3(1,3)& =\frac{361{\pi }^6}{245760\sqrt{2}},\hfill \\ \hfill {\mathsf{\Phi }}_4(1,3)& =\frac{24611{\pi }^8}{165150720\sqrt{2}},\hfill \\ \hfill {\mathsf{\Phi }}_5(1,3)& =\frac{2873041{\pi }^{10}}{190253629440\sqrt{2}}.\hfill \end{array}$$

3.3. $p=1\&q=5$

The corresponding generating function reads as

$$\mathsf{\Phi }\left(x\right|1,5)=\frac{\pi xsin\left(\frac{\pi x}{6}\right)}{\sqrt{3}\left\{2cos\left(\frac{\pi x}{3}\right)-1\right\}}.$$

According to (12), the following power series expansion holds

$$\begin{array}{cc}\hfill \frac{siny}{2cos\left(2y\right)-1}& =\frac{y}{cosy}\times \frac{sin\left(2y\right)}{2y\left\{2cos\left(2y\right)-1\right\}}\hfill \\ & =\sum _{i=0}^{\infty }{(-1)}^i\frac{E_{2i}}{\left(2i\right)!}{y}^{2i+1}\sum _{k=1}^{\infty }{(-1)}^k\frac{4^k(1-4^k)(9^k-1)}{8\left(2k\right)!}{B}_{2k}{y}^{2k-2}\hfill \\ & =\sum _{n=0}^{\infty }{(-1)}^n{y}^{2n-1}\sum _{k=1}^n\frac{4^k(1-4^k)(9^k-1)}{8\left(2k\right)!}{B}_{2k}\frac{E_{2n-2k}}{(2n-2k)!}.\hfill \end{array}$$

Hence, we find the explicit formula

$${\mathsf{\Phi }}_m(1,5)={(-1)}^m\frac{{\pi }^{2m}\sqrt{3}}{4·6^{2m}}\sum _{k=1}^m\frac{4^k(1-4^k)(9^k-1)}{\left(2k\right)!}\frac{B_{2k}{E}_{2m-2k}}{(2m-2k)!}.$$

The first five series are highlighted as examples:

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_1(1,5)& =\frac{{\pi }^2}{6\sqrt{3}},\hfill \\ \hfill {\mathsf{\Phi }}_2(1,5)& =\frac{23{\pi }^4}{1296\sqrt{3}},\hfill \\ \hfill {\mathsf{\Phi }}_3(1,5)& =\frac{1681{\pi }^6}{933120\sqrt{3}},\hfill \\ \hfill {\mathsf{\Phi }}_4(1,5)& =\frac{257543{\pi }^8}{1410877440\sqrt{3}},\hfill \\ \hfill {\mathsf{\Phi }}_5(1,5)& =\frac{67637281{\pi }^{10}}{3656994324480\sqrt{3}}.\hfill \end{array}$$

3.4. $p=1\&q=4$

The corresponding generating function becomes

$$\mathsf{\Phi }\left(x\right|1,4)=\frac{2\pi xcos\left(\frac{\pi }{5}\right)sin\left(\frac{\pi x}{5}\right)}{5\left\{cos\left(\frac{2\pi x}{5}\right)-cos\left(\frac{2\pi }{5}\right)\right\}},$$

where

$$cos\left(\frac{\pi }{5}\right)=\frac{1+\sqrt{5}}{4}\mathrm{and}cos\left(\frac{2\pi }{5}\right)=\frac{\sqrt{5}-1}{4}.$$

Then, by extracting the initial five coefficients from the above generating function, we deduce the following infinite series evaluations:

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_1(1,4)& =\frac{(5+3\sqrt{5}){\pi }^2}{125},\hfill \\ \hfill {\mathsf{\Phi }}_2(1,4)& =\frac{(91+45\sqrt{5}){\pi }^4}{18750},\hfill \\ \hfill {\mathsf{\Phi }}_3(1,4)& =\frac{(4805+2211\sqrt{5}){\pi }^6}{9375000},\hfill \\ \hfill {\mathsf{\Phi }}_4(1,4)& =\frac{(516763+232845\sqrt{5}){\pi }^8}{9843750000},\hfill \\ \hfill {\mathsf{\Phi }}_5(1,4)& =\frac{(472560025+211741263\sqrt{5}){\pi }^{10}}{88593750000000}.\hfill \end{array}$$

3.5. $p=2\&q=3$

The corresponding generating function can be written as

$$\mathsf{\Phi }\left(x\right|2,3)=\frac{2\pi x}{5}\frac{cos\left(\frac{2\pi }{5}\right)sin\left(\frac{\pi x}{5}\right)}{cos\left(\frac{2\pi x}{5}\right)-cos\left(\frac{4\pi }{5}\right)},$$

where

$$cos\left(\frac{2\pi }{5}\right)=\frac{\sqrt{5}-1}{4}\mathrm{and}cos\left(\frac{4\pi }{5}\right)=\frac{-\sqrt{5}-1}{4}.$$

Extracting the first five coefficients from the above $\mathsf{\Phi }\left(x\right|2,3)$, we find the related infinite series evaluations:

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_1(2,3)& =\frac{(3\sqrt{5}-5){\pi }^2}{125},\hfill \\ \hfill {\mathsf{\Phi }}_2(2,3)& =\frac{(45\sqrt{5}-91){\pi }^4}{18750},\hfill \\ \hfill {\mathsf{\Phi }}_3(2,3)& =\frac{(2211\sqrt{5}-4805){\pi }^6}{9375000},\hfill \\ \hfill {\mathsf{\Phi }}_4(2,3)& =\frac{(232845\sqrt{5}-516763){\pi }^8}{9843750000},\hfill \\ \hfill {\mathsf{\Phi }}_5(2,3)& =\frac{(211741263\sqrt{5}-472560025){\pi }^{10}}{88593750000000}.\hfill \end{array}$$

4. Evaluation of ${\psi }_m(p,q)$-Series

The preceding approach to determine the generation functions for the series ${\varphi }_m(p,q)$ and ${\mathsf{\Phi }}_m(p,q)$ does not work directly for ${\psi }_m(p,q)$. We have to rewrite first the bilateral series ${\psi }_m(p,q)$ as two unilateral ones by making the replacement $k\to -1-k$ for the terms with negative summation index k:

$${\psi }_m(p,q)=\sum _{k=0}^{\infty }\left\{\frac{1}{{(p+pk+qk)}^{2m-1}}-\frac{1}{{(q+pk+qk)}^{2m-1}}\right\}.$$

Then, we can deal with the generating function

$$\begin{array}{cc}\hfill \psi \left(x\right|p,q)& =\sum _{m=1}^{\infty }{x}^{2m-2}{\psi }_m(p,q)\hfill \\ & =\sum _{m=1}^{\infty }{x}^{2m-2}\sum _{k=0}^{\infty }\left\{\frac{1}{{(p+pk+qk)}^{2m-1}}-\frac{1}{{(q+pk+qk)}^{2m-1}}\right\}\hfill \\ & =\sum _{k=0}^{\infty }\sum _{m=1}^{\infty }{x}^{2m-2}\left\{\frac{1}{{(p+pk+qk)}^{2m-1}}-\frac{1}{{(q+pk+qk)}^{2m-1}}\right\}\hfill \\ & =\sum _{k=0}^{\infty }\left\{\frac{p+pk+qk}{{(p+pk+qk)}^2-x^2}-\frac{q+pk+qk}{{(q+pk+qk)}^2-x^2}\right\}.\hfill \end{array}$$

By further rewriting the summand

$$\begin{array}{cc}\hfill 2& \left\{\frac{p+pk+qk}{{(p+pk+qk)}^2-x^2}-\frac{q+pk+qk}{{(q+pk+qk)}^2-x^2}\right\}\hfill \\ \hfill =& \left\{\frac{1}{p+pk+qk+x}-\frac{1}{q+pk+qk-x}\right\}\hfill \\ \hfill -& \left\{\frac{1}{q+pk+qk+x}-\frac{1}{p+pk+qk-x}\right\},\hfill \end{array}$$

we can express the generation function $\psi \left(x\right|p,q)$ in terms of the ${}_2{H}_2$-series

$$\begin{array}{cc}\hfill \psi \left(x\right|p,q)& =\frac{1}{2}\sum _{k=-\infty }^{\infty }\left\{\frac{1}{p+pk+qk+x}-\frac{1}{q+pk+qk+x}\right\}\hfill \\ & =\frac{q-p}{2}\sum _{k=-\infty }^{\infty }\frac{1}{(p+pk+qk+x)(q+pk+qk+x)}\hfill \\ & =\frac{q-p}{2(p+x)(q+x)}{}_2{H}_2\left[\begin{array}{c}\frac{p+x}{p+q},\frac{q+x}{p+q}\\ \\ 1+\frac{p+x}{p+q},1+\frac{q+x}{p+q}\end{array}|1\right]\hfill \\ & =\frac{1}{2(p+q)}\Gamma \left[\begin{array}{c}\frac{p+x}{p+q},1-\frac{p+x}{p+q},\frac{q+x}{p+q},1-\frac{q+x}{p+q}\\ \\ 1+\frac{p-q}{p+q},\frac{q-p}{p+q}\end{array}\right].\hfill \end{array}$$

After some simplifications, we find the following trigonometric expression:

Theorem 3.

The generating function for${\psi }_m(p,q)$defined in (8) is given by

$$\psi \left(x\right|p,q)=\frac{\pi sin\left(\frac{2p}{p+q}\pi \right)}{2(p+q)sin\left(\frac{p+x}{p+q}\pi \right)sin\left(\frac{p-x}{p+q}\pi \right)}=\frac{\pi sin\left(\frac{2p}{p+q}\pi \right)}{(p+q)\left\{cos\left(\frac{2x}{p+q}\pi \right)-cos\left(\frac{2p}{p+q}\pi \right)\right\}}.$$

When $p=1$ and $q=5$, this gives rise to the generating function (4) illustrated in the introduction for the series proposed by Zhou [1]. Further examples are recorded in the next four subsections.

4.1. $p=1\&q=2$

In this case, the explicit generating function is given by

$$\psi \left(x\right|1,2)=\frac{\pi }{\sqrt{3}\left\{1+2cos\left(\frac{2x}{3}\pi \right)\right\}}.$$

By making use of (11), we can expand the trigonometric function into power series

$$\begin{array}{cc}\hfill \frac{1}{1+2cosy}& =\frac{y}{siny}\times \frac{siny}{y\left\{1+2cosy\right\}}\hfill \\ & =\sum _{i=0}^{\infty }{(-1)}^i\frac{2-2^{2i}}{\left(2i\right)!}{B}_{2i}{y}^{2i}\sum _{k=1}^{\infty }{(-1)}^k\frac{(1-9^k)}{2\left(2k\right)!}{B}_{2k}{y}^{2k-2}\hfill \\ & =\sum _{n=1}^{\infty }{(-1)}^n{y}^{2n-2}\sum _{k=1}^n\frac{(1-9^k)}{\left(2k\right)!}{B}_{2k}\frac{1-2^{2n-2k-1}}{(2n-2k)!}{B}_{2n-2k}.\hfill \end{array}$$

This yields the following explicit formula:

$${\psi }_m(1,2)={(-1)}^m\frac{4^{m-1}{\pi }^{2m-1}}{9^{m-1}\sqrt{3}}\sum _{k=1}^m\frac{1-9^k}{\left(2k\right)!}{B}_{2k}\frac{1-2^{2m-2k-1}}{(2m-2k)!}{B}_{2m-2k}.$$

The values of the first five series are displayed as follows:

$$\begin{array}{cc}\hfill {\psi }_1(1,2)& =\frac{\pi }{3\sqrt{3}},\hfill \\ \hfill {\psi }_2(1,2)& =\frac{4{\pi }^3}{81\sqrt{3}},\hfill \\ \hfill {\psi }_3(1,2)& =\frac{4{\pi }^5}{729\sqrt{3}},\hfill \\ \hfill {\psi }_4(1,2)& =\frac{56{\pi }^7}{98415\sqrt{3}},\hfill \\ \hfill {\psi }_5(1,2)& =\frac{3236{\pi }^9}{55801305\sqrt{3}}.\hfill \end{array}$$

4.2. $p=1\&q=3$

The corresponding generating function reads as

$$\psi \left(x\right|1,3)=\frac{\pi }{4cos\left(\frac{\pi x}{2}\right)}=\sum _{n=0}^{\infty }\frac{{(-1)}^n{\pi }^{2n+1}}{2^{2n+2}\left(2n\right)!}{E}_{2n}{x}^{2n}.$$

Then, we derive the following explicit formula:

$${\psi }_m(1,3)=\frac{{(-1)}^{m-1}{\pi }^{2m-1}}{4^m(2m-2)!}{E}_{2m-2}$$

together with the evaluations of the first five series:

$$\begin{array}{cc}\hfill {\psi }_1(1,3)& =\frac{\pi }{4},\hfill \\ \hfill {\psi }_2(1,3)& =\frac{{\pi }^3}{32},\hfill \\ \hfill {\psi }_3(1,3)& =\frac{5{\pi }^5}{1536},\hfill \\ \hfill {\psi }_4(1,3)& =\frac{61{\pi }^7}{184320},\hfill \\ \hfill {\psi }_5(1,3)& =\frac{277{\pi }^9}{8257536}.\hfill \end{array}$$

The same values were also obtained previously by Elkies [11].

4.3. $p=1\&q=4$

The corresponding generating function is given by

$$\psi \left(x\right|1,4)=\frac{\pi sin\left(\frac{2\pi }{5}\right)}{5\left\{cos\left(\frac{2x}{5}\pi \right)-cos\left(\frac{2}{5}\pi \right)\right\}},$$

where

$$sin\left(\frac{2\pi }{5}\right)=\sqrt{\frac{5+\sqrt{5}}{8}}\mathrm{and}cos\left(\frac{2\pi }{5}\right)=\frac{\sqrt{5}-1}{4}.$$

Extracting the first five coefficients from $\psi \left(x\right|1,4)$ leads us to the following formulae:

$$\begin{array}{cc}\hfill {\psi }_1(1,4)& =\frac{\pi \sqrt{5+2\sqrt{5}}}{5\sqrt{5}},\hfill \\ \hfill {\psi }_2(1,4)& =\frac{8{\pi }^3\sqrt{5+2\sqrt{5}}}{625(\sqrt{5}-1)},\hfill \\ \hfill {\psi }_3(1,4)& =\frac{712{\pi }^5\sqrt{5+2\sqrt{5}}}{46875(31\sqrt{5}-55)},\hfill \\ \hfill {\psi }_4(1,4)& =\frac{27856{\pi }^7\sqrt{5+2\sqrt{5}}}{17578125(109\sqrt{5}-229)},\hfill \\ \hfill {\psi }_5(1,4)& =\frac{905992{\pi }^9\sqrt{5+2\sqrt{5}}}{615234375(3779\sqrt{5}-8315)}.\hfill \end{array}$$

4.4. $p=2\&q=3$

The corresponding generating function reads as

$$\psi \left(x\right|2,3)=\frac{\pi sin\left(\frac{4}{5}\pi \right)}{5\left\{cos\left(\frac{2x}{5}\pi \right)-cos\left(\frac{4}{5}\pi \right)\right\}},$$

where

$$sin\left(\frac{4\pi }{5}\right)=\sqrt{\frac{5-\sqrt{5}}{8}}\mathrm{and}cos\left(\frac{4\pi }{5}\right)=\frac{-\sqrt{5}-1}{4}.$$

By extracting the first five coefficients from the above generating function, we find the following infinite series evaluations:

$$\begin{array}{cc}\hfill {\psi }_1(2,3)& =\frac{\pi \sqrt{5-2\sqrt{5}}}{5\sqrt{5}},\hfill \\ \hfill {\psi }_2(2,3)& =\frac{8{\pi }^3\sqrt{5-2\sqrt{5}}}{625(1+\sqrt{5})},\hfill \\ \hfill {\psi }_3(2,3)& =\frac{712{\pi }^5\sqrt{5-2\sqrt{5}}}{46875(55+31\sqrt{5})},\hfill \\ \hfill {\psi }_4(2,3)& =\frac{27856{\pi }^7\sqrt{5-2\sqrt{5}}}{17578125(229+109\sqrt{5})},\hfill \\ \hfill {\psi }_5(2,3)& =\frac{905992{\pi }^9\sqrt{5-2\sqrt{5}}}{615234375(8315+3779\sqrt{5})}.\hfill \end{array}$$

5. Evaluation of ${\mathsf{\Psi }}_m(p,q)$-Series

Analogously, by writing first the bilateral series ${\mathsf{\Psi }}_m(p,q)$ as two unilateral ones

$${\mathsf{\Psi }}_m(p,q)=\sum _{k=0}^{\infty }\left\{\frac{{(-1)}^k}{{(p+pk+qk)}^{2m-1}}+\frac{{(-1)}^k}{{(q+pk+qk)}^{2m-1}}\right\},$$

we can proceed with the generating function

$$\begin{array}{cc}\hfill \mathsf{\Psi }\left(x\right|p,q)& =\sum _{m=1}^{\infty }{x}^{2m-2}{\mathsf{\Psi }}_m(p,q)\hfill \\ & =\sum _{m=1}^{\infty }{x}^{2m-2}\sum _{k=0}^{\infty }\left\{\frac{{(-1)}^k}{{(p+pk+qk)}^{2m-1}}+\frac{{(-1)}^k}{{(q+pk+qk)}^{2m-1}}\right\}\hfill \\ & =\sum _{k=0}^{\infty }{(-1)}^k\sum _{m=1}^{\infty }{x}^{2m-2}\left\{\frac{1}{{(p+pk+qk)}^{2m-1}}+\frac{1}{{(q+pk+qk)}^{2m-1}}\right\}\hfill \\ & =\sum _{k=0}^{\infty }{(-1)}^k\left\{\frac{p+pk+qk}{{(p+pk+qk)}^2-x^2}+\frac{q+pk+qk}{{(q+pk+qk)}^2-x^2}\right\}.\hfill \end{array}$$

By reformulating the summand

$$\begin{array}{cc}\hfill 2& \left\{\frac{{(-1)}^k(p+pk+qk)}{{(p+pk+qk)}^2-x^2}+\frac{{(-1)}^k(q+pk+qk)}{{(q+pk+qk)}^2-x^2}\right\}\hfill \\ \hfill =& \left\{\frac{{(-1)}^k}{p+pk+qk+x}+\frac{{(-1)}^k}{q+pk+qk-x}\right\}\hfill \\ \hfill +& \left\{\frac{{(-1)}^k}{q+pk+qk+x}+\frac{{(-1)}^k}{p+pk+qk-x}\right\},\hfill \end{array}$$

we can express $\mathsf{\Psi }\left(x\right|p,q)$ in terms of bilateral series

$$\begin{array}{cc}\hfill \mathsf{\Psi }\left(x\right|p,q)& =\frac{1}{2}\sum _{k=-\infty }^{\infty }\left\{\frac{{(-1)}^k}{p+pk+qk+x}+\frac{{(-1)}^k}{q+pk+qk+x}\right\}\hfill \\ & =(p+q)\sum _{k=-\infty }^{\infty }\frac{{(-1)}^k(k+\frac{p+q+2x}{2(p+q)})}{(p+pk+qk+x)(q+pk+qk+x)}\hfill \\ & =\frac{p+q+2x}{2(p+x)(q+x)}{}_3{H}_3\left[\begin{array}{c}1+\frac{p+q+2x}{2(p+q)},\frac{p+x}{p+q},\frac{q+x}{p+q}\\ \\ \frac{p+q+2x}{2(p+q)},1+\frac{p+x}{p+q},1+\frac{q+x}{p+q}\end{array}|-1\right].\hfill \end{array}$$

To evaluate the above ${}_3{H}_3$-series, letting $b=\frac{1+a}{2}$ and $e\to -\infty$ in (10) yields

$${}_3{H}_3\left[\begin{array}{c}1+\frac{a}{2},c,d\\ \\ \frac{a}{2},1+a-c,1+a-d\end{array}|-1\right]=\Gamma \left[\begin{array}{c}\frac{1+a}{2},\frac{1-a}{2},1-c,1-d,1+a-c,1+a-d\\ \\ 1+a,1-a,\frac{1+a}{2}-c,\frac{1+a}{2}-d,1+a-c-d\end{array}\right].$$

Then, the generating function can explicitly be calculated as

$$\begin{array}{cc}\hfill \mathsf{\Psi }\left(x\right|p,q)& =\frac{1}{2(p+q)}\Gamma \left[\begin{array}{c}1+\frac{x}{p+q},\frac{-x}{p+q},\frac{p+x}{p+q},1-\frac{p+x}{p+q},\frac{q+x}{p+q},1-\frac{q+x}{p+q}\\ \\ 1+\frac{2x}{p+q},\frac{-2x}{p+q},\frac{p}{p+q},1-\frac{p}{p+q}\end{array}\right].\hfill \end{array}$$

Therefore, we obtain the following trigonometric expression.

Theorem 4.

The generating function for${\mathsf{\Psi }}_m(p,q)$defined in (9) is given by

$$\begin{array}{c}\hfill \mathsf{\Psi }\left(x\right|p,q)=\frac{\pi }{(p+q)}\frac{sin\left(\frac{\pi p}{p+q}\right)cos\left(\frac{\pi x}{p+q}\right)}{sin\left(\frac{p+x}{p+q}\pi \right)sin\left(\frac{q+x}{p+q}\pi \right)}=\frac{2\pi }{(p+q)}\frac{sin\left(\frac{\pi p}{p+q}\right)cos\left(\frac{\pi x}{p+q}\right)}{cos\left(\frac{2x}{p+q}\pi \right)+cos\left(\frac{p-q}{p+q}\pi \right)}.\end{array}$$

Particular cases of this theorem are presented in the following examples.

5.1. $p=1\&q=2$

In this case, the generating function reads as

$$\mathsf{\Psi }\left(x\right|1,2)=\frac{2\pi cos\left(\frac{\pi x}{3}\right)}{\sqrt{3}\left\{1+2cos\left(\frac{2\pi x}{3}\right)\right\}}.$$

Recalling (11), we can expand the trigonometric function into power series

$$\begin{array}{cc}\hfill \frac{cosy}{1+2cos\left(2y\right)}& =\frac{y}{siny}\times \frac{sin\left(2y\right)}{2y\left\{1+2cos\left(2y\right)\right\}}\hfill \\ & =\sum _{i=0}^{\infty }{(-1)}^i\frac{2-2^{2i}}{\left(2i\right)!}{B}_{2i}{y}^{2i}\sum _{k=1}^{\infty }{(-1)}^k\frac{2^{2k-3}(1-9^k)}{\left(2k\right)!}{B}_{2k}{y}^{2k-2}\hfill \\ & =\sum _{n=1}^{\infty }{(-1)}^n{y}^{2n-2}\sum _{k=1}^n\frac{2^{2k-3}(1-9^k)}{\left(2k\right)!}{B}_{2k}\frac{2-2^{2n-2k}}{(2n-2k)!}{B}_{2n-2k}.\hfill \end{array}$$

From this, we obtain the following analytical formula:

$${\mathsf{\Psi }}_m(1,2)=\frac{{(-1)}^m{\pi }^{2m-1}}{2·9^{m-1}\sqrt{3}}\sum _{k=1}^m\frac{4^k(1-9^k)}{\left(2k\right)!}{B}_{2k}\frac{1-2^{2m-2k-1}}{(2m-2k)!}{B}_{2m-2k}.$$

The values of the first five infinite series are given as follows:

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}_1(1,2)& =\frac{2\pi }{3\sqrt{3}},\hfill \\ \hfill {\mathsf{\Psi }}_2(1,2)& =\frac{5{\pi }^3}{81\sqrt{3}},\hfill \\ \hfill {\mathsf{\Psi }}_3(1,2)& =\frac{17{\pi }^5{x}^4}{2916\sqrt{3}},\hfill \\ \hfill {\mathsf{\Psi }}_4(1,2)& =\frac{91{\pi }^7}{157464\sqrt{3}},\hfill \\ \hfill {\mathsf{\Psi }}_5(1,2)& =\frac{207913{\pi }^9}{3571283520\sqrt{3}}.\hfill \end{array}$$

5.2. $p=1\&q=3$

In this case, we obtain the following generating function:

$$\mathsf{\Psi }\left(x\right|1,3)=\frac{\sqrt{2}}{4}\pi cos\left(\frac{\pi x}{4}\right)sec\left(\frac{\pi x}{2}\right).$$

By employing the power series expansion

$$\begin{array}{cc}\hfill cosysec\left(2y\right)& =\sum _{i=0}^{\infty }{(-1)}^i\frac{y^{2i}}{\left(2i\right)!}\sum _{k=0}^{\infty }{(-4)}^k\frac{E_{2k}}{\left(2k\right)!}{y}^{2k}\hfill \\ & =\sum _{n=0}^{\infty }{(-1)}^n{y}^{2n}\sum _{k=0}^n\frac{4^k{E}_{2k}}{\left(2k\right)!(2n-2k)!},\hfill \end{array}$$

we derive the following analytical formula:

$${\mathsf{\Psi }}_m(1,3)={(-1)}^{m-1}\frac{{\pi }^{2m-1}\sqrt{2}}{4^{2m}}\sum _{k=1}^m\frac{4^k{E}_{2k-2}}{(2k-2)!(2m-2k)!}.$$

Hence, the closed form evaluations for the first five series follow immediately:

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}_1(1,3)& =\frac{\pi }{2\sqrt{2}},\hfill \\ \hfill {\mathsf{\Psi }}_2(1,3)& =\frac{3{\pi }^3}{64\sqrt{2}},\hfill \\ \hfill {\mathsf{\Psi }}_3(1,3)& =\frac{19{\pi }^5}{4096\sqrt{2}},\hfill \\ \hfill {\mathsf{\Psi }}_4(1,3)& =\frac{307{\pi }^7}{655360\sqrt{2}},\hfill \\ \hfill {\mathsf{\Psi }}_5(1,3)& =\frac{83579{\pi }^9}{1761607680\sqrt{2}}.\hfill \end{array}$$

5.3. $p=1\&q=5$

In this case, the generating function is given by

$$\mathsf{\Psi }\left(x\right|1,5)=\frac{\pi cos\left(\frac{\pi x}{6}\right)}{3\left\{2cos\left(\frac{\pi x}{3}\right)-1\right\}}.$$

Noting that

$$\frac{cosy}{2cos2y-1}=\frac{3}{4}sec3y+\frac{1}{4}secy=\sum _{n=0}^{\infty }\frac{{(-1)}^n(3^{2n+1}+1)}{4\left(2n\right)!}{E}_{2n}{y}^{2n},$$

we find the following explicit formula

$${\mathsf{\Psi }}_m(1,5)=\frac{{(-1)}^{m-1}(1+3^{2m-1})}{2(2m-2)!}{\left(\frac{\pi }{6}\right)}^{2m-1}{E}_{2m-2}$$

as well as the evaluations of the first five series:

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}_1(1,5)& =\frac{\pi }{3},\hfill \\ \hfill {\mathsf{\Psi }}_2(1,5)& =\frac{7{\pi }^3}{216},\hfill \\ \hfill {\mathsf{\Psi }}_3(1,5)& =\frac{305{\pi }^5}{93312},\hfill \\ \hfill {\mathsf{\Psi }}_4(1,5)& =\frac{33367{\pi }^7}{100776960},\hfill \\ \hfill {\mathsf{\Psi }}_5(1,5)& =\frac{194731{\pi }^9}{5804752896}.\hfill \end{array}$$

5.4. $p=1\&q=4$

The corresponding explicit generating function reads as

$$\mathsf{\Psi }\left(x\right|1,4)=\frac{2\pi }{5}\frac{cos\left(\frac{\pi x}{5}\right)sin\left(\frac{\pi }{5}\right)}{cos\left(\frac{2\pi x}{5}\right)+cos\left(\frac{3\pi }{5}\right)},$$

where

$$sin\left(\frac{\pi }{5}\right)=\sqrt{\frac{5-\sqrt{5}}{8}}\mathrm{and}cos\left(\frac{3\pi }{5}\right)=\frac{1-\sqrt{5}}{4}.$$

Then, the first five coefficients of $\mathsf{\Psi }\left(x\right|1,4)$ give rise to the following formulae:

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}_1(1,4)& =\frac{2\pi \sqrt{2}}{5{(5-\sqrt{5})}^{1/2}},\hfill \\ \hfill {\mathsf{\Psi }}_2(1,4)& =\frac{{\pi }^3\sqrt{2}(11+\sqrt{5})}{125{(5-\sqrt{5})}^{3/2}},\hfill \\ \hfill {\mathsf{\Psi }}_3(1,4)& =\frac{{\pi }^5\sqrt{2}(383+75\sqrt{5})}{18750{(5-\sqrt{5})}^{5/2}},\hfill \\ \hfill {\mathsf{\Psi }}_4(1,4)& =\frac{{\pi }^7\sqrt{2}(3583+982\sqrt{5})}{703125{(5-\sqrt{5})}^{7/2}},\hfill \\ \hfill {\mathsf{\Psi }}_5(1,4)& =\frac{{\pi }^9\sqrt{2}(261063+85885\sqrt{5})}{196875000{(5-\sqrt{5})}^{9/2}}.\hfill \end{array}$$

5.5. $p=2\&q=3$

The corresponding generating function reads as

$$\mathsf{\Psi }\left(x\right|2,3)=\frac{2\pi }{5}\frac{cos\left(\frac{\pi x}{5}\right)sin\left(\frac{2\pi }{5}\right)}{cos\left(\frac{2\pi x}{5}\right)+cos\left(\frac{\pi }{5}\right)},$$

where

$$sin\left(\frac{2\pi }{5}\right)=\sqrt{\frac{5+\sqrt{5}}{8}}\mathrm{and}cos\left(\frac{\pi }{5}\right)=\frac{\sqrt{5}+1}{4}.$$

Extracting the initial five coefficients from $\mathsf{\Psi }\left(x\right|2,3)$ yields the following infinite series evaluations:

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}_1(2,3)& =\frac{2\pi \sqrt{2}}{5{(5+\sqrt{5})}^{1/2}},\hfill \\ \hfill {\mathsf{\Psi }}_2(2,3)& =\frac{{\pi }^3\sqrt{2}(11-\sqrt{5})}{125{(5+\sqrt{5})}^{3/2}},\hfill \\ \hfill {\mathsf{\Psi }}_3(2,3)& =\frac{{\pi }^5\sqrt{2}(383-75\sqrt{5})}{18750{(5+\sqrt{5})}^{5/2}},\hfill \\ \hfill {\mathsf{\Psi }}_4(2,3)& =\frac{{\pi }^7\sqrt{2}(3583-982\sqrt{5})}{703125{(5+\sqrt{5})}^{7/2}},\hfill \\ \hfill {\mathsf{\Psi }}_5(2,3)& =\frac{{\pi }^9\sqrt{2}(261063-85885\sqrt{5})}{196875000{(5+\sqrt{5})}^{9/2}}.\hfill \end{array}$$

Concluding comments For the four series defined in (6)–(9), their generating functions and evaluations are determined under the conditions “$p,q\in \mathbb{N}$ and $gcd(p,q)=1$”. However, all the theorems are still valid if these conditions are released by “$p,q\in \mathbb{C}$ with p and q being non-proportional: $p/q,q/p\notin \mathbb{Z}$”. For instance, Theorem 1 can be restated as follows: Let $\alpha ,\beta \in \mathbb{C}$ subject to $\alpha /\beta ,\beta \alpha \notin \mathbb{Z}$. Then, the following generating function holds

$$\sum _{m=1}^{\infty }{x}^{2m}{\varphi }_m(\alpha ,\beta )=\frac{\pi xsin\left(\frac{2\pi x}{\alpha +\beta }\right)}{2(\alpha +\beta )sin\left(\frac{\alpha +x}{\alpha +\beta }\pi \right)sin\left(\frac{\alpha -x}{\alpha +\beta }\pi \right)}.$$

By expanding the above function into power series, it is possible to evaluate theoretically the corresponding infinite series. However, the resulting expressions are too involved to be practically presented.

In addition, the identities presented in this paper suggest that one may further examine, in general, the following bilateral Hurwitz zeta function and its alternating counterpart:

$$\begin{array}{cc}\hfill \sum _{n=-\infty }^{\infty }\frac{1}{{(n+z)}^m}& ={\zeta }^{+}(m,z)+{(-1)}^m{\zeta }^{+}(m,1-z),\hfill \\ \hfill \sum _{n=-\infty }^{\infty }\frac{{(-1)}^n}{{(n+z)}^m}& ={\zeta }^{-}(m,z)-{(-1)}^m{\zeta }^{-}(m,1-z).\hfill \end{array}$$

Then, any closed formula of the above bilateral series will result in a reciprocal identity about ${\zeta }^{+}(m,z)$ or ${\zeta }^{-}(m,z)$. The interested reader is enthusiastically encouraged to make exploration.