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It can be verified that $$ \sum_{n=1}^{\infty}\frac{1}{\cosh(\pi n)^3\sinh(\pi n)^2} =\frac{11}{12}-\frac{3K}{2\pi}+\frac{K^2}{2\pi^2}-\frac{K^3}{\pi^3}$$ where $K=\frac{\Gamma\left ( \frac14 \right )^2 }{4\sqrt{\pi}}$(here and below) and $\Gamma(z)$ is the Euler Gamma function.

Proof. The proof relies on complex analysis. Let us directly consider the function $$ f(z)=\frac{\operatorname{cs}\left(z,\frac{1}{\sqrt{2}}\right)}{\cosh(\frac{\pi z}{2K})^3\sinh(\frac{\pi z}{2K})^2}. $$ Where $\operatorname{cs}(z,k)$ denotes one of Jacobi's elliptic functions. Integrate the function along a rectangular contour with vertices $-L,L,L+2Ki,-L+2Ki$ counterclockwise, where $L$ is a positive real number such that there is no pole on the left(or right) edge of the contour. To avoid poles on the upper and lower edges, construct infinite semicircles around the poles inside the rectangle with radius $r$.
Since $\operatorname{cs}$ has double periodicity and poles at $z=\{ (2m+2ni)K\mid (m,n)\in\mathbb{Z}^2\} $, we have $$ \mathcal{P.V.}\int_{-L}^{L}[f(z)-f(z+2Ki)]\text{d}z+ \sum_{L_n\in\text{semicircles}}\int_{L_n}f(z)\text{d}z +\int_{L}^{L+2Ki}f(z)\text{d}z+\int_{-L+2Ki}^{-L}f(z)\text{d}z =2\pi i\sum_{z_k\in\text{poles inside the contour}}\operatorname{Res}(f(z),z_k). $$ As $L\rightarrow+\infty$, the first, third and fourth integrals obviously vanish because $f(z)$ is odd. So we have $$ \sum_{L_n\in\text{semicircles}}\int_{L_n}f(z)\text{d}z=2\pi i\sum_{z_k\in\text{poles inside the contour}}\operatorname{Res}(f(z),z_k). $$ The rest is easy. Note that $$ \sum_{L_n\in\text{semicircles}}\int_{L_n}f(z)\text{d}z= -\pi i\sum_{z_k\in2K\mathbb{Z},2K\mathbb{Z}+2Ki}\operatorname{Res}(f,z_k)=-\pi i\left ( 4\sum_{n=1}^{\infty}\frac{1}{\cosh(\pi n)^3\sinh(\pi n)^2}+2\operatorname{Res}(f(z),0)\right). $$ Then we get $$ \sum_{n=1}^{\infty}\frac{1}{\cosh(\pi n)^3\sinh(\pi n)^2}=\frac12\left(-\operatorname{Res}(f,0)-\operatorname{Res}(f,iK)\right). $$ The residues are easy to compute, then we prove the sum. We can compute the values of $S(2n+1,2k)=\sum_{m=1}^{\infty} \frac{1}{\cosh(\pi m)^{2n+1} \sinh(\pi m)^{2k}},(n,k)\in\mathbb{Z}^2$ if the sum exists. For example, $$ S(3,4)=-\frac{943}{720}+\frac{5K}{2\pi} -\frac{13K^2}{12\pi^2}+\frac{K^3}{\pi^3}+\frac{K^4}{20\pi^4},$$ $$ \sum_{n=1}^{\infty} \frac{1}{\cosh(\pi n)^{11}} =-\frac{1}{2}+\frac{63K^{}}{256\pi^{}} +\frac{117469K^{3}}{201600\pi^{3}}+ \frac{17281K^{5}}{10080\pi^{5}}+\frac{3553K^{7}}{1200\pi^{7}} +\frac{869K^{9}}{240\pi^{9}}+\frac{1381K^{11}}{600\pi^{11}}. $$

Question 1: Can we prove the sum in an alternative way, or is there a simpler method to compute these $S(n,k)$?
Question 2: Is there a complex method to compute $S(2n,2k)$ for integers $n,k$?

Setness Ramesory
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1 Answers1

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With regard to Question 2, the infinite series $$\sum_{n=1}^{\infty} \frac{1}{\sinh^{2} (\pi n)}$$ can actually be evaluated in the typical manner of exploiting the poles of $\pi \cot(\pi z)$. See this question.

I don't know how to use contour integration to evaluate $\sum_{n=1}^{\infty} \frac{1}{\cosh^{2}(\pi n)}, $ but what can be shown using contour integration is that $$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{\cosh^{2}(\pi n)} = \frac{4 K^{2}}{\sqrt{2} \pi^{2}} , $$ where $K$ is the complete elliptic integral of the first kind with modulus $\frac{1}{\sqrt{2}}$.

(Mathematica for some reason uses the parameter $m$ instead of the modulus $k$.)

We can exploit the fact that the Jacobi elliptic function $$\operatorname{ns} \left(z, \frac{1}{\sqrt{2}} \right) = \frac{1}{\operatorname{sn} \left(z,\frac{1}{\sqrt{2}} \right)}$$ has simple poles at $z=2m K \pm 2 n i K$, $(m, n) \in \mathbb{Z}^{2}$, with residues that alternate between $1$ and $-1$ in the real direction.

(For anyone interested, there's a table here that shows special values of the Jacobi elliptic functions.)

Integrating the function $$f(z) = \frac{\operatorname{ns}\left(z, \frac{1}{\sqrt{2}} \right)}{\cosh^{2} (\frac{\pi z}{2 K})} $$ around the same contour that was used in the question, we get $$-\pi i \sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{\cosh^{2}(\pi n)} - \pi i\sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{\cosh^{2}(\pi n)} = 2 \pi i \operatorname{Res} \left[f(z), iK \right].$$

(The integral vanishes on the left and right sides of the contour because $\cosh (z)$ grows exponentially as $\Re(z) \to \pm \infty$.)

Since $ \operatorname{ns} \left(z, \frac{1}{\sqrt{2}} \right)$ is analytic at $z= iK$ with value $0$ and first derivative $\frac{1}{\sqrt{2}}$ (see here), and $$\frac{1}{\cosh^{2} (\frac{\pi z}{2 K})}= - \frac{4K^{2}}{\pi^{2}} \frac{1}{(z-iK)^{2}} + \mathcal{O}(1),$$

we have $$\operatorname{Res} \left[f(z), iK \right] = - \frac{4K^{2}}{\sqrt{2} \pi^{2}}. $$

Therefore, $$ \sum_{n=-\infty}^{\infty} \frac{(-1)^{n}}{\cosh^{2}(\pi n)} = \frac{4 K^{2}}{\sqrt{2} \pi^{2}} = \frac{\left(\Gamma \left(\frac{1}{4} \right)\right)^{4}}{4 \sqrt{2} \pi^{3}}.$$

This approach can also be used to show that $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sinh^{2}(\pi n)} = \frac{1}{2} \left(\frac{K^{2}}{\pi^{2}}-\frac{1}{3} \right) = \frac{1}{2} \left(\frac{\left(\Gamma \left(\frac{1}{4} \right)\right)^{4}}{16 \pi^{3}}- \frac{1}{3} \right). $$

Random Variable
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