hyperbolic sum and elliptic integral I am found the following sum
$$\sum _{n=1}^{\infty } \frac{n}{\sinh \left(\pi ^2 n\right)}=-\frac{1}{4 \pi ^2}+\frac{-\psi _e^{(1)}\left(1+\frac{i \pi }{2}\right)-\psi _e^{(1)}\left(1-\frac{i \pi }{2}\right)+2}{2 \pi ^2}
$$ it will be possible to get the similar value using elliptic theory 
 A: Let's give it a try using elliptic integrals. We know that
\begin{align}
P(q) &= 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}} = \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6E}{K} + k^{2} - 5\right)\tag{1}\\
P(q^{2}) &=  1 - 24\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}} = \left(\frac{2K}{\pi}\right)^{2}\left(\frac{3E}{K} + k^{2} - 2\right)\tag{2}
\end{align}We have
\begin{align}
S &= \sum_{n = 1}^{\infty}\frac{n}{\sinh \pi^{2}n}\notag\\
&= 2\sum_{n = 1}^{\infty}\frac{n}{e^{\pi^{2}n} - e^{-\pi^{2}n}}\notag\\
&= 2\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{2n}}\text{ (where }q = e^{-\pi^{2}})\notag\\
&= 2\sum_{n = 1}^{\infty}\frac{nq^{n}(1 + q^{n} - q^{n})}{1 - q^{2n}}\notag\\
&= 2\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}} - \frac{nq^{2n}}{1 - q^{2n}}\notag\\
&= \frac{1 - P(q)}{12} - \frac{1 - P(q^{2})}{12}\notag\\
&= \frac{P(q^{2}) - P(q)}{12}\notag\\
&= \left(\frac{K}{\pi}\right)^{2}\left(1 - \frac{E}{K}\right)\notag
\end{align}
I doubt it is possible to go further in this manner to reach your expression. Also the value of $K, E$ seem intractable for $q = e^{-\pi^{2}}$.
