Evaluating $\sum_{n=1}^\infty \frac{n^3}{e^{2\pi n}-1}$ using inverse Mellin transform inspiration on the post Evaluating $\sum_{n=1}^{\infty} \frac{n}{e^{2 \pi n}-1}$ using the inverse Mellin transform
it is possible to calculate in close form 
$$\sum _{k=1}^{\infty } -\frac{k^3}{e^{2 \pi  k}-1}=\frac{3840 \pi ^4 \psi _{e^{2 \pi }}^{(1)}(1)+480 \pi ^2 \psi _{e^{2 \pi }}^{(3)}(1)-704 \pi ^6-5760 \pi ^5+3 \Gamma \left(\frac{1}{4}\right)^8}{23040 \pi ^6}$$ using euler sum I appreciatte some comment.
I like to give another series it will interesting using elliptic theta function theory to prove it
$$\sum _{k=1}^{\infty } \frac{\left(k \left(-\log \left(\frac{\pi }{2}\right)\right)\right)^3}{e^{2 \pi  \left(k \log \left(\frac{\pi }{2}\right)\right)}+1}$$
$$\frac{\log ^4\left(\frac{\pi }{2}\right) \psi _{e^{-\frac{\pi }{\log \left(\frac{\pi }{2}\right)}}}^{(3)}(1)-\log ^4\left(\frac{\pi }{2}\right) \psi _{e^{-\frac{\pi }{\log \left(\frac{\pi }{2}\right)}}}^{(3)}\left(-\frac{\left(i \pi -\frac{\pi }{\log \left(\frac{\pi }{2}\right)}\right) \log \left(\frac{\pi }{2}\right)}{\pi }\right)}{16 \pi ^4 \log \left(\frac{\pi }{2}\right)}-\frac{1}{240} \log ^3\left(\frac{\pi }{2}\right)-\frac{7}{1920 \log \left(\frac{\pi }{2}\right)}$$
sorry for the latex type i do not to improve
 A: This is an alternative approach which is too long for comment. If we put $q = e^{-\pi}$ then the desired sum is $$-\sum_{n = 1}^{\infty}\frac{n^{3}}{q^{-2n} - 1} = -\sum_{n = 1}^{\infty}\frac{n^{3}q^{2n}}{1 - q^{2n}} = \frac{1 - Q(q^{2})}{240}$$ and we know that $$Q(q^{2}) = \left(\frac{2K}{\pi}\right)^{4}(1 - k^{2} + k^{4})$$ For $q = e^{-\pi}$ we have $k = 1/\sqrt{2}$ so that $(1 - k^{2} + k^{4}) = 1 - 1/2 + 1/4 = 3/4$ and $$K(k) = \frac{1}{4\sqrt{\pi}}\Gamma\left(\frac{1}{4}\right)^{2}$$ so that $$Q(q^{2}) = \frac{3}{4}\left(\frac{2K}{\pi}\right)^{4} = \frac{3}{64\pi^{6}}\Gamma\left(\frac{1}{4}\right)^{8}$$ and therefore the desired sum is equal to $$\frac{64\pi^{6} - 3\Gamma(1/4)^{8}}{15360\pi^{6}}$$ The expression for $Q(q^{2})$ in terms of $K, k$ is derived here. Using the same technique and expression for $R(q^{2})$ we can get the surprisingly simple and beautiful result $$\sum_{n=1}^{\infty}\frac{n^{5}}{e^{2\pi n} - 1} = \frac{1}{504}$$ For the second sum mentioned in the question we let $$q = \exp(-2\pi\log(\pi/2)) = \left(\frac{2}{\pi}\right)^{2\pi}$$ and then the desired sum is equal to $$S = -\left(\log\left(\frac{\pi}{2}\right)\right)^{3}\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 + q^{n}} = -\left(\log\left(\frac{\pi}{2}\right)\right)^{3}\cdot A$$ where the sum
\begin{align}
A &= \sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 + q^{n}}\notag\\
&= \sum_{n = 1}^{\infty}n^{3}q^{n}\left(\frac{1}{1 + q^{n}} - \frac{1}{1 - q^{n}} + \frac{1}{1 - q^{n}}\right)\notag\\
&= \sum_{n = 1}^{\infty}n^{3}q^{n}\left(\frac{1}{1 - q^{n}} - \frac{2q^{n}}{1 - q^{2n}}\right)\notag\\
&= \sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}} - 2\sum_{n = 1}^{\infty}\frac{n^{3}q^{2n}}{1 - q^{2n}}\notag\\
&= \frac{Q(q) - 1}{240} - \frac{Q(q^{2}) - 1}{120}\notag\\
&= \frac{1 + Q(q) - 2Q(q^{2})}{240}\notag\\
&= \frac{1}{240} + \frac{1}{240}\left(\frac{2K}{\pi}\right)^{4}(1 + 14k^{2} + k^{4} - 2 + 2k^{2} - 2k^{4})\notag\\
&= \frac{1}{240} - \frac{1}{240}\left(\frac{2K}{\pi}\right)^{4}(1 - 16k^{2} + k^{4})\notag\\
\end{align}
and hence $$S = \frac{1}{240}\left(\log\left(\frac{\pi}{2}\right)\right)^{3}\left(\frac{2K}{\pi}\right)^{4}(1 - 16k^{2} + k^{4}) - \frac{1}{240}\left(\log\left(\frac{\pi}{2}\right)\right)^{3}$$ I doubt if it can be put into a closed form which is as simple as that for the previous sum.
