Ramanujan-type sum Could you show that
$$\sum _{k=0}^{\infty } \frac{k}{e^{\frac{\pi  k}{2}}+1}=\frac{7 \pi ^2+6 \left(\psi _{e^{2 \pi }}^{(1)}(1)+\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{2}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1+\frac{i}{4}\right)-\psi _{e^{2 \pi }}^{(1)}\left(1-\frac{i}{4}\right)\right)}{24 \pi ^2}$$ using the Mellin transform?
 A: If we formally apply Abel–Plana's summation formula (see, for example, http://www.jstor.org/stable/2008442):
$$\sum\limits_{k=0}^\infty f(k)=\frac{1}{2}f(0)+\int\limits_0^\infty f(x)dx+i\int\limits_0^\infty\frac{f(ix)-f(-ix)}{e^{2\pi x}-1}dx,$$
we get
$$\sum\limits_{k=0}^\infty \frac{k}{e^{\frac{\pi}{2}k}+1}\;?=?\;
\int\limits_0^\infty\frac{x}{e^{\frac{\pi}{2}x}+1}dx-\int\limits_0^\infty\frac{x}{e^{2\pi x}-1}dx=\frac{1}{3}-\frac{1}{24}=7/24.$$
Similarly, in the case of Ramanujan sum type we get 
$$\sum\limits_{k=0}^\infty \frac{k}{e^{2k}+1}\;?=?\;
\int\limits_0^\infty\frac{x}{e^{2x}+1}dx-\int\limits_0^\infty\frac{x}{e^{2\pi x}-1}dx=\frac{\pi^2}{48}-\frac{1}{24}.$$
Numerically it seems we got almost correct results (in the first case relative accuracy is about $10^{-4}$ and in the second case about $10^{-3}$). Why Abel–Plana's summation formula cannot be applied in these cases and how it can be amended to produce correct results?
As for the Mellin transform method, the following references might be useful: 
http://www.tandfonline.com/doi/abs/10.1080/14786444908521717 (The application of Mellin Transforms to the summation of slowly convergent series, by G.G. Macfarlane) and http://www.scielo.org.ve/scielo.php?script=sci_arttext&pid=S0254-07702010000100011 (A summation formula concerning the Mellin Transform, by M.L. Glasser, N. Bagis and 
G. Majchrowska).
P.S. Abel–Plana's summation formula cannot be applied in these cases because $f(z)$ is not bounded in the right-half complex plane.
$\psi_q^{(n)}$ is, I think, $q$-Polygamma Function: http://mathworld.wolfram.com/q-PolygammaFunction.html
Similar sum $$\sum\limits_{k=1,3,5,...}^\infty \frac{k}{e^{k\pi}+1}=\frac{1}{24}$$ is calculated by the Mellin transform method in the answer to the question https://math.stackexchange.com/questions/389146/proof-of-frac1e-pi1-frac3e3-pi1-frac5e5-pi1-ldots/389168#389168  See also the answer to the question https://math.stackexchange.com/questions/1482918/evaluating-sum-n-1-infty-fracn3e2-pi-n-1-using-inverse-mellin-tra where the sum 
$$\sum\limits_{k=1}^\infty \frac{k^5}{e^{2\pi k}-1}=\frac{1}{504}$$ is calculated (I found these links thanks to Antonio's comment).
