An alternating series identity with a hidden hyperbolic tangent How does one use the inverse Mellin transform to prove that the following identity holds?
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n(e^{n\pi} + 1)} = \frac{1}{8}(\pi - 5\log(2))$$
The identity follows from MO189199 and the penultimate identity on this page (for $x=\frac{1}{2}$).

Scratch-work: I computed the Mellin transform of 
$$f(x) = \frac{(-1)^x}{x(e^{x\pi} + 1)}$$
and re-wrote the function in terms of its inverse Mellin Transform as (substituting $x = n$)
$$\sum_{n=1}^{\infty}\frac{(-1)^n}{n(e^{n\pi} + 1)} = \frac{1}{2\pi i}\int_{C} (2 \pi)^{1-s} \Gamma(s-1)\big(\zeta(s-1, \frac{1}{2} - \frac{i}{2}) - \zeta(s-1, 1-\frac{i}{2})\big)\zeta(s)ds$$
But I am not altogether confident that this has been computed correctly, and I am still not sure how to find the poles and compute the residues. For example, checking for a residue at $s=0$ gives
$$\frac{1}{8}(\pi - 2\pi i)$$
There seems to be a nontrivial contribution at $s = 1$ and $s = -1$ as well, but already the mathematics has exceeded what I understand; I am especially unclear with regard to how one accounts for residues with respect to the generalized Riemann zeta function $\zeta(s,a)$.
Clarity on using this approach or another to demonstrate the main identity would be appreciated!

The following approach does not use the Mellin transform, but it is worth mentioning. 
It is quite easy to prove that:
$$\sum_{n=1}^{+\infty}\frac{\cos(\pi n)}{n} e^{-nk\pi} = -\log\left(1+e^{-k\pi}\right)$$
hence the original sum equals:
$$-\log\prod_{k=1}^{+\infty}\frac{1+e^{-(2k-1)\pi}}{1+e^{-2k\pi}}.$$
The last product is clearly related with the Jacobi triple product.
In particular, since the values of the Jacobi theta functions $\vartheta_3$ and $\vartheta_4$ in $q=e^{-\pi}$ are known, the claim simply follows.
