I'm calculating this double sum:
$$
\sum _{m=1}^{\infty } \sum _{k=0}^{\infty } \frac{(-1)^m}{(2 k+1)^2+m^2}
$$
I know the answer is 
$$
\frac{ \pi  \log (2)}{16}-\frac{\pi ^2}{16}
$$
which can be verified by numerical calculations. I used the Taylor expansions of $log (1+x)$ and $arcsin (x)$ at x=1 to replace $log (2)$ and $\pi$. I got
$$
\sum _{m=1}^{\infty } \sum _{k=0}^{\infty } \frac{(2 m+1) (-1)^{k+m}}{4 m(2 k+1) (2 m-1)}
$$
I tried to play with the dummy variables but failed.
Any idea?