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?