TITLE: How to calculate the infinite sum of this double series?
QUESTION [12 upvotes]: 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?

REPLY [9 votes]: Here is a proof based on Hachino's idea. We have agreed it's enough to 
prove that
$$S=\sum_{m=1}^\infty\frac{(-1)^m\tanh(m\pi /2)}{m}=\frac{\log 2-\pi}{4}. $$
Plug in the expansion
$$\tanh(m\pi /2)=\frac{1-e^{-m\pi }}{1+e^{-m\pi }}=1+2\sum_{n=1}^\infty(-1)^ne^{-nm\pi }. $$
Changing order of summation and using the Taylor series of $\log(1+x)$ gives
$$S=-\log 2+2\sum_{n=1}^\infty(-1)^{n+1}\log(1+e^{-n\pi })=\log\left(\frac{\prod_{n=1}^\infty(1+e^{-(2n-1)\pi})^2}{2\prod_{n=1}^\infty(1+e^{-2n\pi})^2}\right).$$
The infinite products can be recognized as theta values. Recall that
$$\theta_2(z,q)=2q^{1/4}\cos z\prod_{n=1}^\infty(1-q^{2n})(1+2q^{2n}\cos 2z+q^{4n}), $$
$$\theta_3(z,q)=\prod_{n=1}^\infty(1-q^{2n})(1+2q^{2n-1}\cos 2z+q^{4n-2}). $$
There are lots of conflicting notations for theta functions; I use the
conventions of Whittaker and Watson, A course in modern analysis.
It follows that
$$S=\log\left(e^{-\pi/4}\frac{\theta_3(0,e^{-\pi})}{\theta_2(0,e^{-\pi})}\right). $$
It remains to show that
$$\frac{\theta_3(0,e^{-\pi})}{\theta_2(0,e^{-\pi})}=2^{1/4}. $$
This is clear from generalities on elliptic functions. Again in the notation of
Whittaker and Watson, the modulus $k$ is given by
$$k=\frac{\theta_2(0,q)^2}{\theta_3(0,q)^2}. $$
The value $q=e^{-\pi}$ corresponds to $k=k'$, where $k'$ is the complementary modulus defined by $k^2+(k')^2=1$. Thus, $k=1/\sqrt 2$, which gives the desired result.
If you had a tough childhood, you might worry that I changed the order of summation in series that are not absolutely convergent. You might want to plug in a convergence factor $x^m$, repeat the argument above and prove that both sides of the resulting identity are left continuous at $x=1$.