<s>[INST] Is there a way to evaluate analytically the following infinite double sum? Consider the following double sum
$$
S = \sum_{n=1}^\infty \sum_{m=1}^\infty
\frac{1}{a (2n-1)^2 - b (2m-1)^2} \, ,
$$
where $a$ and $b$ are both positive real numbers given by
\begin{align}
 a &= \frac{1}{2} - \frac{\sqrt{2}}{32}  \, , \\
 b     &= \frac{1}{4} - \frac{3\sqrt{2}}{32}  \, .
\end{align}
It turns out that one of the two sums can readily be calculated and expressed in terms of the tangente function.
Specifically,
$$
S = \frac{\pi}{4\sqrt{ab}} \sum_{m=1}^\infty 
\frac{\tan \left( \frac{\pi}{2} \sqrt{\frac{b}{a}} (2m-1) \right)}{2m-1} \, .
$$
The latter result does not seem to be further simplified. i was wondering whether someone here could be of help and let me know in case there exists a method to evaluate the sum above. Hints and suggestions welcome.
Thank you
PS From numerical evaluation using computer algebra systems, it seems that the series is convergent. This apparently would not be the case if $b<0$. 
 [/INST] Let 
$$q=\sqrt{\dfrac{b}{a}},$$
then attack via digamma-function leads to
$$\begin{align}
&S = \dfrac1a\sum\limits_{n=1}^\infty\sum\limits_{m=1}^\infty\dfrac1{(2n-1)^2-q^2(2m-1)^2}  \\[4pt]
&= \dfrac1a\sum\limits_{n=1}^\infty\sum\limits_{m=1}^\infty\dfrac1{2(2n-1)}\left(\dfrac1{2n-1-q(2m-1)}+\dfrac1{2n-1+q(2m-1)}\right)\\[4pt]
&= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\sum\limits_{m=1}^\infty \left(-\dfrac1{m-\frac{2n-1+q}{2q}}+\dfrac1{m+\frac{2n-1-q}{2q}}\right)\\[4pt]
&= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\left(\psi\left(1-\frac{2n-1+q}{2q}\right)-\psi\left(1+\frac{2n-1-q}{2q}\right)\right)\\[4pt]
&= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\left(\psi\left(\frac{q-2n+1}{2q}\right)-\psi\left(\frac{q+2n-1}{2q}\right)\right)\\[4pt]
&= \dfrac1{4aq}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\cdot\pi\cot\left(\pi\frac{q-2n+1}{2q}\right)\\[4pt]
&= \dfrac\pi{4\sqrt{ab}}\sum\limits_{n=1}^\infty\dfrac1{2n-1}\tan\left(\frac\pi 2 \sqrt{\frac ab}(2n-1)\right)\\[4pt]
&= \dfrac\pi{4\sqrt{ab}}\sum\limits_{m=1}^\infty\dfrac1{2m-1}\tan\left(\frac\pi 2 \sqrt{\color{red}{\frac ab}}(2m-1)\right).
\end{align}$$
</s>