TITLE:  series involving Catalan and Zeta
QUESTION [9 upvotes]: I ran across another challenging and interesting series, and I am wondering if someone could shed some light on how to evaluate it.
$$ \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(n^{2}+k^{2})^{2}}=\zeta(2)\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^{2}}-\zeta(4)$$
This has turned out to be rather challenging. At first glance, I thought it may be somewhere along the lines of the famous $$ \sum_{n=1}^{\infty}\frac{1}{n^{2}+k^{2}}=\frac{\pi}{2k}\coth(\pi k)-\frac{1}{2k^{2}}$$ that is often seen in Complex Analysis.
So, I ran the first sum through Maple and it gave me:
$$ \sum_{n=1}^{\infty} \sum_{k=1}^{\infty}\frac{1}{(n^{2}+k^{2})^{2}}=\frac{{\pi}^{2}}{4}\sum_{n=1}^{\infty}\frac{\coth^2(\pi n)}{n^{2}}+\frac{\pi}{4}\sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^{3}}-\frac{{\pi}^{2}}{4}\sum_{n=1}^{\infty}\frac{1}{n^{2}}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^{4}}$$
Of course, two of these are the very familiar $\zeta(2), \;\ \zeta(4)$.  I managed to evaluate $\displaystyle \sum_{n=1}^{\infty}\frac{\coth(\pi n)}{n^{3}}=\frac{7{\pi}^{3}}{180}$ using Complex Analysis.
The one that has given me the fit is $\displaystyle\sum_{n=1}^{\infty}\frac{\coth^{2}(\pi n)}{n^{2}}$.
This evaluates to $\displaystyle\frac{2}{3}K+\frac{19{\pi}^{2}}{180}$.  But, how?.
I tried a Complex Analysis method, but had trouble finding the residues at $ni$, which are the zeroes of $\sinh^{2}(\pi n)$ .
Anyone have any good ideas on how to evaluate the original sum at the top or even just this 'coth-squared' one?. Complex analysis or other wise.  I thought maybe a clever use of Fourier would work, but maybe not.

REPLY [2 votes]: I managed to evaluate the series by beginning with:
$\displaystyle \sum_{k=1}^{\infty}\frac{1}{n^{2}+k^{2}}=\frac{\pi n\coth(\pi n)-1}{2n^{2}}$, and using some known sums/identities. 
Then, differentiate, do some manipulating and it works out to what Maple gave.
Then, sum these four series separately.
It works out to $
\frac{-{\pi}^{2}}{4}\zeta(2)+\frac{{\pi}^{2}}{4}\left(\frac{2}{3}K+\frac{19{\pi}^{2}}{180}\right)-\frac{1}{2}\zeta(4)+\frac{\pi}{4}\cdot \frac{7{\pi}^{3}}{180}$
This then simplifies down to $\zeta(2)K-\zeta(4)$.  K being the Catalan constant represented by the series in the problem. 
The trickiest sum was $\displaystyle \sum_{n=1}^{\infty}\frac{\coth^{2}(\pi n)}{n^{2}}$.
$=\displaystyle\sum_{n=1}^{\infty}\frac{csch^{2}(\pi n)}{n^{2}}+\sum_{n=1}^{\infty}\frac{1}{n^{2}}$
$=\displaystyle\frac{2}{3}K-\frac{11{\pi}^{2}}{180}+\frac{{\pi}^{2}}{6}=\frac{2}{3}K+\frac{19{\pi}^{2}}{180}$