Prove $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^2(m^2+n^2)}=\frac{\pi^4}{72}$ How may I prove that
$$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^2(m^2+n^2)}=\frac{\pi^4}{72}?$$
I also discussed the problem in the chat, but no solution so far.  Some hints? Thanks!
 A: I can derive the second half of your question.  To do this, rewrite the double sum as
$$\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2} \left ( \sum_{m=-\infty}^{\infty} \frac{1}{m^2+n^2} - \frac{1}{n^2} \right )$$ 
Use the fact that 
$$\sum_{m=-\infty}^{\infty}  \frac{1}{m^2+n^2} = \frac{\pi}{n} \coth{\pi n}$$
Now the sum is
$$\frac{1}{2} \left ( \pi \sum_{n=1}^{\infty} \frac{\coth{\pi n}}{n^3} - \sum_{n=1}^{\infty}\frac{1}{n^4} \right )$$ 
Now use the analysis here:
sum of series involving coth using complex analysis
to derive the following result:
$$ \pi \sum_{n=1}^{\infty} \frac{\coth{\pi n}}{n^3} = \frac{7 \pi^4}{180} $$
The result follows from the well-known result that $\sum_{n=1}^{\infty}\frac{1}{n^4} = \pi^4/90$.
A: For now, here is how we can prove the second equality. Let the second sum be $S.$ Note that by symmetry we also have $$S= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2(m^2+n^2)}.$$ Now adding the two forms gives: $$2S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2n^2}= \left( \sum_{m=1}^{\infty} \frac{1}{m^2} \right)\left( \sum_{n=1}^{\infty} \frac{1}{n^2}\right)= \frac{\pi^4}{36}.$$
As Fabian alludes to in the comments, it appears the first equality does not hold, since the difference between the two sums is $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2-n}{n^3}\frac{1}{(m^2+n^2)}>0.$$
