Is there a way to get a closed expression for the double sum: $$\sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \left( \frac{m}{(n^2 + m^2 - E_1)(n^2 + m^2 - E_2)} \right)^2$$ , where $E_1$ and $E_2$ are some complex parameters, which is not equal to $n^2 + m^2$ for any $n, m \ge 1$?

While struggling to evaluate this sum I tried evaluating a simpler sum by setting $E_1 = 0, E_2 = 0$:

$$\sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \frac{m^2}{(n^2 + m^2)^4}$$

I noticed that:

$$\sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \frac{m^2}{(n^2 + m^2)^4} = \sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \frac{n^2}{(n^2 + m^2)^4}$$, so

$$2 \sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \frac{m^2}{(n^2 + m^2)^4} = \sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \frac{m^2}{(n^2 + m^2)^4} + \sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \frac{n^2}{(n^2 + m^2)^4} = \sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \frac{n^2 + m^2}{(n^2 + m^2)^4} = \sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \frac{1}{(n^2 + m^2)^3}$$

But I still have no idea how to evaluate even $\sum\limits_{n = 1}^\infty \sum\limits_{m = 1}^\infty \frac{1}{(n^2 + m^2)^3}$. I tried using Wolfram Mathematica, and it just didn't give anything after 10 minutes of computation.

It is known we can evaluate infinite sums (indexed by one variable) of rational functions and get closed expressions in terms of the polygamma function. Is there any similar result for double sums? Is there some general method of evaluating double sums? It seems not, since evaluating even simpler sum $\frac{1}{(n^2 + m^2)^2}$ involves combinatorics, but I still hope to find something.

UPD: I was not very careful while reading this and seems the sum $\frac{1}{(n^2 + m^2)^3}$ can be evaluated the same way as for $\frac{1}{(n^2 + m^2)^2}$. However, this still leaves the question of evaluating the original sum.

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    $\begingroup$ do you absolutely want an exact solution? It looks it can be approximated using Euler-Maclaurin method, or by finding some lower/upper bounds $\endgroup$ – Alex May 21 '14 at 20:13
  • $\begingroup$ @Alex I hoped to find exact solution, but now I'm not sure it is possible :) $\endgroup$ – karlicoss May 21 '14 at 20:18
  • $\begingroup$ The sums seem to be of digamma form, and (if I'm not too wrong) approximated with $n \coth (n)$. Also try approximating with the corresponding integral. I think you should be able to get the asymptotic order. wolframalpha.com/input/…\ $\endgroup$ – Alex May 21 '14 at 20:21

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