sequences and series - Insight about $sum_{m=1}^infty sum_{n=1}^infty frac{cos(nx)cos(my)}{n^2+m^2}$

Can someone give me some insight about the following double sum? I would be deeply appreciated.

$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{\cos(nx)\cos(my)}{n^2+m^2},$$

where $x,y\in[-\pi,\pi]$.

I don't even know if it converges for $(x,y)\neq(0,0)$... For the first sum Mathematica gives me some sum of Hypergeometric functions but it can't do the second one and I don't even know how to tackle this beast...

Answer

The double sum only converges when $x$ and $y$ are not multiples of $2 \pi$. To see this, evaluate the inner sum over $n$ by extending the summation range to $-\infty$ and using the residue theorem. That is, write

$$\begin{align}\sum_{n=-\infty}^{\infty} \frac{\cos{n x}}{n^2+m^2} &= -\sum \text{Res}_{z=\pm i m} \frac{\pi \cot{\pi z}\, \cos{x z}}{z^2+m^2}\\ &= \frac{\pi}{m} \text{coth}\,{\pi m}\, e^{-|m| x} + \text{exponentially small error}\end{align}$$

The double sum then takes the form

$$\frac12 \sum_{m=1}^{\infty} \left [ \frac{\pi}{m} \,e^{-m x}\text{coth}\,{\pi m} - \frac{1}{m^2}\right] \cos{m y}$$

The sum will converge unless both $x$ and $y$ are zero.