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...
 A: 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.  
A: The point of this answer is just to elaborate on what I wrote in the comments, that this sum isn't absolutely convergent. Intuitively, $|\cos (n x) \cos(m y)|$ is spread out equally between $0$ and $1$. If we approximate it as a constant $c>0$, then the sum is
$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{c}{m^2+n^2}.$$
We have
$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{c}{m^2+n^2} \approx \int_{t^2+u^2 \geq 1} \frac{c \ dt\  du}{t^2+u^2}= \int_{r=1}^{\infty} \frac{(\pi/2) \ c\  r\  dr}{r^2} = \frac{\pi c}{2} \int_{r=1}^{\infty} \frac{dr}{r}$$
which diverges.
To be rigorous, one would need to bound $|\cos(nx) \cos(my)|$ from below. I'll do that if you need it; but I just wanted to point out what behavior you should expect.
A: Below is just a hunch (which is a bit too long for a comment), which can probably made rigorous. We have
$$S(x,y) = \sum_{m,n} \dfrac{e^{i(mx+ny)}}{m^2+n^2} = \sum_{r} \dfrac1{r^2}\sum_{m^2+n^2=r^2} e^{i(mx+ny)}$$
I would expect $$\sum_{m^2+n^2=r^2} e^{i(mx+ny)} \approx 2 \pi r  \times e^{i f(r)}$$ such that $\sum_{r\leq R} e^{if(r)}$ is bounded independent of $R$.
Hence,
$$S(x,y) \approx 2 \pi \sum_{r} \dfrac{e^{if(r)}}r,$$which converges for $(x,y) \neq (0,0)$.
Also, note that $S(0,0)$ clearly diverges.
