TITLE: Insight about $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{\cos(nx)\cos(my)}{n^2+m^2}$
QUESTION [10 upvotes]: 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...

REPLY [5 votes]: 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.