# 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...

• I don't know how to handle the denominator yet, but maybe this identity is a start. I can transform it to $$\sum_{k=1}^{\infty}\frac{\sin\left(k\alpha\right)\sin\left(k\beta\right)}{k^{2}}=-\frac{1}{8}\left(\alpha+\beta\right)\left(\alpha+\beta-2\pi\right)$$ and hopefully it is related to your double sum. – Ian Mateus Apr 21 '13 at 23:17
• To @IanMateus: It certainly seems like a start. Thank you very much – PML Apr 21 '13 at 23:18
• Warning: the identity I posted in the comment is probably wrong, give me a second and I'll correct it. – Ian Mateus Apr 21 '13 at 23:19
• Interestingly, I got $\alpha(\pi-\beta)/2$ for the sum. It is too simple to be right! It requires some restrictions in the domain, though. – Ian Mateus Apr 21 '13 at 23:40
• @IanMateus I feel dumb. I only got a sum of Dilogarithms... But if correct it may be a good start. Thank you – PML Apr 21 '13 at 23:46

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.

• This is just a hunch. I believe the sum converges everywhere except $(x,y) \neq (0,0)$. The hunch is based on generalized alternating test. – user17762 Apr 21 '13 at 23:26
• @user17762: if you can sort a a reason why my cosh term is wrong, please speak up. But I do not see a way around it. – Ron Gordon Apr 21 '13 at 23:29
• The double sum is expected to diverge at $(x,y)\to(0,0)$. But if diverges the way you say that will come as a surprise to me. I'll take a look at your proof. Thank you – PML Apr 21 '13 at 23:29
• @PML: I fixed and it agrees numerically. – Ron Gordon Apr 21 '13 at 23:58
• @PML: sum from 1 to infinity = (1/2) (sum from -infinity to infinity - (n=0)). The $1/m^2$ term is the $n=0$ term. – Ron Gordon Apr 22 '13 at 0:41

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.

• You might be right about funny behavior at the zeroes of the cosines - this is to be expected because, as you point out and I mention in my solution, the sum diverges without any oscillations. I will amend my convergence statement - thanks. – Ron Gordon Apr 22 '13 at 14:11

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.