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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...

PML
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  • 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
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    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
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    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
  • This sum isn't absolutely convergent, so order of summation matters. Before we spend too much time on it, is this the order you want to sum it in? – David E Speyer Apr 22 '13 at 00:40
  • @DavidSpeyer Well that's wierd. The order of summation shouldn't matter (maybe it helps to know that the domain of $x$ and $y$ is the same). – PML Apr 22 '13 at 00:45

3 Answers3

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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.

Ron Gordon
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  • 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. –  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
  • @user17762: my fix works out numerically – Ron Gordon Apr 21 '13 at 23:58
  • @RonGordon I'm sorry for my dumbness (I believe it's from the lateness of the hour...) but I don't understand from where the term $-1/m^2$ comes from... – PML Apr 22 '13 at 00:35
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    @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 00:41
  • Something must be slightly wrong here; the sum definitely diverges if $x$ and $y$ are of the form $2 \pi k$. – David E Speyer Apr 22 '13 at 13:48
  • @RonGordon: Sorry, your first sum does not look correct to me. E.g. for $x=\pi$ it evaluates exactly to $\frac{\pi}{m\sinh\pi m}$, which does not coincide with your expression (though it can well be numerically close, especially for large $m$). The reason is that when you rewrite the sum as a contour integral over $C_+\cup C_-$ where $C_+$ runs slightly above and $C_-$ slightly below the real axis, you cannot simply remove these contours to infinity - since $\cos zx$ (or $e^{izx}$, whatever) will not allow to apply Jordan's lemma to at least one of them. – Start wearing purple Apr 22 '13 at 16:30
  • O.L.: You caught me. Actually, I am a bit surprised by this, but you are correct. That said, whatever errors are present are exponentially small - the numerics I did verifies this. So while we need to find something exactly analytical - and I suspect that it involves nasty hypergeometrics of complex argument - what I have is not bad over the range of values used here. – Ron Gordon Apr 22 '13 at 16:49
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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.

David E Speyer
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  • 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
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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.