Can anybody help to tell how to sum $$\sum_{-\infty}^{\infty} \sum_{-\infty}^{\infty} e^{-\frac{x^2+y^2}{2}}$$ in other words I want to sum $e^{-\frac{x^2+y^2}{2}}$ on all the integer coordinate pairs of an infinite grid. Sorry for bothering, and I have done some "homework", how to calculate the similar two-dimensional integral is understood. Thanks in advance.
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2Do you have some reason to think a closed form exists? That would surprise me. You might note that since $e^{a+b}=e^ae^b$ your sum is equal to $(\sum_{n=-\infty}^\infty e^{-n^2/2})^2$. – David C. Ullrich Jan 22 '16 at 15:19
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Thank you very much for the quick reply. I have noticed that but still have no clue for $\sum_{-\infty}^{\infty} e^{-\frac{n^2}{2}}$. The original term should be equal to $2 \pi$, if I am not wrong. – user307181 Jan 22 '16 at 15:31
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You might answer the question I asked! Do you have some reason to think that a closed form for this sum exists? Like an exercise in some book asked you to calculate the sum or something? (I have no idea what you mean by "the original term"...) – David C. Ullrich Jan 22 '16 at 15:39
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(Sorry I didn't study math in university, had another subject) I need this to evaluate some kernel-based estimator on a (rectangular) grid. By computer calculation it ("the original term", that is, the two-dimensional sum in the question) seems to converge to $2 \pi$. Thank you. – user307181 Jan 22 '16 at 15:43
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ANSWER THE QUESTION! Do you have some reason to think that a closed-form expression for the sum exists? Hint: the answer is yes or no. Hint: You know the answer to this question. (If your comment about the computer calculation was meant as an answer to this question: That's no evidence at all - my guess is that there simply is no closed form for the sum.) – David C. Ullrich Jan 22 '16 at 15:44
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Thank you for your hint. I knew calculation is no evidence, that is why I am here. On the other hand, if the machine calculation "converges" to some value, it appears to me that there is a closed term exists. I find another similar question somewhere else, I need to do some other "homework" first. – user307181 Jan 22 '16 at 16:04
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I doubt that a closed form exists. If the sum is in fact not $2\pi$ it's very likely that you can prove numerically that it's not $2\pi$ - see the answer I posted. – David C. Ullrich Jan 22 '16 at 16:06
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It's 6.28318537..., so it's more than $2 \pi$ (calculated with Mathematica) – Paul Jan 22 '16 at 16:31
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1It's so close to $2 \pi$ because if you integrate instead of sum you get exactly $2 \pi$: $\int_{-\infty}^\infty e^{-x^2/2} \mathrm{d}x = \sqrt{2\pi}$ – Paul Jan 22 '16 at 16:43
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1@Paul It's clear the sum is an approximation to the integral. One reason it's closer than you might expect is that the integrand is alternately convex and concave, making the terms in the sum alternately too small and too large – David C. Ullrich Jan 22 '16 at 19:33
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The search term is "theta function". – Gerry Myerson Jan 22 '16 at 22:15
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@GerryMyerson That's the search term all right. Do you find a closed form for the sum via that search? I don't, but knowing nothing about these things it could be I saw it but didn't recognize it as such. – David C. Ullrich Jan 22 '16 at 23:27
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@David, not having done the search, I don't know what one would find. But I suspect that if there is a closed form, that search would find it. – Gerry Myerson Jan 23 '16 at 04:07
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I doubt that a closed form exists. You say you've done computer calculations that suggest the sum is $2\pi$. It's impossible to show that the sum is $2\pi$ by just calculating, but here's a way that could work to show the sum is not exactly $2\pi$ using a computer (Edit: It turns out that the sum is greater than $2\pi$, which is almost too bad, because it's clear how that could be established numerically; the interesting part of the argument below is showing how we could have established numerically that the sum was less than $2\pi$):
First note that saying the sum is $2\pi$ is the same as saying that $\sum_{-\infty}^\infty e^{-n^2/2}=\sqrt{2\pi}$. Now
$$\sum_{|n|>N}e^{-n^2/2}\le2\int_N^\infty e^{-t^2/2}dt\le\frac2N\int_{N}^\infty te^{-t^2/2}dt=\frac2Ne^{-N^2/2}.$$ So $$\sum_{n=-N}^Ne^{-n^2/2}\le\sum_{n=-\infty}^\infty e^{-n^2/2} \le\sum_{n=-N}^Ne^{-n^2/2}+\frac2Ne^{-N^2/2}.$$ So if you find an $N$ with $\sum_{n=-N}^Ne^{-n^2/2}>\sqrt{2\pi}$ then the sum is greater than $\sqrt{2\pi}$, obviously, and perhaps not so obviously if you find an $N$ with $\sum_{n=-N}^Ne^{-n^2/2}+\frac2Ne^{-N^2/2}<\sqrt{2\pi}$ then the sum is less than $\sqrt{2\pi}$. (My money's on the latter.)
David C. Ullrich
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The original term (for which I had the numerical evaluation, which seemed to converge) was complicated than which I posted in question. The value $2 \pi$ is the result of reversing the simplification. Maybe I made a mistake here. I studed your answer roughly. I need to find out where I made my mistake. Thank you very much indedd for your time and patience. – user307181 Jan 22 '16 at 16:15
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Can this be performed in the Fourier domain? The Fourier transform of a Gaussian is a Gaussian (of course), and sampling function is a two-dimensional comb function. Multiplication in one domain is convolution in the transform domain. – David G. Stork Jan 22 '16 at 23:07
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@DavidG.Stork If you work that out you're led to the "Poisson summation formula", which transforms our sum into another sum of more or less the same sort. I don't see how it helps in finding a closed form for the sum. Nor for getting numerical estimates, in this case. If $a>0$ is small then the series $\sum e^{-an^2/2}$ converges very slowly, and Poisson summation gives that sum in terms of $\sum e^{-bn^2/2}$ where $b$ is large, so the new sum converges much faster. But here $a$ is just medium-sized and so $b$ is going to be medium-sized as well. – David C. Ullrich Jan 22 '16 at 23:22
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Yes... I suppose you're right. In this case the Fourier Transform of a Gaussian is a Gaussian and the Fourier Transform of a Dirac Comb is a Dirac Comb, so Fourier doesn't help us here. Oh well. – David G. Stork Jan 22 '16 at 23:28
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@DavidG.Stork Too bad the question wasn't about $\sum1/(1+n^2)$ - your comment would have saved the day. – David C. Ullrich Jan 22 '16 at 23:30