How to sum Gaussian function on a grid? 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.
 A: 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.)
