# Proving ∑∞n=−∞e−πn2=4√πΓ(34)\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}

Wikipedia informs me that

I tried considering $f(x,n) = e^{-x n^2}$ so that its Mellin transform becomes $\mathcal{M}_x(f)=n^{-2z} \Gamma(z)$ so inverting and summing

However, this last integral (whose integrand has poles at $z=0,\frac{1}{2}$ with respective residues of $-\frac 1 2$ and $\frac 1 2$) is hard to evaluate due to the behavior of the function as $\Re(z)\to \pm\infty$ which makes a classic infinite contour over the entire left/right plane impossible.

How does one go about evaluating this sum?

This one is a direct evaluation of elliptic integrals. Jacobi’s theta function $\vartheta_{3}(q)$ is defined via the equation Let $0 < k < 1$ and $k' = \sqrt{1 - k^{2}}$ and we define elliptic integrals $K, K'$ via Then it is almost a miracle that we can get $k$ in terms of $K, K'$ via the variable $q = e^{-\pi K'/K}$ using equations where $\vartheta_{2}(q)$ is another theta function of Jacobi defined by Also the function $\vartheta_{3}(q)$ is directly related to $K$ via The proofs of $(3)$ and $(5)$ are given in the linked post on my blog.
The sum in the question is $\vartheta_{3}(e^{-\pi})$ so that we have $q = e^{-\pi}$. This implies that $K'/K = 1$ so that $k = k'$ and from $k^{2} + k'^{2} = 1$ we get $k^{2} = 1/2$. And then Now using $\Gamma(1/4)\Gamma(3/4) = \pi/\sin(\pi/4) = \pi\sqrt{2}$ we get The value of $K = K(1/\sqrt{2})$ in terms of $\Gamma(1/4)$ is evaluated in this answer.