# Proving $\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt \pi}{\Gamma\left(\frac 3 4\right)}$

Wikipedia informs me that

$$S = \vartheta(0;i)=\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$$

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

$$\frac{1}{2}(S-1)=\sum_{n=1}^\infty f(\pi,n)=\sum_{n=1}^\infty \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}n^{-2z} \Gamma(z)\pi^{-z}\,dz = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\zeta(2z) \Gamma(z) \pi^{-z}\,dz$$

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?

#### Solutions Collecting From Web of "Proving $\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt \pi}{\Gamma\left(\frac 3 4\right)}$"

This one is a direct evaluation of elliptic integrals. Jacobi’s theta function $\vartheta_{3}(q)$ is defined via the equation $$\vartheta_{3}(q) = \sum_{n = -\infty}^{\infty}q^{n^{2}}\tag{1}$$ Let $0 < k < 1$ and $k’ = \sqrt{1 – k^{2}}$ then we define elliptic integral $$K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 – k^{2}\sin^{2}x}}, K = K(k), K’ = K(k’)\tag{2}$$ 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 $$k = \frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}\tag{3}$$ where $\vartheta_{2}(q)$ is another theta function of Jacobi defined by $$\vartheta_{2}(q) = \sum_{n = -\infty}^{\infty}q^{(n + (1/2))^{2}}\tag{4}$$ Also the function $\vartheta_{3}(q)$ is directly related to $K$ via $$\vartheta_{3}(q) = \sqrt{\frac{2K}{\pi}}\tag{5}$$ 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 $$\vartheta_{3}(q) = \sqrt{\frac{2K}{\pi}} = \sqrt{\frac{2}{\pi}\cdot\frac{\Gamma^{2}(1/4)}{4\sqrt{\pi}}} = \frac{\Gamma(1/4)}{\pi^{3/4}\sqrt{2}}$$ Now using $\Gamma(1/4)\Gamma(3/4) = \pi/\sin(\pi/4) = \pi\sqrt{2}$ we get $$\sum_{n = -\infty}^{\infty}e^{-\pi n^{2}} = \vartheta_{3}(e^{-\pi}) = \frac{\sqrt[4]{\pi}}{\Gamma(3/4)}$$ The value of $K = K(1/\sqrt{2})$ in terms of $\Gamma(1/4)$ is evaluated in this answer.

see: Ramanujan’s Notebooks Volume 3, Chapter 17, Example(i). pp 103.

see also: Ramanujan’s Notebook Volume 5 chapter 35. Values of Theta-Functions P. 325.

(seems like many of the previous comments mention what I have.)

I am not sure if it will ever help, but the following identity can be proved:

$$S^2 = 1 + 4 \sum_{n=0}^{\infty} \frac{(-1)^n}{\mathrm{e}^{(2n+1)\pi} – 1}.$$

Maybe you can use this relationship:

If $$\vartheta(x)=\sum_{n\in \mathbb{Z}}e^{-\pi n^2 x},$$

then:

$$\pi^{-s/2}\Gamma(s/2)\zeta(s)=\int_{0}^{\infty}x^{s/2-1}\frac{\vartheta(x)-1}{2}dx.$$