Proving $\sum_{n=-\infty}^\infty n^2e^{-\pi n^2}=\frac{\Gamma (1/4)}{4\sqrt{2}\pi^{7/4}}$

Question

I conjecture that $$\sum_{n=-\infty}^\infty n^2e^{-\pi n^2}=\frac{\Gamma (1/4)}{4\sqrt{2}\pi^{7/4}}$$ because the left-hand side and right-hand side agree to at least $50$ decimal places. Is the identity true?

This series popped up when I was trying to write down the Hadamard factorization of a certain function related to an elliptic function with a square period lattice.

The series in question is $(-1/4)$ times the second derivative of a Jacobi theta function evaluated at zero: $$\sum_{n=-\infty}^\infty n^2e^{-\pi n^2}=-\frac{1}{4}\theta_3''(0,e^{-\pi}),$$ in the notation of DLMF. I was trying to look it up in DLMF (https://dlmf.nist.gov/20.4). I found that $$\frac{\theta_3''(0,e^{-\pi})}{\theta_3(0,e^{-\pi})}=-8\sum_{n=1}^\infty \frac{e^{-(2n-1)\pi}}{(1+e^{-(2n-1)\pi})^2},$$ while $\theta_3(0,e^{-\pi})$ is known (Proving $\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$).

This translates the series in question to an alternative "Lambert-like" series (https://en.wikipedia.org/wiki/Lambert_series), but I'm unable to proceed.

Answer 1

You can reduce your problem to $\sum_{n\in \mathbb{Z}} e^{-n^2\pi}$ using the Fourier transform and Poisson summation formula. First a few observations: setting $f(x) = e^{-\pi x^2}$, we have $\hat{f}(k)=f(k)$, and also $x^2 f(x)= \frac{1}{4\pi^2}(f’’(x)+2\pi f(x))$. By the Poisson summation formula,

$$\sum_{n\in\mathbb{Z}} n^2 f(n) = \sum_{n\in\mathbb{Z}} \mathcal{F}(x^2f(x))(n)= \sum_{n\in\mathbb{Z}} \frac{1}{4\pi^2}[(2\pi i n)^2\hat{f}(n)+2\pi \hat{f}(n)] $$

Moving the first term to the other side shows that

$$ \sum_{n\in\mathbb{Z}}n^2 f(n) = \frac{1}{4\pi}\sum _{n\in\mathbb{Z}} f(n) $$

Answer 2

Let us define $$f(s) =\sum_{n\in\mathbb{Z}} e^{-\pi sn^2},s>0\tag{1}$$ then it can be proved using Poisson summation formula that $$f(1/s)=\sqrt{s}f(s)$$ Differentiating the above we get $$f'(1/s)(-1/s^2)=\frac{f(s)}{2\sqrt{s}}+\sqrt{s}f'(s)$$ and putting $s=1$ we get $$f'(1)=-\frac{f(1)}{4}\tag{2}$$ Now differentiating $(1)$ we get $$f'(s) =-\pi\sum_{n\in\mathbb{Z}} n^2e^{-\pi sn^2}$$ Thus $f'(1)=-\pi S$ where $S$ is the sum in question. And then we get $$S=\frac{f(1)}{4\pi}$$ using $(2)$. The value of $f(1)$ is well known to be $\sqrt [4]{\pi}/\Gamma(3/4)$ and hence we get the desired value of $S$.