Closed-form expression for the Gaussians on infinite lattices $\sum_{n=-\infty}^\infty e^{-\frac{(x-Ln)^2}{2\sigma^2}}$ Does this sum have a closed-form expression?
$$\sum_{n=-\infty}^\infty e^{-\frac{(x-Ln)^2}{2\sigma^2}}$$
I'm trying to find the closed-form expression for the function composed of the Gaussians on infinite lattices of which interval is $L$.
Any suggestions or advice is greatly appreciated. Thanks in advance!
 A: Expanding upon Graham Hesketh's comment, we can solve this in terms of the Jacobi theta function.
The original expression is:
$$\sum_{n=-\infty}^{\infty} e^{-\frac{(x-Ln)^2}{2\sigma^2}}$$
The Jacobi theta function can be defined by:
$$\vartheta(z;\tau) = \sum_{n=-\infty}^{\infty}e^{\pi in^2\tau + 2\pi inz }$$
First we'll expand the original expression a bit.
$$\sum_{n=-\infty}^{\infty} e^{-\frac{(x-Ln)^2}{2\sigma^2}} = 
\sum_{n=-\infty}^{\infty} e^{-\frac{x^2 - 2Lnx + L^2n^2}{2\sigma^2}}$$
$$= \sum_{n=-\infty}^{\infty} e^{-\frac{x^2}{2\sigma^2}} e^{-\frac{L^2n^2}{2\sigma^2} + \frac{Lnx}{\sigma^2}}$$
$$= e^{-\frac{x^2}{2\sigma^2}} \sum_{n=-\infty}^{\infty} e^{-\frac{L^2n^2}{2\sigma^2} + \frac{Lnx}{\sigma^2}}$$
Now equating terms, first for $\tau$:
$$\pi i n^2\tau = -\frac{L^2n^2}{2\sigma^2}$$
$$\tau = -\frac{L^2}{2\pi i \sigma^2}$$
And for $z$:
$$2\pi i nz = \frac{Lnx}{\sigma^2}$$
$$z = \frac{Lx}{2\pi i \sigma^2}$$
Combining all of the parts:
$$\sum_{n=-\infty}^{\infty} e^{-\frac{(x-Ln)^2}{2\sigma^2}} =
e^{-\frac{x^2}{2\sigma^2}}\vartheta\left(\frac{Lx}{2\pi i \sigma^2};\frac{L^2}{2\pi i \sigma^2}\right)$$
