Closed-form expression for the Gaussians on infinite lattices $\sum_{n=-\infty}^\infty -\frac{x-Ln}{\sigma^2}e^{-\frac{(x-Ln)^2}{2\sigma^2}}$

Question

Does there exist a closed-form expression for this sum?

$$\sum_{n=-\infty}^\infty -\frac{x-Ln}{\sigma^2}e^{-\frac{(x-Ln)^2}{2\sigma^2}}$$

I need to find the closed-form expression for the function composed of the Gaussian derivatives on infinite lattices of which interval is $L$.

Any help is appreciated. Thanks in advance!

Answer 1

As @lcv commented $$S=\sum_{n=-\infty}^\infty e^{-\frac{(x-Ln)^2}{2\sigma^2}}=\frac{\sqrt{2 \pi }\, |\sigma |}{|L|}\,\vartheta _3\left(-\frac{\pi x}{L},e^{-\frac{2 \pi ^2 \sigma ^2}{L^2}}\right)$$ $$T=-\sum_{n=-\infty}^\infty \frac{(x-Ln)}{\sigma^2}e^{-\frac{(x-Ln)^2}{2\sigma^2}}=\frac {\partial S}{\partial x}$$ $$T=\frac {\partial S}{\partial x}=-\frac{\pi\sqrt{2\pi} |\sigma |}{L |L|}\vartheta _3^{\prime }\left(-\frac{\pi x}{L},e^{-\frac{2 \pi ^2 \sigma ^2}{L^2}}\right)$$

I do not think about any other possibility.