I have a summation of the form:
$$\sum_{m\in\mathbb{Z}}m\exp\left\{-\frac{B}{2}\left(z-\frac{2\pi iv}{B}-\frac{2\pi m}{BL_2}\right)^2\right\}$$ where $B, v, L_2$ are constants and $z$ is a variable. I've been trying to look for a solution via Gauss sums, but have found anything of help yet (my background is in theoretical physics so I have to read and learn number theory as I go). I was maybe thinking about using a Fourier expansion of a function which would produce the series above, but haven't found anything yet...
Any suggestions?
This is related to a derivative of a Jacobi theta-function: if we write $$ \theta(z,\tau) = \sum_{m = -\infty}^{\infty} \exp{(\pi i m^2 \tau + 2\pi i n z)}, $$ then $$ \sum_{m = -\infty}^{\infty} \exp{\left(-\frac{B}{2}(z-2\pi i \nu/B)^2+\frac{2\pi m}{L_2}(z-2\pi i \nu/B) - \frac{2\pi^2}{BL_2^2}m^2 \right)} = e^{-B(z-2\pi i \nu/B)^2/2} \theta\left(\frac{z-2\pi i \nu/B}{iL_2},\frac{2\pi i}{BL_2^2}\right), $$ and then your sum is $$ e^{-B(z-2\pi i \nu/B)^2/2} \frac{L_2}{2\pi}\frac{\partial}{\partial z}\theta\left(-i\frac{z-2\pi i \nu/B}{L_2},\frac{2\pi i}{BL_2^2}\right) $$ (Notice that the derivative brings down a $2\pi m/L_2$ inside the sum.) There is in general no way to write this more simply.