$\sum_{n=-\infty}^\infty e^{-\alpha n^2+\beta n}$ Hi I am trying to calculate the sum given by
$$
\sum_{n=-\infty}^\infty e^{-\alpha n^2+\beta n}=\ = \sqrt{\frac{\pi}{\alpha}} e^{\beta^2/(4\alpha)} \vartheta_3\big(-\frac{\pi\beta}{2\alpha},e^{-\pi^2/(2\alpha)}\big),\quad \alpha>0,\beta \in \mathbb{R}.
$$
I know we can express this in terms of elliptic theta functions as shown.  But I am looking for a method to prove this.  Thanks.  I haven't studied summations in 30 years, however I am looking for methods possibly using Poisson Summation formula, residue analysis, etc.
Thanks. 
 A: Let $f(x) = e^{-a x^2 + b x}$.
Using the Poisson summation formula (the sums and integrals run over all integers/reals):
$$\sum \limits_{n} f(n)= \sum \limits_{n} \hat{f}(n) = \sum \limits_{n} \int f(t) e^{-2\pi i n t} dt \\
= \sum \limits_{n} \int e^{-a \left( t-\frac{b-2\pi in}{2a} \right)^2}e^{\frac{(b-2\pi in)^2}{4a}} dt\\
$$ (completing the square) 
$$
=\sum \limits_{n} e^{\frac{(b-2\pi in)^2}{4a}} \int e^{-a \left( t-\frac{b-2\pi in}{2a} \right)^2} dt \\
= \sqrt{\frac{\pi}{a}}\sum \limits_{n} e^{\frac{(b-2\pi in)^2}{4a}} 
$$
Here I used the general Gaussian integral, a nontrivial result from complex analysis. A relevant question is this one.
So we obtain
$$
\sum \limits_{n} f(n) =
\sqrt{\frac{\pi}{a}} e^{\frac{b^2}{4a}}  \sum \limits_{n} e^{\frac{-\pi^2 n^2}{a}} e^{\frac{-i \pi n b}{a}} = \sqrt{\frac{\pi}{a}} e^{\frac{b^2}{4a}} \vartheta_3 \left(-\frac{\pi b}{2a},e^{-\pi^2/a}\right) $$
as desired.
The last line follows by comparing our expression with Mathematica's definition 
$ \vartheta_3(u,q) = \sum \limits_{n} q^{n^2} e^{2 i n u}.$ This seems to differ  somewhat from your answer, but that is probably a difference in definitions.
