$\sum_{n=-\infty}^\infty e^{-\alpha n^2+\beta n}$

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.

Solution 1:

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.