Use Poisson summation formula to prove Gaussian sum formula

Question

The Poisson summation formula states that for any Schwartz function $f$, $\sum\limits_{k\in\mathbb{Z}}f(k)=\sum\limits_{k\in\mathbb{Z}}\hat{f}(k)$, where $\hat{f}$ is the Fourier transform of $f$. The question that I am trying to solve asks me to use this fact to show that $G(t)=\sum\limits_{k\in\mathbb{Z}}e^{-\pi tk^2}$ satisfies $G(t)=t^{-1/2}G(t^{-1})$ and in particular $G(t)\sim t^{-1/2}$ as $t\to 0$ (i.e. $\lim\limits_{t\to 0} t^{1/2}G(t)=1$). I tried computing the Fourier transform of $e^{-\pi tk^2}$ but I can't seem to go anywhere. I'm fairly certain that this is supposed to be an easy question but I just can't seem to find my way to the solution. Any help would be much appreciated.

Answer 1

I'm going to assume that you are using the normalization $\widehat{f}(s) := \int_{-\infty}^{\infty} f(x) e^{-2 \pi i s x} dx$ (otherwise, the desired identity will have a $(2\pi)^{-1/2}$ floating around in it somewhere). One can show that if $f(x) = e^{-\pi t x^2}$, then $$ \widehat{f}(s) = \frac{1}{\sqrt{t}}e^{-\pi s^2 / t}. $$ If you don't remember how this is shown or haven't seen this trick before, it is well explained here (though be careful - the normalization in this link is slightly different from the above).

Next, we will apply the Poisson Summation Formula to the function $G(t)$, which gives that $$ G(t) = \sum_{k \in \mathbb{Z}} e^{-\pi t k^2} = \sum_{k \in \mathbb{Z}} \frac{1}{\sqrt{t}} e^{-\pi k^2 / t} = t^{-1/2} \sum_{k \in \mathbb{Z}} e^{-k^2 /t} = t^{-1/2} G(t^{-1}), $$ as desired.