question on identity of sums with exponent.

Question

I want to show that for $x>0$: $$\sum_{n=-\infty}^\infty e^{-n^2\pi x}= \frac{1}{\sqrt{x}}\sum_{n=-\infty}^\infty e^{-n^2\pi / x}$$

It doesn't seem that a simple change of variables will do, like $n\mapsto n^2 x$, so how to show this identity?

Answer 1

This is related to Jacobi Theta functions since $$\sum_{n=-\infty}^\infty e^{-n^2\pi x}=\vartheta _3\left(0,e^{-\pi x}\right)$$ $$\sum_{n=-\infty}^\infty e^{-n^2\pi / x}=\vartheta _3\left(0,e^{-\frac{\pi }{x}}\right)$$ and the function $$\Psi(x)=\frac 12 \Big(\vartheta _3\left(0,e^{-\pi x}\right)-1\Big)$$ satisfies $$\frac{1+2\Psi(x)}{1+2\Psi(\frac 1x)}=\frac 1{\sqrt x}$$ "which Jacobi attributes to Poisson and follows from the Poisson sum formula" as already mentioned by uranix (I just quote WolframMathWorld page here).