Showing $\sum_{n=-\infty}^{\infty}\exp\left(-\pi an^2+2\pi ibn\right)=a^{-\frac{1}{2}}\sum_{m=-\infty}^{\infty}\exp\left(-\frac{\pi(m-b)^2}{a}\right)$

Question

How do I show that

\begin{align} \sum_{n=-\infty}^{\infty} \exp\left(-\pi a n^2 + 2 \pi i bn\right) = a^{-\frac{1}{2}} \sum_{m=-\infty}^{\infty} \exp\left(-\frac{\pi(m-b)^2}{a}\right) \end{align}

is a valid identity?

It is quite a long shot, @MhenniBenghorbal. Such identity can be seen as a consequence of the residue theorem, too, but Poisson summation formula is the standard way. — Jack D'Aurizio, Jan 22 '15 at 19:03

score 2 · Accepted Answer · answered Jan 22 '15 at 18:16

It is a consequence of Poisson summation formula. You just have to prove that, if $$ f(x) = \exp\left(-\pi a x^2+2\pi i b x\right), $$ then its Fourier transform is: $$ \widehat{f}(s) = \frac{1}{\sqrt{2\pi a}}\exp\left(-\frac{(2b\pi+s)^2}{4a\pi}\right).$$

Showing $\sum_{n=-\infty}^{\infty}\exp\left(-\pi an^2+2\pi ibn\right)=a^{-\frac{1}{2}}\sum_{m=-\infty}^{\infty}\exp\left(-\frac{\pi(m-b)^2}{a}\right)$

1 Answers1