Exponential series property: $\alpha (z)=\sum_{n=-\infty }^{\infty }e^{-zn^{2}\pi}$

Question

Good day, please a dude, show that if $$\alpha (z)=\sum_{n=-\infty }^{\infty }e^{-zn^{2}\pi}$$ then $\alpha(z^{-1})=z^{\frac{1}{2}}\alpha (z)$ for $\Re(z)>0$.

I thought for properties of $e$, or maybe dividing the sum in 2 parts when $n = 0$?

Thanks.

Answer 1

Hint. One may apply the Poisson summation formula

$$ \sum_{n=-\infty}^\infty f(n)=\sum_{n=-\infty}^\infty \hat f\left(n\right) $$

with $f(u)=e^{-z\pi u^2}$, observing that $f$ is continuous and satisfies $|f(u)|\le\frac C{\left(1+|u|\right)^2}$ as $u \to \pm \infty$ and observing that, by the use of the gaussian result, $$ \hat f\left(t\right)=\int_{-\infty}^{+\infty}e^{-itu}f(u)\:du=\int_{-\infty}^{+\infty}e^{-itu}e^{-z\pi u^2}\:du=\frac{e^{\large -\frac{t^2}{4\pi z}}}{\sqrt{z}},\quad \Re z>0. $$