TITLE: if $w(x)=\sum \limits_{n=1}^\infty e^{-n^2\pi x}$ then how to show that $\frac {1+2w(x)} {1+2w(\frac{1}{x})}=\frac{1}{\sqrt{x}}$
QUESTION [6 upvotes]: $w(x)=\sum \limits_{n=1}^\infty e^{-n^2\pi x}$ 
$\frac {1+2w(x)} {1+2w(\frac{1}{x})}=\frac{1}{\sqrt{x}}$  
I wonder how can be proved such a beautiful relation as shown in wolfram page 
I need to learn which technics be used to prove such relation above.
Thanks for answers and links.

REPLY [7 votes]: We set the theta function
$$\theta(\tau)=\sum_{n\in\mathbb{Z}} e^{-\pi n^2\tau}=2w(\tau)+1.$$
Write the Gaussian $f(t)=e^{-\pi t^2}$ so that $f=\hat{f}$. We also write $h(x)=f(x\sqrt{\tau})$, so that we have the rescaled (ordinary, unitary) Fourier transform $\hat{h}(x)=\tau^{-1/2}\hat{f}(x\tau^{-1/2})$. Thus
$$\theta(\tau)=\sum_{n\in\mathbb{Z}} h(n)=\sum_{n}\hat{h}(n)=\sum_n \frac{1}{\sqrt{\tau}} \hat{f}\left(\frac{n}{\sqrt{\tau}}\right)=\frac{1}{\sqrt{\tau}}\theta\left(\frac{1}{\tau}\right),$$
by the Poisson summation formula applied to $h$.

The functional equation for the theta function is the canonical route to deriving the functional equation and analytic continuation of the Riemann zeta function. Viz.
$$\xi(s):=\pi^{-s/2}\Gamma(s/2)\zeta(s)=\sum_{n=1}^\infty \int_0^\infty \left(\frac{x}{\pi n^2}\right)^{s/2}e^{-x}\frac{dx}{x} $$ $$ = \int_0^\infty x^{s/2} \left(\sum_{n=1}^\infty e^{-\pi n^2}\right)\frac{dx}{x} =\int_0^\infty x^{s/2} \frac{\theta(x)-1}{2}\frac{dx}{x}
.$$
With the functional equation, we find that $\xi(s)+s^{-1}+(1-s)^{-1}$ is symmetrical under $s\mapsto 1-s$, whence we have $\xi(s)=\xi(1-s)$. (See these notes or these notes or this article for a little more.)

REPLY [5 votes]: Hint: As Bruno said, this follows by the Poisson Summation Formula:
$$
\sum_{n \in \mathbb{Z}} g(n) = \sum_{m \in \mathbb{Z}} \widehat{g}(m)
$$
(assuming some conditions on the function $g$ which your function definitely enjoys).  Note that to properly apply Poisson Summation, you should probably write
$$
\sum_{n = 1}^{\infty} e^{-n^2 \pi x} = \frac{1}{2} \sum_{n \in \mathbb{Z}} e^{-n^2 \pi x } - \frac{1}{2} := \frac{1}{2} f(x) - \frac{1}{2}.
$$
Thus, $f(x) = 2 w(x) + 1$.  It is straight-forward to find $\widehat{f}(m)$ using the fact that $\int_{\mathbb{R}} e^{- \pi y^2} dy = 1$.  I will leave you to fill out the details (which are quite straight-forward).