if $w(x)=\sum \limits_{n=1}^\infty e^{-n^2\pi x}$ then how to show that $\frac {1+2w(x)} {1+2w(\frac{1}{x})}=\frac{1}{\sqrt{x}}$

$w(x)=\sum \limits_{n=1}^\infty e^{-n^2\pi x}$

$\frac {1+2w(x)} {1+2w(\frac{1}{x})}=\frac{1}{\sqrt{x}}$

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.

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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.)

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This is a standard identity from the theory of theta functions. It is usually proved by the Poisson summation formula, but you can find another proof in Riemann's paper on primes. It is used to establish the functionnal equation of the Riemann zeta function. For a more modern reference, check Montgomery & Vaughan's book on multiplicative number theory.

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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).

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