Proving the functional equation $\theta (x) = x^{-\frac{1}{2}} \theta (x^{-1})$ from the Poisson summation formula

We have the relationship $\theta (x) = x^{-\frac{1}{2}} \theta (x^{-1})$

Now I know one uses the Poisson summation formula to prove this. The Poisson summation formula comes from Fourier Transform and series and satisfies, \begin{equation} \sum_{n=-\infty}^{+\infty} f(n) = \sum_{n=-\infty}^{+\infty} \hat f (n) \end{equation}

So do we say $f(n)= \theta(x)$? (where $\theta(x) = \sum_{-\infty}^\infty e^{-\pi n^2 x}$)

So filling in we have, \begin{equation} \sum_{n= - \infty}^{+\infty} e^{-\pi n^2 x}= \sum_{m= - \infty}^{+\infty} \int_{- \infty}^{\infty} e^{-\pi y^{2}x}e^{-2 \pi i y m} dy \end{equation}

Now what is the best way to keep going? Do I rearrange with the goal of getting a Gaussian integral (and hence the relationship)? Or is there a more elegant way of achieving the desired result?

What I've done so far is, after completing the square in the power of the integrand, we have \begin{equation} =\sum_{m=-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-\pi x((y+i\frac{m}{x})^{2}-i^{2} \frac{m^{2}}{x^{2}})}dy \end{equation}

Then separating the integrand, breaking it up into a product of exponentials again we have,

\begin{equation} =\sum_{m=-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-\pi m^{2} \frac{1}{x}} e^{-\pi x (y+i \frac{m}{x})^{2}}dy \end{equation}

The left factor of the integrand is a constant as it does not contain $y$, so it can be brought outside of the integral.

$$ \sum_{n = -\infty}^{+\infty} e^{-\pi n^{2} x} = \sum_{m=-\infty}^{+\infty}e^{- \pi m^{2} \frac{1}{x}} \int_{- \infty}^{+ \infty} e^{ -\pi x (y+i \frac{m}{x})^{2}} dy $$

What would be the correct approach from here? Should I make a substitution? Any help or suggestions will be appreciated!

Taking $u=t\sqrt{x} $ we have $$\int_{-\infty}^{\infty}e^{-\left(\pi t^{2}/x\right)-2\pi int}dt=\sqrt{x}\int_{-\infty}^{\infty}e^{-\pi u^{2}-2\pi in\sqrt{x}u}du=\sqrt{x}\int_{-\infty}^{\infty}e^{-\pi u^{2}-2\pi in\sqrt{x}u}du= $$ $$=\sqrt{x}\int_{-\infty}^{\infty}e^{-\pi\left(u+in\sqrt{x}\right)^{2}+\pi n^{2}x}du=\sqrt{x}e^{\pi n^{2}x}\int_{-\infty}^{\infty}e^{-\pi\left(u+in\sqrt{x}\right)^{2}}du. $$ Now observe that $$\int_{-\infty}^{\infty}e^{-\pi\left(u+in\sqrt{x}\right)^{2}}du=\int_{-\infty}^{\infty}e^{-\pi v^{2}}dv=2\int_{0}^{\infty}e^{-\pi v^{2}}dv $$ and using the substitution $s=\pi v^{2} $ $$=\pi^{-1/2}\int_{0}^{\infty}e^{-s}s^{-1/2}ds=\pi^{-1/2}\Gamma\left(\frac{1}{2}\right)=1 $$ hence $$\sum_{m=-\infty}^{\infty}e^{-\pi m^{2}/x}=\lim_{N\rightarrow\infty}\sum_{n=-N}^{N}\int_{-\infty}^{\infty}e^{-\left(\pi t^{2}/x\right)-2\pi int}dt=\lim_{N\rightarrow\infty}\sum_{n=-N}^{N}\sqrt{x}e^{\pi n^{2}x}= $$ $$=\sqrt{x}\sum_{n=-\infty}^{\infty}e^{\pi n^{2}x}. $$