Given that $\vartheta(x) = \sum_{n = -\infty}^\infty e^{-\pi n^2 x}$, I am trying to finish a derivation that $\vartheta(x) = \frac{1}{\sqrt{x}}\vartheta(1/x)$.
I believe that I am very close. I have derived that $\int_{-\infty}^\infty e^{-\pi x^2 - 2\pi i x\xi} dx = e^{-\pi \xi^2}$, which means that $e^{-\pi x^2}$ is its own Fourier transform.
Using the Poisson Summation Formula, $\sum_{n= -\infty}^{\infty} f(n) = \sum_{n= -\infty}^{\infty} \hat{f}(n)$, I am hoping to conclude the desired transformation.
My complex-analysis text tells me it is supposed to be a simple u-substitution with $u\sqrt{t} = x$. When I do that then $\sqrt{t} du = dx$ and so $$ \frac{e^{-\pi \xi^2}}{\sqrt{t}} = \int_{-\infty}^\infty e^{-\pi tu^2}e^{-2\pi i (\sqrt{t}u)\xi} du $$ but this doesn't tell me anything useful besides that I do not have the right answer. Any help is appreciated!
Let $f(t) = e^{-\pi t^2}$. Then $\hat{f}(\xi) = e^{-\pi \xi^2}$. Hence, by Poisson's summation formula,
$$\sum_{n = -\infty}^\infty e^{-\pi n^2 x} = \sum_{n = -\infty}^\infty f(n\sqrt{x}) = \sum_{n = -\infty}^\infty \frac{1}{\sqrt{x}}\hat{f}\left(\frac{n}{\sqrt{x}}\right) = \frac{1}{\sqrt{x}} \sum_{n = -\infty}^\infty e^{-\pi n^2/x} = \frac{1}{\sqrt{x}}\vartheta\left(\frac{1}{x}\right).$$
Therefore,
$$\vartheta(x) = \frac{1}{\sqrt{x}} \vartheta\left(\frac{1}{x}\right).$$