How to prove the following transformation formula: $$ \theta(x)=\frac{1}{\sqrt{x}} \theta\left(\frac{1}{x}\right), $$ where $\theta$ is the Jacobi theta function $\theta(x)=\sum_{n\in \mathbb{Z}} e^{-\pi n^2 x}$?


1 Answer 1


One can use the Poisson summation formula: $$ \sum_{n\in\mathbb{Z}}f(n)=\sum_{k\in\mathbb{Z}}\hat{f}(k),$$ where $\hat{f}(\nu)$ denotes the Fourier transform of $f(t)$, $$ \hat{f}(\nu)=\int_{-\infty}^{\infty}f(t)e^{-2\pi i \nu t}dt.$$ Namely, setting $f(t)=e^{-\pi x t^2}$ in the above, we obtain $$\theta(x)=\sum_{n\in\mathbb{Z}}f(n)=\sum_{k\in\mathbb{Z}}\underbrace{\int_{-\infty}^{\infty}e^{-\pi x t^2-2\pi i k t}dt}_{\hat{f}(k)}= \sum_{k\in\mathbb{Z}}\frac{e^{-\pi k^2/x}}{\sqrt{x}}=\frac{\theta\left(x^{-1}\right)}{\sqrt{x}}.$$

  • $\begingroup$ How did you solve $\hat{f}(k) = \frac{e^{-\pi k^2/x}}{\sqrt{x}}$? I understand that if $g(t) = e^{-\pi t^2},$ then $\hat{g}(v) = g(v).$ The text I am referencing tells me to do a substitution, but I am having trouble. $\endgroup$
    – Nick
    Jan 10, 2015 at 6:08
  • $\begingroup$ @nick Try the substitution $t=\alpha t'+\beta$ and find out for which $\alpha$ and $\beta$ the function to integrate reduces to $e^{-\pi t'^2}$. $\endgroup$ Jan 10, 2015 at 12:31

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