Prove the property

\(\displaystyle \vartheta (0; i x^2) = \frac{1}{x} \, \vartheta \! \! \left (0; \frac{i}{x^2} \right )\)

It's a bit easy, but worths a try.

Balarka

.

Board index **‹** Special Functions **‹** Jacobi theta function
## Jacobi theta function

Prove the property

\(\displaystyle \vartheta (0; i x^2) = \frac{1}{x} \, \vartheta \! \! \left (0; \frac{i}{x^2} \right )\)

It's a bit easy, but worths a try.

Balarka

.

If anyone is interested, I can post the solution. Let me know otherwise if you want to try it by yourself...

I don't know how Poisson integral formula helps at this point, perhaps you can show?

What I exactly had in mind is to relate a infinite integral of theta to Riemann xi function to continue theta along the symmetry axis 1/2 slowly.

**Moderator:** Shobhit

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\(\displaystyle \vartheta (0; i x^2) = \frac{1}{x} \, \vartheta \! \! \left (0; \frac{i}{x^2} \right )\)

It's a bit easy, but worths a try.

Balarka

.

I know not much about that function, but I think that the result follows from the Poisson summation formula.

Every integral harbors a story as fascinating as a comic book!

What I exactly had in mind is to relate a infinite integral of theta to Riemann xi function to continue theta along the symmetry axis 1/2 slowly.

The proposed formula is equivalent to

\(\displaystyle \sum_{n=-\infty}^{\infty} e^{-\pi \tau n^2} = \vartheta(0; i\tau) = \frac{1}{\sqrt{\tau}} \vartheta\left(0, \frac{i}{\tau} \right) = \frac{1}{\sqrt{\tau}} \sum_{n=-\infty}^{\infty} e^{-\pi n^2 / \tau}.\)

As both sides define analytic functions, it suffices to show that they coincide only for \(\displaystyle \tau > 0\), which we assume hereafter.

Let \(\displaystyle f(x) = e^{-\pi \tau x^2}\). It is well-known that \(\displaystyle \hat{f}(\xi) = \tau^{-1/2} e^{-\pi \xi^2 / \tau}\). Thus by the Poisson summations formula,

\(\displaystyle \begin{align*} \sum_{n=-\infty}^{\infty} e^{-\pi \tau n^2} &= \sum_{n=-\infty}^{\infty} f(n) \\ &= \sum_{n=-\infty}^{\infty} \hat{f}(n) = \frac{1}{\sqrt{\tau}} \sum_{n=-\infty}^{\infty} e^{-\pi n^2 / \tau} \end{align*}\)

as desired.

\(\displaystyle \sum_{n=-\infty}^{\infty} e^{-\pi \tau n^2} = \vartheta(0; i\tau) = \frac{1}{\sqrt{\tau}} \vartheta\left(0, \frac{i}{\tau} \right) = \frac{1}{\sqrt{\tau}} \sum_{n=-\infty}^{\infty} e^{-\pi n^2 / \tau}.\)

As both sides define analytic functions, it suffices to show that they coincide only for \(\displaystyle \tau > 0\), which we assume hereafter.

Let \(\displaystyle f(x) = e^{-\pi \tau x^2}\). It is well-known that \(\displaystyle \hat{f}(\xi) = \tau^{-1/2} e^{-\pi \xi^2 / \tau}\). Thus by the Poisson summations formula,

\(\displaystyle \begin{align*} \sum_{n=-\infty}^{\infty} e^{-\pi \tau n^2} &= \sum_{n=-\infty}^{\infty} f(n) \\ &= \sum_{n=-\infty}^{\infty} \hat{f}(n) = \frac{1}{\sqrt{\tau}} \sum_{n=-\infty}^{\infty} e^{-\pi n^2 / \tau} \end{align*}\)

as desired.

Every integral harbors a story as fascinating as a comic book!

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