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## Jacobi theta function

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### Jacobi theta function

Tue May 14, 2013 4:23 pm

Posts: 38
Location: India, West Bengal
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
.

### Re: Jacobi theta function

Sun May 19, 2013 4:34 pm

Posts: 38
Location: India, West Bengal
If anyone is interested, I can post the solution. Let me know otherwise if you want to try it by yourself...

### Re: Jacobi theta function

Fri May 24, 2013 7:56 am
sos440
Integration Guru

Posts: 124
Location: California, US

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!

### Re: Jacobi theta function

Fri May 24, 2013 12:44 pm

Posts: 38
Location: India, West Bengal
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.

### Re: Jacobi theta function

Sat May 25, 2013 12:50 am
sos440
Integration Guru

Posts: 124
Location: California, US

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