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## JAcobi Theta

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### JAcobi Theta

Sun Aug 18, 2013 6:42 pm
zaidalyafey
Global Moderator

Posts: 354
Can we proof

$$\vartheta(0;i)=\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\sqrt[4] \pi}{\Gamma\left(\frac 3 4\right)}$$

By Mellin transform ?
$\displaystyle \sum_{k\geq 0}\frac{f^{(k)} (s)}{(s+1)_k} (-s)^{k}= \int^{1}_0x^{s} f(s x) \left( \frac{x}{s}\right) \, dx$

Wanna learn what we discuss , see tutorials

### Re: JAcobi Theta

Mon Aug 19, 2013 9:44 am

Posts: 850
Location: Jaipur, India

This can be derived using the fact that

$\displaystyle \varphi(q) = \sqrt{\frac{2K(k)}{\pi}}$

where $\displaystyle \varphi (q) = \sum_{n=-\infty}^\infty q^{-n^2}$

It is well known that when $q=e^{-\pi}$, $k$ becomes $\frac{1}{\sqrt{2}}$. Now,

\displaystyle \begin{align*} K \left(\frac{1}{\sqrt{2}} \right) &= \sqrt{2} \int_0^1 \frac{dt}{\sqrt{(1-t^2)(2-t^2)}} \\ &= \sqrt{2} \int_0^1 \frac{dx}{\sqrt{1-x^4}} \quad x^2 :=t^2/(2-t^2) \\ &= \frac{\sqrt{2}}{4} \beta \left(\frac{1}{4},\frac{1}{2} \right) \\ &= \frac{1}{2\sqrt{2}} \frac{\Gamma \left(\frac{1}{4} \right)\Gamma \left(\frac{1}{2} \right)}{\Gamma \left(\frac{3}{4} \right)} \\ &= \frac{\pi^{\frac{3}{2}}}{2\Gamma^2 \left(\frac{3}{4} \right)} \end{align*}

So, we get

$\displaystyle \varphi(e^{-\pi}) = \frac{\sqrt[4]{\pi}}{\Gamma \left(\frac{3}{4} \right)}$

### Re: JAcobi Theta

Tue Aug 20, 2013 1:12 pm
Here >>. I evaluated these for $\displaystyle k_1,k_2,k_3,k_4$ and $\displaystyle k_{25}$ but the others are really tough. There are some sophisticated number theoretic evaluations available but they are hard to understand. I believe Ramanujan's modular identities will prove to be useful for proving them all.