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JAcobi Theta

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Post Sun Aug 18, 2013 6:42 pm
zaidalyafey User avatar
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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

Post Mon Aug 19, 2013 9:44 am
Shobhit Site Admin
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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

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

Post Tue Aug 20, 2013 1:12 pm
Shobhit Site Admin
Site Admin

Posts: 850
Location: Jaipur, India

Hey Z, you should also see about Elliptic Integral Evaluation at Singular values.

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.


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