Prove this equality

**usagi_killer** · April 4th 2011, 03:50 AM

I actually have no idea how to go about this question, in fact I weren't even taught what the Gamma function is/does in my classes nor have I ever seen the Jacobi theta-function in my life! So I am really clueless on this question. Any help would be appreciated!

Cheers!

**usagi_killer** · April 5th 2011, 11:16 PM

Hmm after a while of thinking i've came up with something, however im not sure if its a correct method, i'm mostly worried about splitting the integrals up into parts such as $\int_1^{\infty}t^{s-1}dt$ which are divergent...

$\displaystyle{RHS = \frac{1}{2s-1} + \frac{1}{2}\int_0^1\left(\theta(it)-t^{\frac{-1}{2}}\right)t^{s-1}dt-\frac{1}{2s}+\frac{1}{2}\int_1^{\infty}\left(\thet a(it)-1\right)t^{s-1}dt}$
$<br /> \displaystyle{= \frac{1}{2s-1}-\frac{1}{2s}+\frac{1}{2}\int_0^1\theta(it)t^{s-1}-t^{s-\frac{3}{2}}dt+\frac{1}{2}\int_1^{\infty}\theta(it )t^{s-1}-t^{s-1}dt}$
$<br /> \displaystyle{= \frac{1}{2s-1}-\frac{1}{2s}+\frac{1}{2}\int_0^{\infty}\theta(it)t ^{s-1}dt-\frac{1}{2}\int_0^1t^{s-\frac{3}{2}}dt-\frac{1}{2}\int_1^{\infty}t^{s-1}dt}$

Now $\displaystyle{\theta(it) = \sum_{n \in \mathbb{Z}} \exp(-\pi n^2t)}$

Because of the symmetric property of $n^2$ we have:

$\displaystyle{\theta(it) = 1 + 2\sum_{n=1}^{\infty} \exp(-\pi n^2t)}$ where the $n$ is the set of natural numbers summed to infinity.

Now $\displaystyle{\frac{1}{2}\int_0^{\infty}\theta(it) t^{s-1}dt = \frac{1}{2}\int_0^{\infty}\left(1 + 2\sum_{n=1}^{\infty} \exp(-\pi n^2t)\right)t^{s-1}dt}$
$<br /> \displaystyle{= \frac{1}{2}\int_0^{\infty}t^{s-1}dt + \sum_{n=1}^{\infty}\int_0^{\infty}\exp(-\pi n^2t)t^{s-1}dt}$

Substituting this result back into our original equation yields:

$\displaystyle{\frac{1}{2s-1}-\frac{1}{2s}+\frac{1}{2}\int_0^{\infty}t^{s-1}dt+\sum_{n=1}^{\infty}\int_0^{\infty}\exp(-\pi n^2t)t^{s-1}dt-\frac{1}{2}\int_0^1t^{s-\frac{3}{2}}dt - \frac{1}{2}\int_1^{\infty}t^{s-1}dt}$

$\displaystyle{= \frac{1}{2s-1}-\frac{1}{2s}+\frac{1}{2}\int_0^1t^{s-1}dt-\frac{1}{2}\int_0^1t^{s-\frac{3}{2}}dt+\sum_{n=1}^{\infty}\int_0^{\infty}\ exp(-\pi n^2t)t^{s-1}dt}$

$\displaystyle{= \frac{1}{2s-1}-\frac{1}{2s}+\frac{1}{2}\int_0^1t^{s-1}-t^{s-\frac{3}{2}}dt+\sum_{n=1}^{\infty}\int_0^{\infty}\ exp(-\pi n^2t)t^{s-1}dt}$

Now $\displaystyle{\frac{1}{2}\int_0^1t^{s-1}-t^{s-\frac{3}{2}}dt = \frac{1}{2}\left[\frac{t^s}{s}-\frac{2t^{s-\frac{1}{2}}}{2s-1}\right]^1_0}$

$\displaystyle{= \frac{1}{2s}-\frac{1}{2s-1}}$

Substituting this back in leaves us with only:

$\displaystyle{\sum_{n=1}^{\infty}\int_0^{\infty}\e xp(-\pi n^2t)t^{s-1}dt}$

Now let $\displaystyle{u = \pi n^2t \implies t=\frac{u}{\pi n^2}}$

$\displaystyle{\frac{du}{dt} = \pi n^2 \implies dt = \frac{du}{\pi n^2}}$

Substituting these results back in yields:

$\displaystyle{\sum_{n=1}^{\infty}\int_0^{\infty}\e xp(-u)\left(\frac{u}{\pi n^2}\right)^{s-1}\frac{du}{\pi n^2}}$

$\displaystyle{= \sum_{n=1}^{\infty}\frac{1}{(\pi n^2)^s} \int_0^{\infty}\exp(-u)u^{s-1}du}$

But $\displaystyle{\int_0^{\infty}\exp(-u)u^{s-1}du = \Gamma(s)}$

Also $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{(\pi n^{2})^s} = \sum_{n=1}^{\infty} \frac{1}{\pi^s n^{2s}} = \pi^{-s}\sum_{n=1}^{\infty}\frac{1}{n^{2s}} = \pi^{-s}\zeta(2s)}$

Therefore $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{\pi n^{2s}} \int_0^{\infty}\exp(-u)u^{s-1}du = \pi^{-s}\zeta(2s)\Gamma(s) = \pi^{-s}\Gamma(s) \times \zeta(2s) = LHS}$

The proof is complete.

so yeah, if anyone could confirm the validity of this, that'd be much appreciated!

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