# Abstract Nonsense

## Interesting Integral

Problem: Compute $\displaystyle I=\int_0^{\infty}\frac{x^\alpha}{\sinh(x)}dx,\text{ }\alpha>0$

Solution: First note that

$\displaystyle I=2\int_0^{\infty}\frac{e^{-x}x^\alpha}{1-e^{-2x}}dx$

But, since for every $x\in(0,\infty)$ we have that $|e^{-x}|<1$ we see that

$\displaystyle I=\int_0^{\infty}e^{-x}x^\alpha\sum_{n=0}^{\infty}e^{-2nx}dx=\int_0^{\infty}\sum_{n=0}^{\infty}x^\alpha e^{-(2n+1)x}$

Now, using the fact that a power series is uniformly convergent on it’s disk of convergence it follows that

$\displaystyle I=2\sum_{n=0}^{\infty}\int_0^{\infty}x^\alpha e^{-(2n+1)x}dx$

So, letting $\xi=(2n+1)x$ we see that

$\displaystyle I=2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha+1}}\int_0^\infty \xi^\alpha e^{-\xi}d\xi$

And thus remembering the definition of $\Gamma(z)$ this becomes

$\displaystyle I=2\Gamma(\alpha+1)\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha+1}}$

To compute this sum we note that (ignoring convergence issues)

$\displaystyle \zeta(\alpha+1)=\sum_{n=1}^{\infty}\frac{1}{n^{\alpha+1}}=\sum_{n\in2\mathbb{N}}\frac{1}{n^{\alpha+1}}+\sum_{n\in2\mathbb{N}+1}\frac{1}{n^{\alpha+1}}$

which upon inspection is equal to

$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n)^{\alpha+1}}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha+1}}=\frac{1}{2^{\alpha+1}}\zeta(\alpha+1)+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{\alpha+1}}$

Thus, solving for the desired sum we get that

$\displaystyle \sum_{n=0}^{\infty}\frac{1}{(2n+1)^\alpha}=\left(1-\frac{1}{2^{\alpha+1}}\right)\zeta(\alpha+1)=\frac{1}{2}\left(2-\frac{1}{2^\alpha}\right)\zeta(\alpha+1)$

So, finally

$\displaystyle I=\left(2-\frac{1}{2^\alpha}\right)\Gamma(\alpha+1)\zeta(\alpha+1)$
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