Abstract Nonsense

Crushing one theorem at a time

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)


May 15, 2010 - Posted by | Analysis, Computations, Fun Problems, Uncategorized | , , , , ,

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