## JamesJ 4 years ago Definite integral. Here's a fun one.

1. JamesJ

$\int_0^\infty \frac{x}{e^x - 1}$

2. anonymous

I was working on this one for a bit earlier, pretty sure it isn't solvable by standard methods. I wolframed it to find that the answer is $\frac{\pi^2}{6}$ so I'm assuming that it is possible to convert it to the form of Basel's problem. I also tried converting it to one of the integrals I know is equal to that, like $\int_0^1 \int_0^1 \frac{dxdy}{1-xy}$ I also found several different Taylor Series for the integrand, but that didn't lead anywhere. I didn't attempt to use complex analysis yet, since I would rather that be my final effort, haha. I have a page and a half of scribbles but nothing that jumps out at me as convertible to Basel's problem. Any particular hints that you would suggest? It seems like a really interesting question and I'd like to solve it.

3. JamesJ

write the integrand as $\frac{xe^{-x}}{1-e^{-x}}$ and now write that as a geometric series.

4. anonymous

Ah, I see now. I tried to convert it to a geometric series several times, but for some reason never tried that one (even though I have that fraction scribbled in a corner of the page!). Very very nice problem. For sake of completeness: $\int_0^{\infty}\frac{xdx}{e^x-1}=\int_0^{\infty}\frac{xe^{-x}dx}{1-e^{-x}}=\int xe^{-x}(1+e^{-x}+e^{-2x}+ \cdots)dx$ $=\int x(e^{-x}+e^{-2x}+\cdots)dx$ $=-x(e^{-x}+\frac12e^{-2x}+\frac13e^{-3x}|_0^{\infty} + \cdots)+\int e^{-x}+\frac12e^{-2x}+\frac13e^{-3x} +\cdots dx$ The left goes off to zero, leaving $e^{-x}+\frac14e^{-2x}+\frac19e^{-3x}+ \cdots |_0^{\infty}$ $=\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ Beautiful :)

5. JamesJ

Yes, nicely done.

6. JamesJ

here's another one... http://openstudy.com/study#/updates/4f0e4b24e4b04f0f8a91269e which points to the solution to the more general problem: $\int_0^\infty \frac{x^n}{e^x -1} dx$