How to compute $\int_0^\infty \frac{t}{e^t-1}dt$? I know how to do this by using series, but I'm interested in other solutions. For example, is this possible to do with complex analysis? If it is, what is contour of integration?
Any help is welcome. Thanks in advance.
Best Answer
This is a very special integral. Let, $$\Gamma(s)=\int_{0}^{\infty}t^{s-1}e^{-t}dt$$ Substitute $t=nu\implies dt=ndu$ $$\frac{\Gamma(s)}{n^{s}}=\int_{0}^{\infty}u^{s-1}e^{-nu}du$$ Run sum on both sides with $n$ from $0$ to $\infty$. $$\Gamma(s)\zeta(s)=\int_{0}^{\infty}\frac{u^{s-1}}{e^{u}-1}du$$ Setting $s=2$ we have that, $$\Gamma(2)\zeta(2)=\int_0^\infty \frac{t}{e^t-1}dt$$ Which is $\frac{\pi^{2}}{6}$