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.
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 $\frac{\pi^{2}}{6}$
Substitute $x=e^{-t}$ to rewrite the integral as $$\int_0^\infty \frac{t}{e^t-1}dt=\int_0^1 \frac{\ln x}{x-1}dx $$ Then, refer to the non-series solution of $\int_0^1 \frac{\ln x}{x-1}dx=\frac{\pi^2}6$
One way in terms of the dilogarithm function:
$$\int_0^\infty \frac{t}{e^t-1}dt= \int_0^\infty \frac{e^{-t}t}{1-e^{-t}}dt= \int_1^0 \frac{\ln(w)}{1-w}dw = \int_1^0 \frac{\ln(1-r)}{r}dr= \operatorname{Li}_{2}(1)= \frac{\pi^2}{6}$$
Another method, which applies in quite a few other situations, is to use the Hankel or keyhole contour: a path coming in from $+\infty$ "just below" the real axis, traversing a small circle clockwise, and going out to $+\infty$ "just above" the positive real axis. The details are documented in many places. This method gives a way to evaluate $\zeta(s)$ for $s$ non-positive integer, and appeared in B. Riemann's 1858 paper on $\zeta$.
$$\begin{align*} & \color{blue}{I = \int_0^1 \frac {\ln x}{x-1}dx}\\ \end{align*}$$
Now, $$\color{red}{\psi_0 (z) = -\gamma + \int_0^{1} \frac {x^{z-1}-1}{x-1}dx}$$ We'll differentiate and that is Trigamma function $\psi_1(z)$
$$\implies \frac {\partial\psi_0}{\partial z}= \psi_1 (z) = \int_0^1 \frac {x^{z-1}\ln x}{x-1}dx$$ $$\color{green}{I = \left[\psi_1(z)\right]_{z=1} = \frac {\pi^2}{6}}$$