My friend and I have been trying weeks to evaluate the integral
$$\int^{\infty }_{0}\frac{x \,dx}{e^{x} -1} .$$
We have together tried 23 contours, and all have failed.
We already know how to solve this with infinite sums (i.e., using the zeta function and the Basel problem), but we can't figure out how to solve it using contour integration methods.
We already know the answer is $\frac{\pi^{2}}{6}$.
Best Answer
I would have guessed this was a duplicate, but I wasn't able to find another instance of this question during a cursory search.
Hint The denominator has period $2 \pi i$, which suggests using the following contour $\Gamma_{\epsilon, R}$, $0 < \epsilon < \pi$, $\epsilon < R$, (for which an illustration was already drawn for an answer to the similar question linked by Zacky in the comments):
The key trick here, which we apply with the benefit of hindsight, is to evaluate instead the similar integral $$\int_{\Gamma_{\epsilon, R}} \frac{z^2 \,dz}{e^z - 1} .$$ The interior of $\Gamma_{\epsilon, R}$ contains no poles, so this integral vanishes. Thus, parameterizing the constituent arcs of the contour gives \begin{multline} 0 = \underbrace{\int_\epsilon^R \frac{x^2 \,dx}{e^x - 1}}_{A} + \underbrace{\int_0^{2 \pi} \frac{(R + i y)^2 \cdot i \,dy}{e^{R + i y} - 1}}_{B} + \underbrace{\int_R^\epsilon \frac{(x + 2 \pi i)^2 \,dx}{e^x - 1}}_{C} \\ + \underbrace{\int_0^{-\pi / 2} \frac{(2 \pi i + \epsilon e^{i\theta})^2 \cdot i \epsilon e^{i \theta} d \theta}{e^{\epsilon e^{i \theta}} - 1}}_{D} + \underbrace{\int_{2 \pi - \epsilon}^\epsilon \frac{(i y)^2 \cdot i \,dy}{e^{i y} - 1}}_{E} + \underbrace{\int_{\pi / 2}^0 \frac{(\epsilon e^{i\theta})^2 \cdot i \epsilon e^{i \theta} d \theta}{e^{\epsilon e^{i \theta}} - 1}}_{F} . \qquad (\ast) \end{multline}