Calculate $$\int_0^\infty\frac{\sin^3{x}}{e^x-1}\,\mathrm dx.$$
It seems that the integral cannot be solved in terms of elementary functions, so I try to use the Cauchy (residue) theorem to evaluate it. However, I couldn't find a complex function $$f(z) = \frac{?}{e^z-1}$$ to evaluate this real integral. If the numerator were $\sin{x}$, we can consider $e^{iz}$ since $$ \sin{z} = \frac{e^{iz}-e^{-iz}}{2i}.$$
Is there any hint or method to solve this problem?
Best Answer
Note that $\sin^3 x = \frac34 \sin x- \frac14 \sin(3x)$, and for any $a\in\mathbb{R}$, $$\begin{align} \int_{0}^{\infty} \frac{\sin(ax)}{e^x - 1} dx &= \int_{0}^{\infty} \frac{\sin (ax) e^{-x}}{1 - e^{-x}} dx = \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} \sin(ax) \, e^{-nx} \right) dx \\ &=\sum_{n=1}^{\infty} \int_{0}^{\infty} \sin (ax) \, e^{-nx} \; dx = \sum_{n=1}^{\infty} \frac{a}{n^2+a^2}. \end{align}.$$ Hence $$\int_0^\infty\frac{\sin^3 x}{e^x-1}dx=\frac{3}{4}\sum_{n=1}^{\infty} \left(\frac{1}{n^2+1}-\frac{1}{n^2+9}\right).$$ In order to find a closed formula see How to sum $\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}$?