Problem - I. Mezo (China)
Prove
\[\int_0^\infty\frac{\cos(x)-1}{x(e^x-1)}\,\mathrm dx=\frac{1}{2}\ln\left(\frac{\pi}{\sinh(\pi)}\right).\]Solution
First we acquire two preliminary identities. In Euler’s product for the sine
\[\frac{\sin(x)}{x}=\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2\pi^2}\right)\]we let $x=\pi i$ to get
\[\frac{\sinh(\pi)}{\pi}=\prod_{n=1}^\infty\left(1+\frac{1}{n^2}\right).\]We are also interested in getting the value of
\[I(s)=\int_0^\infty\frac{\cos(x)-1}{x}e^{-sx}\,\mathrm dx,\]which we find using the Feynman trick.
\[\begin{align*} I'(s) &= \int_0^\infty(1-\cos(x))e^{-sx}\,\mathrm dx \\ &= \frac{1}{s}-\frac{s}{1+s^2}.\end{align*}\]Then we have
\[\begin{align*} I(s) &= \int_\infty^s I'(u) \,\mathrm du \\ &= \left[\ln(u)-\frac{1}{2}\ln(1+u^2)\right]\bigg\vert_\infty^s \\ &= \ln(s)-\frac{1}{2}\ln(1+s^2)-\lim_{u\to \infty}I(u) \\ &= -\frac{1}{2}\ln\left(1+\frac{1}{s^2}\right).\end{align*}\]Finally,
\[\begin{align*} \int_0^\infty\frac{\cos(x)-1}{x(e^x-1)}\,\mathrm dx &= \int_0^\infty\frac{(\cos(x)-1)e^{-x}}{x(1-e^{-x})}\,\mathrm dx \\ &= \sum_{n=1}^\infty\int_0^\infty\frac{\cos(x)-1}{x}e^{-nx}\,\mathrm dx,\;\;\text{monotone convergence} \\ &= -\frac{1}{2}\sum_{n=1}^\infty\ln\left(1+\frac{1}{n^2}\right) \\ &= \frac{1}{2}\ln\left(\prod_{n=1}^\infty\left(1+\frac{1}{n^2}\right)^{-1}\right) \\ &= \frac{1}{2}\ln\left(\frac{\pi}{\sinh(\pi)}\right).\end{align*}\]