Work out the value of $\int_{0}^{\infty}{\cos{x}\over x}[1-\cos(nx)]\mathrm dx$

\begin{align} 2\int_{0}^{\infty}{\cos{x}\over x}[1-\cos(nx)]\mathrm dx &= \int_0^\infty\dfrac{2\cos x-\cos(n+1)x-\cos(n-1)x}{x}\mathrm dx \\ &= \int_0^\infty\dfrac{2s}{s^2+1}-\dfrac{s}{s^2+(n+1)^2}-\dfrac{s}{s^2+(n-1)^2}\mathrm ds \\ &= \ln\dfrac{s^2+1}{\sqrt{(s^2+(n+1)^2)(s^2+(n-1)^2)}}\Big|_0^\infty \\ &= \color{blue}{\ln(n^2-1)} \end{align}