 Solution

Cauchy’s formula is $f(w)=(2\pi i{)}^{-1}{\int }_{\gamma }f(z)/(z-w)dz$, proved by removing the removable singularity of $[f(z)-f(w)]/(z-w)$ and applying Cauchy’s theorem; similarly $f\prime (w)=(2\pi i{)}^{-1}\int f(z)/(z-w{)}^{2}dz$. The resulting estimates prove Liouville’s theorem. Finally ${\cos }^{2}x=(1+\cos 2x)/2$ and residues in the upper half-plane give ${\int }_{-\infty }^{\infty }{\cos }^{2}x/(x^2+1)dx=\frac{\pi }{2}(1+e^{-2})$.