Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 3 / 5E (Further Analysis)

The contour has winding number one around the single pole at zero. By the Cauchy integral formula, or the residue of $1/z$,Rotation by $z\mapsto iz$ carries each oriented edge to the next oriented edge and leaves $dz/z$ unchanged. The four edge integrals are therefore equal, each $\pi i/2$.

On the right edge use $z=1+it$, $-1\le t\le 1$, soThe real integrand $t/(1+t^2)$ is odd and integrates to zero. Dividing by $i$ gives ${\int }_{-1}^1(1+t^2{)}^{-1}dt=\pi /2$ without needing its real antiderivative.