Integrate $(1+z^p)^{-1}$ around the wedge $0\le\arg z\le2\pi/p$. The radial <integrals> differ by the factor $e^{2\pi i/p}$, the arc vanishes, and the wedge contains the simple pole $e^{i\pi/p}$ with residue $-[p e^{-i\pi/p}]^{-1}$. Solving the resulting identity gives $I=(\pi/p)\csc(\pi/p)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 3 / 3C / Solution

 Solution (source code)