Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 4 / 17B / c

Let $\alpha =\pi (m+1)/n$, which lies strictly between zero and $\pi$. On the lower ray $z=x$, the integral is $I_R={\int }_0^R{x}^m/(1+x^n)dx$. On the upper ray $z=x{e}^{2\pi i/n}$, the positive boundary orientation runs from $R$ down to zero. Since $z^n=x^n$, that contribution is $-e^{2i\alpha }{I}_R$. The circular contribution vanishes by part (b); henceUse $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$ to obtainThe result is positive as expected. Convergence near zero follows from $m\ge 0$, so both endpoint limits used in the contour argument are justified.