Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 4 / 17B (Complex Methods)

The sector contains exactly one simple pole, at $z_0=e^{i\pi /n}$. All roots of $1+z^n$ are simple, and its residue issince $z_0^n=-1$. For the positively oriented boundary, the residue theorem givesThe vertex at zero is harmless: $f$ is analytic there for nonnegative integer $m$, or a small indent may be removed because its integral is $O({\rho }^{m+1})$.

On the circular arc, $|z|=R$, the reverse triangle inequality gives $|1+z^n|\ge R^n-1$. The arc length is $2\pi R/n$, so the ML inequality givesThe strict assumption $n>m+1$ is exactly what makes the exponent negative; it also ensures convergence of the positive-real-axis integral at infinity.

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.