Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 8E / Solution

Choose $z=e^{it}$ on the right semicircle, with the continuous branch $\log z=it$. ThenHere the integral is along that semicircle. Since $|A|>1$, the pole lies outside the unit disc and the geometric series converges uniformly on the contour.

PutThe Euler integral for the hypergeometric function, initially for $\mathrm{Re}u>-1$, givesThe prefactor follows from $\Gamma (u+2)/\Gamma (u+1)=u+1$. Hence an antiderivative of the contour integrand before multiplication by $1/i$ isEvaluating at the endpoints with the chosen branch yieldsThis is first obtained where the individual expressions are defined. The original finite-interval integral is entire in $u$, so analytic continuation gives the result elsewhere, taking removable limits at the apparent exceptional negative integers. That convention matters: the two separate hypergeometric terms may be singular even though their combination is finite.