Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 14B / iii / Solution

Because $b$ is not an integer, the coefficient $\sin (\pi b)$ is nonzero. The expression of $B$ in terms of the entire function $J$ can therefore have poles only at integral $z$, and at most simple ones. Positive integers are removable: $B(z,b)$ is already analytic there for $\mathrm{Re}b>0$, and analytic continuation in $b$ givesThus $J$ vanishes at those integers, cancelling the denominator zero. The Gamma function formula $B(z,b)=\Gamma (z)\Gamma (b)/\Gamma (z+b)$ identifies the remaining singularities:For nonintegral $b$, these residues are nonzero, so every listed singularity is a genuine simple pole.