= 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 $\operatorname{Re}b>0$, and <analytic continuation> in $b$ gives
$$
B(n,b)=\frac{(n-1)!}{b(b+1)\cdots(b+n-1)}\qquad(n\ge1).
$$
Thus $J$ 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:
$$
\boxed{z=-m\ (m=0,1,\ldots),\qquad
\operatorname{Res}_{z=-m}B(z,b)=\frac{(-1)^m\Gamma(b)}{m!\Gamma(b-m)}.}
$$
For nonintegral $b$, these residues are nonzero, so every listed singularity is a genuine simple pole.
Back to article page
