 Solution

For the beta function, split $1=t+(1-t)$ inside the integral to obtain $B(z,q)=B(z,q+1)+B(z+1,q)$. Integrating the derivative of $t^q(1-t{)}^z$ over $(0,1)$ gives $qB(z+1,q)=zB(z,q+1)$, since both endpoint terms vanish in the initial domain. Combining these identities yieldsRepeatedly apply this recurrence until $\mathrm{Re}(z+n)>0$. Each resulting expression agrees with the original integral on overlap, so it gives its unique analytic continuation. Possible singularities are only $z=0,-1,-2,\dots$.

Near $z=0$, $zB(z,q)\to qB(1,q)=1$. Continuing the recurrence downwards giveswhere the empty product is $1$. The recurrence shows that each singularity is at most a simple pole. If $q$ is not a positive integer, all these residues are nonzero and all the listed points are poles. If $q=m$ is a positive integer, poles occur only for $0\le n<m$; the later candidates are removable, as is also clear from $B(z,m)=(m-1)!/[z(z+1)\cdots (z+m-1)]$. This qualification matters when claiming a pole at every nonpositive integer.