Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 7 / b

Substituting $z=1$ into the defining Gamma integral givesBoth endpoints are convergent, so no continuation of the Gamma function is needed here.

For real $z>0$, apply integration by parts to the Gamma integral with $u=t^z$ and $dv=e^{-t}dt$:The boundary term vanishes: $t^z\to 0$ at zero, and exponential decay dominates the power at infinity. This proves the Gamma function recurrence

Iterating the Gamma function recurrence gives a meromorphic continuationNear $z=-m$, the numerator is analytic and tends to $\Gamma (1)=1$. Exactly one denominator factor, $z+m$, vanishes, while all other factors tend to the nonzero product $(-1{)}^{m}m!$. ThereforeThus $\boxed{z=0,-1,-2,\dots \text{ are all simple poles}}$. More precisely, the Gamma function residue at a nonpositive integer is $\boxed{{\mathrm{Res}}_{z=-m}\Gamma (z)=(-1{)}^m/m!}$. Taking $m$ large enough to put $z+m+1$ in the original right half-plane also shows that this continuation introduces no other finite poles.