Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 7 / ii

For $\mathrm{Re}z>0$, the defining Gamma integral converges at zero and at infinity and defines a holomorphic function. The Gamma function recurrence then supplies its meromorphic continuation to the rest of the complex plane, as used below.

For $\mathrm{Re}z>0$, integration by parts givesThe boundary term vanishes because $t^{\mathrm{Re}z}\to 0$ at zero and exponential decay dominates any fixed power at infinity. ThusThis also gives $\Gamma (n+1)=n!$ for nonnegative integers, linking the Gamma function to the factorial.

Iterate the Gamma function recurrence to obtain, near $z=-m$ for $m\ge 0$,The numerator is holomorphic and equals $\Gamma (1)=1$ at $z=-m$. Exactly one denominator factor, $z+m$, vanishes; the product of the remaining factors there is $(-m)(-m+1)\cdots (-1)=(-1{)}^{m}m!$. Thus there is a simple pole, with Gamma function residue at a nonpositive integerFor $m=0$ the empty product is one. The same recurrence with a sufficiently large shift defines the continuation at every point away from these integers, so no further poles occur. At the poles, the recurrence is an identity of meromorphic functions, rather than an equality of finite values.