the gamma functional equation

The single most useful identity the gamma function obeys is its functional equation: Gamma(z+1) = z Gamma(z), valid for every z where both sides make sense. In words, shifting the input up by one multiplies the output by the old input. This is exactly the rule the factorial follows — (n+1)! = (n+1) * n! — now stated for all complex z, not just integers. It is the gamma function's heartbeat, and almost every calculation with Gamma starts by applying it.

It is easy to see where the equation comes from: one integration by parts of Euler's integral gives it directly, as shown by Gamma(z+1) = integral of t^z e^(-t) dt = [-t^z e^(-t)] from 0 to infinity + z integral of t^(z-1) e^(-t) dt = z Gamma(z) (the boundary term vanishes for Re z > 0). But the equation earns its real keep as the engine of analytic continuation. Suppose you only know Gamma on Re z > 0. Rewrite the equation as Gamma(z) = Gamma(z+1)/z. The right-hand side makes sense for Re z > -1 (except at z = 0), so this defines Gamma one strip further left, and the apparent blow-up at z = 0 is exactly a simple pole. Repeat: Gamma(z) = Gamma(z+2)/(z(z+1)) reaches Re z > -2 and exposes the pole at z = -1, and so on across the whole plane.

This recursive continuation also pins down the poles and their residues. Near z = -n the dominant factor is 1/(z+n), so Gamma has a simple pole there; the residue works out to (-1)^n / n!. So the gamma function is meromorphic on all of the plane, with simple poles precisely at 0, -1, -2, ... and no others. A frequent slip: students try to evaluate Gamma at a negative integer and expect a number — there is no value there, the function genuinely blows up, which is why 1/Gamma is the cleaner entire function with zeros at exactly those points.