Euler's integral for the gamma function
/ OY-ler /
Where does the gamma function come from concretely? The cleanest starting point is an integral discovered by Euler: for any complex z with positive real part, Gamma(z) = integral from 0 to infinity of t^(z-1) e^(-t) dt. Read the integrand as a tug-of-war. The factor t^(z-1) wants to grow as t increases (when Re z > 0 it also stays integrable near t = 0), while e^(-t) decays so violently that it tames any power of t out at infinity. The integral converges and defines a finite number for every z in the right half-plane Re z > 0.
Why does this integral deserve to be called a factorial? Integrate by parts once. With u = t^z and dv = e^(-t) dt you get Gamma(z+1) = integral of t^z e^(-t) dt = z * integral of t^(z-1) e^(-t) dt = z Gamma(z) — the functional equation drops out of one integration by parts. Combine that with the easy base case Gamma(1) = integral of e^(-t) dt = 1, and induction gives Gamma(n+1) = n! exactly. So Euler's integral is simultaneously a definition of Gamma and a built-in proof that it interpolates the factorial. Differentiating under the integral sign shows Gamma is holomorphic on Re z > 0, since t^(z-1) is holomorphic in z and the integral converges nicely.
The integral itself only converges for Re z > 0, so it is the seed, not the whole tree. To reach the rest of the plane you use the functional equation Gamma(z) = Gamma(z+1)/z repeatedly to push leftward, which both extends the definition and reveals the simple poles at 0, -1, -2, .... Alternatively Weierstrass's product formula gives Gamma on all of the plane in one stroke. A caution worth stating: 'Euler's integral' is the integral of the SECOND kind; the integral of the FIRST kind is the beta function, a close relative but a different integral.
Compute Gamma(3) from the integral. Gamma(3) = integral from 0 to infinity of t^2 e^(-t) dt. Integrating by parts twice (or recalling that the n-th moment of e^(-t) is n!) gives integral of t^2 e^(-t) dt = 2, and indeed 2 = 2! = (3-1)!. The same integral with z = 1 gives integral of e^(-t) dt = 1 = 0!, confirming the base case the whole induction rests on.
One integration by parts turns the integral into the factorial recurrence; the rest is induction.
The integral representation is only valid for Re z > 0; using it blindly at, say, z = -1/2 gives a divergent integral. To work to the left of the imaginary axis you must first continue Gamma via the functional equation or the Weierstrass product.