the beta function
/ BAY-tuh /
The beta function B(x, y) is the gamma function's natural partner. Its definition is an integral over the unit interval: B(x, y) = integral from 0 to 1 of t^(x-1) (1 - t)^(y-1) dt, valid for Re x > 0 and Re y > 0. Picture a probability-like density on [0, 1] that is weighted toward 0 when x is small and toward 1 when y is small; the beta function is the total mass under that weighting. It is symmetric, B(x, y) = B(y, x), as you can see by the substitution t -> 1 - t.
The single fact that makes the beta function so useful is its identity with gamma: B(x, y) = Gamma(x) Gamma(y) / Gamma(x + y). This converts a two-parameter integral into a ratio of gamma values, which we already understand completely. The cleanest proof multiplies Gamma(x) Gamma(y) as a double integral over the first quadrant, switches to polar-like coordinates t = r cos^2 and (1 - t), and watches the radial part factor off as Gamma(x + y) while the angular part is exactly the beta integral. So the beta function is not an independent mystery — it is gamma values in disguise, which is why it is called Euler's integral of the first kind to gamma's second kind.
Because so many integrals can be massaged into beta form, the identity is a master tool for evaluating definite integrals with no elementary antiderivative — integrals of powers of sine and cosine, of x^a (1-x)^b, and many others. It also names the beta probability distribution, central to Bayesian statistics, whose normalising constant is exactly 1/B(x, y). A caveat: the integral form only converges for Re x > 0 and Re y > 0, but the gamma-ratio formula B(x, y) = Gamma(x)Gamma(y)/Gamma(x+y) continues the beta function meromorphically beyond that region wherever the gammas are defined.
Compute integral from 0 to 1 of t^2 (1 - t)^3 dt. This is B(3, 4) = Gamma(3) Gamma(4) / Gamma(7) = (2!)(3!)/6! = (2)(6)/720 = 12/720 = 1/60. Doing the integral directly by expanding (1-t)^3 also gives 1/60, but the beta-gamma identity reads off the answer with no expansion at all.
B(x, y) = Gamma(x)Gamma(y)/Gamma(x+y) turns a messy power integral into a one-line ratio of factorials.
The beta function is not a new transcendental object — it is entirely determined by gamma. If you can evaluate the relevant gamma values, you can evaluate the beta value; the integral is just a convenient packaging.