Gamma Function
y = ∫₀^∞ u^(x − 1) · exp(−u) du
Open in the app The dials and keys named below are the app's.
What it draws
The curve is an integral, computed at every point: y = ∫₀^∞ u^(x−1) · e^(−u) du. That is Euler's integral for the gamma function, Γ(x), and nothing in it depends on the clock, so the curve stands still. Integrating by parts gives Γ(x + 1) = x · Γ(x), and Γ(1) = ∫₀^∞ e^(−u) du = 1, so at the whole numbers Γ(n) = (n − 1)!. Read along the curve: 1, 1, 2, 6, 24 at x = 1, 2, 3, 4, 5. Between them it fills in the factorial for every real argument.
The integral is taken numerically out to a fixed reach, and e^(−u) makes sure the integrand has died away by then. To the left of x = 1 the curve is blank: there u^(x−1) is infinite at u = 0, where the rule samples it, though the true integral converges down to x = 0. The lowest point of the curve lies between the first two integers, Γ(1.4616) ≈ 0.8856.
Between the integers
Infinitely many smooth curves pass through 1, 1, 2, 6, 24. The Bohr–Mollerup theorem singles out Euler's integral: it is the only one with Γ(x + 1) = x · Γ(x) whose logarithm is convex. The growth is faster than any exponential. Γ(6) = 120 lies just off the right edge, and Stirling's approximation Γ(x) ≈ √(2π/x) · (x/e)^x tracks the curve from below, 23.6 against 24 at x = 5.
History
Leonhard Euler found the integral in 1729, in letters to Christian Goldbach, answering the question of how to extend n! to fractions; Daniel Bernoulli had a product formula the same year. Adrien-Marie Legendre gave the function its name and the symbol Γ in 1811. Harald Bohr and Johannes Mollerup proved the uniqueness theorem in 1922, in a textbook.
Try
- Shift by one, integral(u^x·exp(−u), u, 0, inf): now Γ(x + 1) = x!, reading 1, 1, 2, 6, 24, 120 at x = 0 to 5, and the blank stretch ends at x = 0.
- Take the logarithm, log(integral(u^(x − 1)·exp(−u), u, 0, inf)): zero at x = 1 and x = 2, ln 24 ≈ 3.18 at x = 5, and convex throughout, as Bohr and Mollerup require.
- Compare with Stirling, sqrt(2π/x)·x^x·exp(−x): the same rise, below the curve by 8% at x = 1 and under 2% at x = 5, and drawn for every positive x since no integral can fail.
- Grow the upper limit with the clock, integral(u^(x − 1)·exp(−u), u, 0, t), then Rewind and Play: the incomplete gamma function fills in from the left. After five seconds x = 5 has reached only 13.4 of its 24.