Why is Euler’s Gamma function the “best” extension of the factorial function to the reals?

There are lots (an infinitude) of smooth functions that coincide with f(n)=n! on the integers. Is there a simple reason why Euler’s Gamma function Γ(z)=0tz1etdt is the “best”? In particular, I’m looking for reasons that I can explain to first-year calculus students.

Answer

The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties

  • f(1)=1;
  • f(x+1)=xf(x) for every x0;
  • logf is a convex function.

The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function.


By the way, the gamma function is not the only meromorphic function satisfying
f(z+1)=zf(z),f(1)=1,
with no zeroes and no poles other than the points z=n, n=0,1,2. There is a whole family of such functions, which, in general, have the form
f(z)=exp(g(z))1zm=1(1+zm)ez/m,
where g(z) is an entire function such that
g(z+1)g(z)=γ+2kπi,kZ,
(γ is Euler’s constant). The gamma function corresponds to the simplest choice
g(z)=γz.

Edit: corrected index in the product.

Attribution
Source : Link , Question Author : pbrooks , Answer Author : Community

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