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)=∫∞0tz−1e−tdt 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 x≥0;
- 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))1z∞∏m=1(1+zm)e−z/m,

where g(z) is an entire function such that

g(z+1)−g(z)=γ+2kπi,k∈Z,

(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*