# 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 $$f(n)=n!f(n)=n!$$ on the integers. Is there a simple reason why Euler’s Gamma function $$Γ(z)=∫∞0tz−1e−tdt\Gamma (z) = \int_0^\infty t^{z-1} e^{-t} dt$$ is the “best”? In particular, I’m looking for reasons that I can explain to first-year calculus students.

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

• $$f(1)=1f(1)=1$$;
• $$f(x+1)=xf(x)f(x+1)=xf(x)$$ for every $$x≥0x\geq 0$$;
• $$logf\log f$$ 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
$$f(z+1)=zf(z),f(1)=1,f(z+1)=z f(z),\qquad f(1)=1,$$
with no zeroes and no poles other than the points $$z=−nz=-n$$, $$n=0,1,2…n=0,1,2\dots$$. There is a whole family of such functions, which, in general, have the form
$$f(z)=exp(−g(z))1z∞∏m=1(1+zm)e−z/m,f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{m=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$
where $$g(z)g(z)$$ is an entire function such that
$$g(z+1)−g(z)=γ+2kπi,k∈Z,g(z+1)-g(z)=\gamma+2k\pi i,\quad k\in\mathbb Z,$$
($$γ\gamma$$ is Euler’s constant). The gamma function corresponds to the simplest choice
$$g(z)=γzg(z)=\gamma z$$.

Edit: corrected index in the product.