$$\Gamma(x)= \int_0^ \infty e^{-t}t^{x-1}dt \ \ \ \ \ \ \ x>0. $$
Bohr and Mollerup showed that the gamma function is the only positive function $f$ defined on $(0,\infty)$ that has these properties
- $f(x+1)=xf(x)$
- $f(1)=1$
- $f(x)$ is a continuous
- $f(x)$ is log convex
My question is why are these properties matter (I mean 3, 4 the first two are essential )? I get that it is probably useful defining $\Gamma$ to be continuous but why it should be log convex it seem to be a random and absurd thing that has no motivation but I am sure that there is a reason why it matters and why did Euler choosed this property when he gave this integral .
The first condition implies that $f$ has the form $f(x)=f(\{x\})\prod_{n=0}^{\lfloor x\rfloor-1}(n+\{x\})$ (where $\{x\}$ is the fractional part of $x$), and thus, since also $\Gamma(x)=\Gamma(\{x\})\prod_{n=0}^{\lfloor x\rfloor-1}(n+\{x\})$,
$$f(x)=\frac{f(\{x\})}{\Gamma(\{x\})}\Gamma(x).$$
The first factor is periodic; if it isn’t constant, its logarithm must alternate between being convex and concave. Since $(\log\Gamma(x))''\to0$ for $x\to\infty$, that means that $\log f(x)$ eventually alternates between being convex and concave on every interval $[n,n+1]$. That’s not a nice property for an interpolation of $n!$ to have.