Proof that the gamma function has a minimum between $x=1$ and $x=2$?

Question

So I know that the gamma function $\Gamma(x)$ for $x>0$ has a minimum at $x_{min}$ which lies between 1 and 2. Where does this follow from though? I know that the gamma function is defined as below: $$\int_0^{\infty} {exp(-t) t^{x-1} dt}$$

Answer 1

ΟΚ, we have that $\Gamma''(x)=\int_{0}^{+\infty}{(\log t)}^2t^{x-1}e^{-t}dt>0,\ \forall x>0$. By the comment of Arthur above, Rolle's theorem implies that there is an extremum in $(1,2)$ and the positivity of the second derivative guarantees that this is a minimum.

The only thing that you have to do, is to verify the passage of the second derivative inside the integral. This is a technical detail, that I leave up to you. However, if you know measure theory, there is a standard theorem to use for this. Otherwise you can take $\Gamma(s)$ as a function of a complex variable on $\Re(s)>0$ and pass the derivatives in the integral. In Titchmarsh's book The Theory of Functions you can find theorems for passing derivatives with respect to a complex variable into integrals with respect to a real variable.