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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: $$\int_0^{\infty} {exp(-t) t^{x-1} dt}$$

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    You at least know it has an extremum on $[1, 2]$ from just evaluating $\Gamma(1) = \Gamma(2) = 1$. – Arthur Apr 29 '19 at 14:22
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    And then if you evaluate it numerically at, say, $1.5$, you find a value less than $1$. – Robert Israel Apr 29 '19 at 14:40
  • Robert Israel proposed to calculate $\Gamma(1.5)$ numerically. Refraining from numerics we could employ the doubling formula of Legendre $\Gamma(\frac{x}{2})\Gamma(\frac{1+x}{2})=\frac{\sqrt{\pi}}{2^{x-1}}\Gamma(x)$ at $x=2$ giving $\Gamma(\frac{3}{2}) = \frac{\sqrt{\pi}}{2}$ which is $\lt 1$ since $\pi \lt 4$. – Dr. Wolfgang Hintze Apr 29 '19 at 19:35

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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.

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