TITLE: Minimum of the Gamma Function $\Gamma (x)$ for $x>0$. How to find $x_{\min}$?
QUESTION [21 upvotes]: The $\Gamma (x)$ function has just one minimum for $x>0$ . This result uses some
properties of the gamma function:

$\Gamma ^{\prime \prime }(x)>0$ and $\Gamma (x)>0$ for all $x>0$
$\Gamma (1)=\Gamma (2)=1$.

Observing the following graph (created in SWP) of $y=\Gamma (x)$ this minimum is near $x=3/2$, but likely $\min \Gamma (x)\neq \Gamma \left( 3/2\right) =\dfrac{1}{2}\Gamma \left( 1/2\right) =\dfrac{1}{2}\sqrt{\pi }$. 

I think that it is not possible to find analytically the exact value of $x_{\min }$, even by converting to an adequate problem in the interval $]0,1]$ and using the functional equation $\Gamma (x+1)=x\Gamma (x)$ and the reflection formula
$\Gamma (p)\Gamma (p-1)=\dfrac{\pi }{\sin px}\qquad $( $0\lt p\lt 1$)  
Question: 
a) Which is the best way to find $\min_{[1,2]}\Gamma (x)$ and does $x_{\min }$ lay in $[1,3/2]$ or in $[3/2,2]$?
b) Is there some useful series expansion of $\Gamma (x)$?
c) Which numeric method do you suggest?

Edit:  Due to the shape of $\Gamma (x)$ I thought on the one-dimensional Davies-Swann-Campey method of direct search for unconstrained optimization, which approximates a function near a minimum by successive approximating quadratic polynomials.

REPLY [2 votes]: You have
$$\frac d {dx} \Gamma(x)=\Gamma (x)\, \psi (x)$$ Knowing that the solution is close to $\frac 32$, make a Taylor expansion around this value. This would give
$$y=\psi (x)=\psi^{(0)} \left(\frac{3}{2}\right)+\left(\frac{\pi ^2}{2}-4\right)
   \left(x-\frac{3}{2}\right)+\frac{1}{2}  \psi
   ^{(2)}\left(\frac{3}{2}\right)\left(x-\frac{3}{2}\right)^2+O\left(\left(x-\frac{3}{2}\right)^3\right)$$ Now, using series reversion
$$x=\frac{3}{2}+t-\frac{ \psi ^{(2)}\left(\frac{3}{2}\right)}{\pi
   ^2-8}t^2+O\left(t^3\right)\qquad \text{where} \qquad t=\frac{y-\psi ^{(0)}\left(\frac{3}{2}\right)}{\frac{\pi ^2}{2}-4}$$ Making $y=0$  gives $t=-\frac{2 (2-\gamma -2 \log (2))}{\pi ^2-8}$ leading to the estimate
$$x\sim\frac 32 +\frac{2(\gamma +2 \log (2)-2)}{\left(\pi ^2-8\right)^3} A$$ where
$$A=28 \zeta (3) (\gamma +2 \log (2)-2)+128-32 \gamma -16 \pi ^2+\pi ^4-64 \log (2)$$
$$x \approx 1.461632068$$ while the "exact" solution is $1.461632145$.