What is the positive solution to $\Gamma'(x)=0$? When is $\Gamma'(x)=0$, given $x>0$?
There is only one value in this range which is $x≈1.461632$, but I can’t seem to get it in exact form.
I’ve tried taking the derivative of the definition of the Gamma function, but that doesn’t seem to get me very far.
Any help is appreciated.
 A: You can obtain decent approximations of the root.
We have
$$\Gamma'(x)=\Gamma (x)\, \psi (x)$$ which means that we look for the zero of $\psi (x)$ (the digamma function). The solution is "close" to $1.5$.
Developed as a series
$$\psi (x)=(2-\gamma -2 \log (2))+\left(\frac{\pi ^2}{2}-4\right)
   \left(x-\frac{3}{2}\right)+(8-7 \zeta
   (3))\left(x-\frac{3}{2}\right)^2 +$$ $$\left(\frac{\pi ^4}{6}-16\right)
   \left(x-\frac{3}{2}\right)^3+ (32-31 \zeta
   (5))\left(x-\frac{3}{2}\right)^4+\left(\frac{\pi ^6}{15}-64\right)
   \left(x-\frac{3}{2}\right)^5+O\left(\left(x-\frac{3}{2}\right)^6\right)$$
Using series reversion
$$x=\frac{3}{2}+t+\frac{2  (7 \zeta (3)-8)}{\pi ^2-8}t^2+$$ $$\frac{ \left(-2688 \zeta
   (3)+1176 \zeta (3)^2+768+96 \pi ^2+8 \pi ^4-\pi ^6\right)}{3 \left(\pi
   ^2-8\right)^2}t^3+O\left(t^4\right)$$ where $t=-\frac{2 (2-\gamma -2 \log (2))}{\pi ^2-8}$.
Using this very truncated series, an approximation of the root is
$x=1.461632068$ while the solution given by Newton method is
$x=1.461632145$.
