# Is there a nice solution to the equation $\Psi(x)=\ln(\pi)$ with a positive real $x$?

I tried to find a nice solution to the following equation: $$\Psi(x)=\ln(\pi)$$ with $x\in\Bbb R_{\ge0}$ and where $\Psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$. Is there a nice expression for x satisfying this equation?

• This is not an "algebra-precalculus" problem :) – user142299 Jul 5 '14 at 20:44
• You can use $\psi(x)=\frac{d}{dx}\ln\left(\Gamma (x)\right)$. But that doesn't help much. I doubt there are analytical solutions to this. – ClassicStyle Jul 5 '14 at 21:27
• You can do a Newton-Rapson starting with $\large x = \pi$. That yields $\large x \approx 3.62847320$. – Felix Marin Jul 6 '14 at 4:38

Approximates of many real solutions can be computed thanks to numerical methods : http://www.wolframalpha.com/input/?i=solve+digamma%28x%29%3Dln%28pi%29+for+x Only one positive real is $x=3.6284732024302883900664192...$
On a formal viewpoint, one could write : $x=\psi^{-1}(ln(\pi))$ but this is of no interest.