Question Number 153164 by alisiao last updated on 05/Sep/21
$$\int_{−\mathrm{1}} ^{\:\mathrm{1}} \boldsymbol{{ln}}\left(\boldsymbol{\Gamma}\left(\boldsymbol{{x}}\right)\right)\:\boldsymbol{{dx}} \\ $$
Answered by Ar Brandon last updated on 05/Sep/21
$$\Omega=\int_{−\mathrm{1}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left({x}\right)\right){dx}=\underset{\Omega_{\mathrm{1}} } {\underbrace{\int_{−\mathrm{1}} ^{\mathrm{0}} \mathrm{ln}\left(\Gamma\left({x}\right)\right){dx}}}+\underset{\Omega_{\mathrm{2}} } {\underbrace{\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left({x}\right)\right){dx}}} \\ $$$$\Omega_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left({x}\right)\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\mathrm{ln}\left(\Gamma\left({x}\right)\right)+\mathrm{ln}\left(\Gamma\left(\mathrm{1}−{x}\right)\right){dx}\right\}{dx} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right)\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\frac{\pi}{\mathrm{sin}\left(\pi{x}\right)}\right){dx} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{ln}\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{sin}\left(\pi{x}\right)\right){dx}=\frac{\mathrm{ln}\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}\pi}\underset{=\mathrm{0}} {\underbrace{\int_{\mathrm{0}} ^{\pi} \mathrm{ln}\left(\mathrm{sin}\left({u}\right)\right){du}}} \\ $$