Operation Research Questions

Question Number 135127 by Dwaipayan Shikari last updated on 10/Mar/21

$$\int_{\mathrm{0}} ^{\mathrm{1}} {log}^{\mathrm{2}} \left(\Gamma\left({x}\right)\right){dx} \\$$

Answered by mathmax by abdo last updated on 11/Mar/21

$$\mathrm{we}\:\mathrm{have}\:\Gamma\left(\mathrm{x}\right).\Gamma\left(\mathrm{1}−\mathrm{x}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{x}\right)}\:\Rightarrow \\$$$$\mathrm{log}\left(\Gamma\left(\mathrm{x}\right)\right)+\mathrm{log}\left(\Gamma\left(\mathrm{1}−\mathrm{x}\right)\right)=\mathrm{log}\left(\pi\right)−\mathrm{log}\left(\mathrm{sin}\left(\pi\mathrm{x}\right)\right)\:\Rightarrow \\$$$$\mathrm{log}\left(\Gamma\left(\mathrm{x}\right)\right)^{\mathrm{2}} \:+\mathrm{2log}\left(\Gamma\left(\mathrm{x}\right)\right)\mathrm{log}\left(\Gamma\left(\mathrm{1}−\mathrm{x}\right)\right)+\mathrm{log}^{\mathrm{2}} \left(\Gamma\left(\mathrm{1}−\mathrm{x}\right)\right) \\$$$$=\mathrm{log}^{\mathrm{2}} \left(\pi\right)−\mathrm{2log}\pi\:\mathrm{log}\left(\mathrm{sin}\left(\pi\mathrm{x}\right)\right)+\mathrm{log}^{\mathrm{2}} \left(\mathrm{sin}\left(\pi\mathrm{x}\right)\right)\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}^{\mathrm{2}} \left(\Gamma\left(\mathrm{x}\right)\right)\mathrm{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}^{\mathrm{2}} \left(\Gamma\left(\mathrm{1}−\mathrm{x}\right)\right)\mathrm{dx}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\Gamma\left(\mathrm{x}\right)\right)\mathrm{log}\left(\Gamma\left(\mathrm{1}−\mathrm{x}\right)\right)\mathrm{dx} \\$$$$=\mathrm{log}^{\mathrm{2}} \left(\pi\right)−\mathrm{2log}\pi\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{sin}\left(\pi\mathrm{x}\right)\right)\mathrm{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{log}^{\mathrm{2}} \left(\mathrm{sin}\left(\pi\mathrm{x}\right)\right)\mathrm{dx} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}^{\mathrm{2}} \left(\Gamma\left(\mathrm{1}−\mathrm{x}\right)\right)\mathrm{dx}\:=_{\mathrm{1}−\mathrm{x}=\mathrm{t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}^{\mathrm{2}} \left(\Gamma\left(\mathrm{t}\right)\right)\mathrm{dt} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{log}\left(\mathrm{sin}\left(\pi\mathrm{x}\right)\right)\mathrm{dx}\:=_{\pi\mathrm{x}=\mathrm{t}} \:\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \mathrm{log}\left(\mathrm{sint}\right)\mathrm{dt} \\$$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{log}\left(\mathrm{sint}\right)\mathrm{dt}\:+\frac{\mathrm{1}}{\pi}\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{log}\left(\mathrm{sint}\right)\mathrm{dt}\:\left(\rightarrow\mathrm{t}=\frac{\pi}{\mathrm{2}}+\mathrm{u}\right) \\$$$$=\frac{\mathrm{1}}{\pi}\left(−\frac{\pi}{\mathrm{2}}\mathrm{log2}\right)+\frac{\mathrm{1}}{\pi}\left(−\frac{\pi}{\mathrm{2}}\mathrm{log2}\right)\:=−\mathrm{log}\left(\mathrm{2}\right) \\$$$$\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{log}^{\mathrm{2}} \left(\Gamma\left(\mathrm{x}\right)\right)\mathrm{dx}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\Gamma\left(\mathrm{x}\right)\right).\mathrm{log}\left(\Gamma\left(\mathrm{1}−\mathrm{x}\right)\right)\mathrm{dx} \\$$$$=\mathrm{log}^{\mathrm{2}} \pi+\mathrm{2log}\pi\mathrm{log2}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{log}^{\mathrm{2}} \left(\mathrm{sin}\left(\pi\mathrm{x}\right)\right)\mathrm{dx}\:\:...\mathrm{be}\:\mathrm{continued}... \\$$