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Question Number 150103 by mnjuly1970 last updated on 09/Aug/21

$$\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\pi} {sin}^{\:\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right).\:\mathrm{ln}\left(\:{sin}\:\left({x}\right)\:\right){dx}=? \\$$$$\\$$

Answered by Ar Brandon last updated on 09/Aug/21

$$\Omega=\int_{\mathrm{0}} ^{\pi} \mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}\centerdot\mathrm{ln}\left(\mathrm{sin}{x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}\centerdot\mathrm{ln}\left(\mathrm{sin}{x}\right){dx} \\$$$$\:\:\:\:=\mathrm{2}\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha−\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right){dx}=\mathrm{2}\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{1}} \beta\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$$$\:\:\:\:=\mathrm{2}\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{1}} \frac{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\right)}=\mathrm{2}\sqrt{\pi}\frac{\partial}{\partial\alpha}\mid_{\alpha=\mathrm{1}} \frac{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\right)} \\$$$$\:\:\:\:=\sqrt{\pi}\mid_{\alpha=\mathrm{1}} \frac{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma'\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\right)−\Gamma\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma'\left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}\right)} \\$$$$\:\:\:\:=\sqrt{\pi}\left[\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\left(\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{5}}{\mathrm{4}}\right)\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{4}}\right)}\right] \\$$$$\:\:\:\:=\mathrm{4}\sqrt{\pi}\centerdot\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\left(\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{4}\right) \\$$$$\:\:\:\:=\frac{\mathrm{4}\pi\sqrt{\mathrm{2}\pi}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\left(\pi−\mathrm{4}\right)=\frac{\mathrm{4}\pi\left(\pi−\mathrm{4}\right)\sqrt{\mathrm{2}\pi}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)} \\$$$$\\$$