Question Number 123745 by mnjuly1970 last updated on 27/Nov/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:\:{calculus}… \\ $$$$\:\:{evaluation}\:\:{of}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}\left({x}\right){log}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:{by}\:{using}\:{the}\:{euler}\:{beta}\:{and}\:{gamma}\:{function}: \\ $$$$\:\:\:\:\:\:\:\:\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \left({x}\right){dx} \\ $$$$\:\:\:\frac{{d}\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)}{{dp}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{2}{sin}^{\mathrm{2}{p}−\mathrm{1}} \left({x}\right){ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{p}−\mathrm{1}} \left({x}\right){ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{d}\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)}{{dp}}\right]_{{p}=\mathrm{1}} \\ $$$$\:\frac{{d}\left(\beta\left({p},\frac{\mathrm{1}}{\mathrm{2}}\right)\right)}{{dp}}=\sqrt{\pi}\left[\frac{\Gamma'\left({p}\right)\Gamma\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\Gamma'\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({p}\right)}{\Gamma^{\mathrm{2}} \left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)}\right]_{{p}=\mathrm{1}} \\ $$$$\:\Omega=\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\pi}\left[\frac{\Gamma^{'} \left(\mathrm{1}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\Gamma'\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\right]\right) \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}}\left[\frac{−\gamma\left(\frac{\sqrt{\pi}}{\mathrm{2}}\right)−\frac{\sqrt{\pi}}{\mathrm{2}}\left(\mathrm{2}−\gamma−\mathrm{2}{ln}\left(\mathrm{2}\right)\right)}{\frac{\pi}{\mathrm{4}}}\right] \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}}\ast\frac{\sqrt{\pi}}{\mathrm{2}}\left(\frac{−\gamma−\mathrm{2}+\gamma+\mathrm{2}{ln}\left(\mathrm{2}\right)}{\frac{\pi}{\mathrm{4}}}\right) \\ $$$$={ln}\left(\mathrm{2}\right)−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:…\:{m}.{n}.{july}.\mathrm{1970}… \\ $$$$\:\:\: \\ $$