Integration Questions

Question Number 156028 by mnjuly1970 last updated on 07/Oct/21

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

Answered by Ar Brandon last updated on 07/Oct/21

$${f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha} {xdx}=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\sqrt{\pi}}{\mathrm{2}\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)} \\$$$${f}\:'\left(\alpha\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\alpha} {x}\mathrm{ln}\left(\mathrm{sin}{x}\right){dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\left\{\frac{\Gamma\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)\Gamma'\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)−\Gamma\left(\frac{\alpha+\mathrm{1}}{\mathrm{2}}\right)\Gamma'\left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}+\mathrm{1}\right)}\right\} \\$$$${f}\:'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{sin}{x}}\mathrm{ln}\left(\mathrm{sin}{x}\right){dx}=\frac{\sqrt{\pi}}{\mathrm{2}}\left\{\frac{\Gamma\left(\frac{\mathrm{5}}{\mathrm{4}}\right)\Gamma'\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma'\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{5}}{\mathrm{4}}\right)}\right\} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\pi}}{\mathrm{2}}\left\{\frac{\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\left[\psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi\left(\frac{\mathrm{5}}{\mathrm{4}}\right)\right]}{\frac{\mathrm{1}}{\mathrm{16}}\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\right\} \\$$$$... \\$$

Commented by Ar Brandon last updated on 07/Oct/21

$$\Gamma\left({x}\right)\Gamma\left(\mathrm{1}−{x}\right)=\frac{\pi}{\mathrm{sin}\left(\pi{x}\right)} \\$$$$\psi\left({s}+\mathrm{1}\right)=\frac{\mathrm{1}}{{s}}+\psi\left({s}\right) \\$$$$\psi\left({x}\right)−\psi\left(\mathrm{1}−{x}\right)=−\pi\mathrm{cot}\left(\pi{x}\right),\:{x}=\frac{\mathrm{1}}{\mathrm{4}} \\$$

Commented by mnjuly1970 last updated on 07/Oct/21

$${grateful}\:{mr}\:{brandon}.. \\$$