Integration Questions

Question Number 208692 by Ghisom last updated on 21/Jun/24

$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}{x}\mathrm{ln}\:\mathrm{sin}\:{x}\:{dx}=? \\$$

Answered by Berbere last updated on 21/Jun/24

$${ln}\left({sin}\left({x}\right)\right)=−{ln}\left(\mathrm{2}\right)−\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left(\mathrm{2}{kx}\right)}{{k}} \\$$$${proof}\:{ln}\left({sin}\left({x}\right)\right)={Re}\left({ln}\left({sin}\left({x}\right)\right)\right. \\$$$${ln}\left({sin}\left({x}\right)={ln}\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\right)\right. \\$$$${ln}\left({sin}\left({x}\right)\right)={ln}\left({e}^{{ix}} \left(\mathrm{1}−{e}^{−\mathrm{2}{ix}} \right)\right)−{ln}\left(\mathrm{2}\right)=={ix}−{ln}\left(\mathrm{2}\right)−\Sigma\frac{{e}^{−\mathrm{2}{kix}} }{{k}} \\$$$${Reln}\left({sin}\left({x}\right)\right)=−{ln}\left(\mathrm{2}\right)−\Sigma\frac{{cos}\left(\mathrm{2}{kx}\right)}{{k}} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {xln}\left({sin}\left({x}\right)\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\left(−{ln}\left(\mathrm{2}\right)−\Sigma\frac{{cos}\left(\mathrm{2}{kx}\right)}{{k}}\right) \\$$$$=\left[−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{k}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {xcos}\left(\mathrm{2}{kx}\right){d}\bar {{x}} \\$$$$\int{xcos}\left(\mathrm{2}{kx}\right){dx}={x}.\frac{{sin}\left(\mathrm{2}{kx}\right)}{\mathrm{2}{k}}+\frac{{cos}\left(\mathrm{2}{kx}\right)}{\mathrm{4}{k}^{\mathrm{2}} } \\$$$$.−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}{ln}\left(\mathrm{2}\right)−\underset{{k}} {\sum}\frac{\mathrm{1}}{{k}}\left[\frac{{sin}\left(\mathrm{2}{kx}\right)}{\mathrm{2}{k}}+\frac{{cos}\left(\mathrm{2}{kx}\right)}{\mathrm{4}{k}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}{ln}\left(\mathrm{2}\right)−\Sigma\frac{\left(−\mathrm{1}\right)^{{k}} −\mathrm{1}}{\mathrm{4}{k}^{\mathrm{3}} } \\$$$$=−\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{ln}\left(\mathrm{2}\right)+\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{3}} }=−\frac{\pi^{\mathrm{3}} }{\mathrm{8}}{ln}\left(\mathrm{2}\right)+\frac{\mathrm{7}}{\mathrm{16}}\zeta\left(\mathrm{3}\right) \\$$

Commented by Ghisom last updated on 22/Jun/24

$$\mathrm{thank}\:\mathrm{you} \\$$