Differentiation Questions

Question Number 159854 by mnjuly1970 last updated on 21/Nov/21

$$\\$$$$\\$$$$\:\:\:\:\:\Omega\::=\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} {x}.{ln}\left({sin}\left({x}\right)\right){dx}=\:? \\$$$$\\$$$$\\$$

Answered by mindispower last updated on 22/Nov/21

$${ln}\left({sin}\left({x}\right)\right)=−{ln}\left(\mathrm{2}\right)−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}} \\$$$$\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}\left(−{ln}\left(\mathrm{2}\right)−\Sigma\frac{{cos}\left(\mathrm{2}{nx}\right)}{{n}}\right){dx} \\$$$$=−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} −\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cos}\left(\mathrm{2}{nx}\right){xdx} \\$$$$=−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} −\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}\left[\frac{{sin}\left({n}\frac{\pi}{\mathrm{2}}\right)}{\mathrm{2}{n}}.\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {sin}\left(\mathrm{2}{nx}\right){dx}\right. \\$$$$=−\frac{{ln}\left(\mathrm{2}\right)\pi^{\mathrm{2}} }{\mathrm{32}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\frac{\pi}{\mathrm{8}}.\frac{{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{3}} }\left[{cos}\left(\frac{{n}\pi}{\mathrm{2}}\right)−\mathrm{1}\right]\right) \\$$$$=−\frac{{ln}\left(\mathrm{2}\right)\pi^{\mathrm{2}} }{\mathrm{32}}−\frac{\pi}{\mathrm{8}}.\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{8}{n}^{\mathrm{3}} }+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}} \\$$$$=\frac{−{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} −\frac{\pi}{\mathrm{8}}{G}+\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{32}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)\zeta\left(\mathrm{3}\right) \\$$$$=−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{32}}\pi^{\mathrm{2}} −\frac{\pi}{\mathrm{8}}{G}+\frac{\mathrm{35}}{\mathrm{128}}\zeta\left(\mathrm{3}\right) \\$$$$\\$$$$\\$$

Commented by mnjuly1970 last updated on 23/Nov/21

$${thanks}\:{alot}\:\:{sir}\:\mathrm{Power}... \\$$

Commented by mindispower last updated on 23/Nov/21

$${withe}\:{pleasur}\:{sir}\:{have}\:{a}\:{nice}\:{day} \\$$