Question Number 126465 by mnjuly1970 last updated on 20/Dec/20
$$\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:\mathscr{E}{valuate}\:… \\ $$$$\:\:\:\:\:\:\:\:\phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{sin}\left({ln}\left({x}\right)\right)−{ln}\left({x}\right)}{{ln}^{\mathrm{2}} \left({x}\right)}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathscr{A}{ns}\:::\:{ln}\left(\sqrt{\mathrm{2}}\:\right)+\frac{\pi}{\mathrm{4}}\:−\mathrm{1}\:… \\ $$
Commented by talminator2856791 last updated on 22/Dec/20
$$\:\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{more}\:\mathrm{difficult}\:\mathrm{question}\:\mathrm{than}\:\mathrm{this} \\ $$
Answered by mindispower last updated on 23/Dec/20
$$−{ln}\left({x}\right)={t} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{t}−{sin}\left({t}\right)}{{t}^{\mathrm{2}} }{e}^{−{t}} \\ $$$${sin}\left({t}\right)−{t}=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} {t}^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}{t}^{\mathrm{2}{k}−\mathrm{1}} {e}^{−{t}} {dt} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!}\Gamma\left(\mathrm{2}{k}\right) \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}\left(\mathrm{2}{k}+\mathrm{1}\right)}=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}}−\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{\mathrm{2}{k}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}−\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}}\underset{{k}\geqslant\mathrm{1}} {\sum}\left(−{x}\right)^{{k}} {dx}−\underset{{k}\geqslant\mathrm{1}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} {x}^{\mathrm{2}{k}} {dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}}{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} {dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\mathrm{1}+\frac{\pi}{\mathrm{4}}−{ln}\left(\sqrt{\mathrm{2}}\right) \\ $$
Commented by mnjuly1970 last updated on 24/Dec/20
$${thanks}\:{a}\:{lot}.. \\ $$$$ \\ $$