Question Number 159592 by mnjuly1970 last updated on 19/Nov/21
$$ \\ $$$$\:\:\:{calculate}: \\ $$$$ \\ $$$$\:\:\:\mathcal{I}\::=\int_{\mathrm{0}} ^{\:\infty} \left(\frac{{arctan}\left({x}\right)}{{x}}\right)^{\mathrm{3}} {dx}=? \\ $$$$ \\ $$
Answered by mindispower last updated on 19/Nov/21
$${byart}\:=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{arctan}^{\mathrm{2}} \left({x}\right)}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\mathrm{3}\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}−\frac{\mathrm{1}}{\mathrm{2}}.\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$$=\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{xcos}\left({x}\right)}{{sin}\left({x}\right)}−\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$$$−\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx}−\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\ $$
Commented by mnjuly1970 last updated on 19/Nov/21
$${thanks}\:{alot}\:{sir}\:{power} \\ $$
Commented by mindispower last updated on 19/Nov/21
$${withe}\:{pleasur}\:{sir} \\ $$