Question Number 150990 by talminator2856791 last updated on 17/Aug/21
$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}\:=\:\int_{\mathrm{0}\:} ^{\:\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{{n}} }\:=\:? \\ $$$$\: \\ $$$$\: \\ $$
Answered by Ar Brandon last updated on 17/Aug/21
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{{n}} }=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{1}+{u}}{du} \\ $$$$\:\:=\frac{\mathrm{1}}{{n}}\beta\left(\frac{\mathrm{1}}{{n}},\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)=\frac{\pi}{{n}\mathrm{sin}\left(\frac{\pi}{{n}}\right)} \\ $$