Question Number 144499 by mathmax by abdo last updated on 25/Jun/21
$$\mathrm{find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{n}} \right)}{\mathrm{x}^{\mathrm{n}} }\mathrm{dx}\:\:\:\left(\mathrm{n}\geqslant\mathrm{2}\right)\:\mathrm{natural} \\ $$
Answered by mindispower last updated on 27/Jun/21
$${by}\:{part}=\left[−\frac{{arctan}\left({x}^{{n}} \right)}{\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{1}} }\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{n}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }{dx} \\ $$$$=\frac{{n}}{{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }{dx},{x}^{\mathrm{2}{n}} ={u} \\ $$$$=\frac{{n}}{{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}{n}}−\mathrm{1}} }{\mathrm{2}{n}\left(\mathrm{1}+{u}\right)}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}{n}}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}{n}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}} }{du}=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\beta\left(\frac{\mathrm{1}}{\mathrm{2}{n}},\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$