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=[−((arctan(x^n ))/((n−1)x^(n−1) ))]_0 ^∞ +(1/((n−1)))∫_0 ^∞ (n/(1+x^(2n) ))dx =(n/(n−1))∫_0 ^∞ (1/(1+x^(2n) ))dx,x^(2n) =u =(n/(n−1))∫_0 ^∞ (u^((1/(2n))−1) /(2n(1+u)))du =(1/(2(n−1)))∫_0 ^∞ (u^((1/(2n))−1) /((1+u)^((1/(2n))+1−(1/(2n))) ))du=(1/(2(n−1)))β((1/(2n)),1−(1/(2n))) =(1/(2(n−1))).(π/(sin((π/(2n)))))](https://www.tinkutara.com/question/Q144654.png)
$${by}\:{part}=\left[−\frac{{arctan}\left({x}^{{n}} \right)}{\left({n}−\mathrm{1}\right){x}^{{n}−\mathrm{1}} }\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{n}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }{dx} \\ $$$$=\frac{{n}}{{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}{n}} }{dx},{x}^{\mathrm{2}{n}} ={u} \\ $$$$=\frac{{n}}{{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}{n}}−\mathrm{1}} }{\mathrm{2}{n}\left(\mathrm{1}+{u}\right)}{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{\mathrm{2}{n}}−\mathrm{1}} }{\left(\mathrm{1}+{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}{n}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}} }{du}=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}\beta\left(\frac{\mathrm{1}}{\mathrm{2}{n}},\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({n}−\mathrm{1}\right)}.\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)} \\ $$