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I-0-dx-1-x-n-




Question Number 150990 by talminator2856791 last updated on 17/Aug/21
                         I = ∫_(0 ) ^( ∞)  (dx/(1+x^n )) = ?
$$\: \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}\:=\:\int_{\mathrm{0}\:} ^{\:\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{{n}} }\:=\:? \\ $$$$\: \\ $$$$\: \\ $$
Answered by Ar Brandon last updated on 17/Aug/21
I=∫_0 ^∞ (dx/(1+x^n ))=(1/n)∫_0 ^∞ (u^((1/n)−1) /(1+u))du    =(1/n)β((1/n),1−(1/n))=(π/(nsin((π/n))))
$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+{x}^{{n}} }=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{1}+{u}}{du} \\ $$$$\:\:=\frac{\mathrm{1}}{{n}}\beta\left(\frac{\mathrm{1}}{{n}},\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)=\frac{\pi}{{n}\mathrm{sin}\left(\frac{\pi}{{n}}\right)} \\ $$

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