Question Number 131678 by liberty last updated on 07/Feb/21
$$\:\mathrm{Nice}\:\mathrm{calculus}\: \\ $$$$\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}−\mathrm{2}} \:−\mathrm{1}}{\mathrm{x}^{\mathrm{n}} \:+\mathrm{1}}\:\mathrm{dx}\:? \\ $$
Answered by Dwaipayan Shikari last updated on 07/Feb/21
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}−\mathrm{2}} }{{x}^{{n}} +\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{1}+{x}^{{n}} }{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{{n}} ={u} \\ $$$$=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \frac{{x}^{−\mathrm{1}} }{{u}+\mathrm{1}}−\frac{{x}^{\mathrm{1}−{n}} }{{u}+\mathrm{1}}{du} \\ $$$$=\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{−\frac{\mathrm{1}}{{n}}} }{{u}+\mathrm{1}}{du}−\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{\frac{\mathrm{1}−{n}}{{n}}} }{\mathrm{1}+{u}}{du} \\ $$$$=\frac{\mathrm{1}}{{n}}.\frac{\pi}{{sin}\left(\pi\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\right)}−\frac{\mathrm{1}}{{n}}.\frac{\pi}{{sin}\left(\frac{\pi}{{n}}\right)}=\mathrm{0} \\ $$
Answered by Dwaipayan Shikari last updated on 07/Feb/21
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{n}−\mathrm{2}} −\mathrm{1}}{{x}^{{n}} +\mathrm{1}}{dx}\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{1}}{{t}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{2}} −{t}^{{n}} }{{t}^{{n}} +\mathrm{1}}.\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt}=−\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{n}−\mathrm{2}} −\mathrm{1}}{{t}^{{n}} +\mathrm{1}}{dt}={I} \\ $$$$\mathrm{2}{I}=\mathrm{0}\Rightarrow{I}=\mathrm{0} \\ $$