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Question Number 42157 by Tawa1 last updated on 19/Aug/18

$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{(1+{\mathrm{x}}^{\mathrm{n}})}$$

Commented by math khazana by abdo last updated on 19/Aug/18

$$changement{x}^n=tgivex=t^{\frac{1}{n}}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^n}{\int }_0^{\mathrm{\infty }}\frac{1}{1+t}\frac{1}{n}{t}^{\frac{1}{n}-1}dt$$$$=\frac{1}{n}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{n}-1}}{1+t}=\frac{1}{n}\frac{\pi }{sin\left(\frac{\pi }{n}\right)}=\frac{\pi }{nsin\left(\frac{\pi }{n}\right)}$$$$(wecantaken⩾2)andIhaveusedthe$$$$result{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dt=\frac{\pi }{sin\left(\pi a\right)}if0<a<1.$$

Commented by Tawa1 last updated on 19/Aug/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by math khazana by abdo last updated on 20/Aug/18

$$thankssir.$$

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