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Question Number 63662 by mathmax by abdo last updated on 06/Jul/19

$$let{A}_n={\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{1+x^n}dxwithnintegrandn⩾2and0<a<1$$ $$1)calculate{A}_n$$ $$2)findthevaluesof{\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{1+x^2}dxand{\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{1+x^3}dx$$ $$3)calculate{\int }_0^{\mathrm{\infty }}\frac{dx}{\sqrt{x}(1+x^4)}and{\int }_0^{\mathrm{\infty }}\frac{dx}{\left({}^3\sqrt{x^2}\right)(1+x^4)}$$

Commented bymathmax by abdo last updated on 10/Jul/19

$$1)wehave{A}_n={\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{1+x^n}dxchangement{x}^n=tgivex=t^{\frac{1}{n}}\Rightarrow$$ $$A_n={\int }_0^{\mathrm{\infty }}\frac{{\left(t^{\frac{1}{n}}\right)}^{a-1}}{1+t}\frac{1}{n}{t}^{\frac{1}{n}-1}dt=\frac{1}{n}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{a-1}{n}+\frac{1}{n}-1}}{1+t}dt$$ $$=\frac{1}{n}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{a}{n}-1}}{1+t}dt=\frac{1}{n}\frac{\pi }{sin\left(\frac{\pi a}{n}\right)}byuseofresult{\int }_0^{\mathrm{\infty }}\frac{t^{\alpha -1}}{1+t}dt=\frac{\pi }{sin\left(\pi \alpha \right)}$$ $$2){\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{1+x^2}dx=A_2=\frac{\pi }{2sin\left(\frac{\pi a}{2}\right)}$$ $${\int }_0^{\mathrm{\infty }}\frac{x^{a-1}}{1+x^3}dx=A_3=\frac{\pi }{3sin\left(\frac{\pi a}{3}\right)}$$ $$3){\int }_0^{\mathrm{\infty }}\frac{dx}{\sqrt{x}(1+x^4)}={\int }_0^{\mathrm{\infty }}\frac{x^{-\frac{1}{2}}}{1+x^4}dx={\int }_0^{\mathrm{\infty }}\frac{x^{\frac{1}{2}-1}}{1+x^4}(a=\frac{1}{2}andn=4)$$ $$=\frac{\pi }{4sin\left(\frac{\pi }{4}\right)}=\frac{\pi }{4\frac{\sqrt{2}}{2}}=\frac{\pi }{2\sqrt{2}}.$$ $${\int }_0^{\mathrm{\infty }}\frac{dx}{\left({}^3\sqrt{x^2}\right)(1+x^4)}={\int }_0^{\mathrm{\infty }}\frac{x^{-\frac{2}{3}}}{1+x^4}dx={\int }_0^{\mathrm{\infty }}\frac{x^{\frac{1}{3}-1}}{1+x^4}(\to n=4anda=\frac{1}{3})$$ $$=\frac{\pi }{4sin\left(\frac{\pi }{6}\right)}=\frac{\pi }{4.\frac{1}{2}}=\frac{\pi }{2}$$

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