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Question Number 127779 by Bird last updated on 02/Jan/21

$$find{A}_n={\int }_0^{+\mathrm{\infty }}\frac{dx}{{(x^2+1)}^n}$$

Answered by Dwaipayan Shikari last updated on 02/Jan/21

$${\int }_0^{\mathrm{\infty }}\frac{1}{{(1+x^2)}^n}dx{x}^2=u\Rightarrow 2x=\frac{du}{dx}$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{u^{-\frac{1}{2}}}{{(1+u)}^n}du\frac{u}{1+u}=t\Rightarrow \frac{1}{{(1+u)}^2}=\frac{dt}{du}$$$$=\frac{1}{2}{\int }_0^1{\left(\frac{t}{1-t}\right)}^{-\frac{1}{2}}{(1+u)}^{2-n}dt=\frac{1}{2}{\int }_0^1{t}^{-\frac{1}{2}}{(1-t)}^{\frac{1}{2}}{(1-t)}^{n-2}dt$$$$=\frac{1}{2}\beta (\frac{1}{2},n-\frac{1}{2})=\frac{1}{2}.\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(n-\frac{1}{2})}{\mathrm{\Gamma }\left(n\right)}=\frac{\sqrt{\pi }}{2}.\frac{\mathrm{\Gamma }(n-\frac{1}{2})}{\mathrm{\Gamma }\left(n\right)}$$

Answered by Ar Brandon last updated on 02/Jan/21

$${\mathrm{A}}_{\mathrm{n}}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{dx}}{{({\mathrm{x}}^2+1)}^{\mathrm{n}}}$$$${\mathrm{x}}^2=\mathrm{u}\Rightarrow 2\mathrm{xdx}=\mathrm{du}$$$${\mathrm{A}}_{\mathrm{n}}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{-\frac{1}{2}}}{{(1+\mathrm{u})}^{\mathrm{n}}}\mathrm{du}=\frac{1}{2}\beta (\frac{1}{2},\mathrm{n}-\frac{1}{2})$$$$=\frac{1}{2}\cdot \frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(\mathrm{n}-\frac{1}{2})}{\mathrm{\Gamma }\left(\mathrm{n}\right)}=\frac{\sqrt{\pi }}{2(\mathrm{n}-1)!}\mathrm{\Gamma }(\mathrm{n}-\frac{1}{2})$$$$\mathrm{\Gamma }((\mathrm{n}-1)+\frac{1}{2})=\frac{\sqrt{\pi }}{2^{2\mathrm{n}-3}}\cdot \frac{\mathrm{\Gamma }(2\mathrm{n}-2)}{\mathrm{\Gamma }(\mathrm{n}-1)}=\frac{\sqrt{\pi }}{2^{2\mathrm{n}-3}}\cdot \frac{(2\mathrm{n}-3)!}{(\mathrm{n}-2)!}$$$${\mathrm{A}}_{\mathrm{n}}=\frac{\sqrt{\pi }}{2(\mathrm{n}-1)!}\cdot \frac{\sqrt{\pi }}{2^{2\mathrm{n}-3}}\cdot \frac{(2\mathrm{n}-3)!}{(\mathrm{n}-2)!}=\frac{\pi (2\mathrm{n}-3)!}{2^{2\mathrm{n}-2}(\mathrm{n}-1)!(\mathrm{n}-2)!}$$

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