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Question Number 126641 by TITA last updated on 22/Dec/20

Commented by TITA last updated on 22/Dec/20

$$\mathrm{please}\mathrm{help}$$

Answered by Lordose last updated on 22/Dec/20

$$$$$$\mathrm{\Omega }={\int }_0^1\frac{1}{\sqrt{1-\sqrt[\mathrm{n}]{\mathrm{x}}}}\mathrm{dx}\stackrel{\mathrm{x}={\mathrm{u}}^{\mathrm{n}}}{=}\mathrm{n}{\int }_0^1{\mathrm{u}}^{\mathrm{n}-1}{(1-\mathrm{u})}^{-\frac{1}{2}}\mathrm{du}$$$$\mathrm{\Omega }=\mathbf{nB}(\mathrm{n},\frac{1}{2})=\frac{\mathbf{n}\mathbf{\Gamma }\left(\mathbf{n}\right)\mathbf{\Gamma }\left(\frac{1}{2}\right)}{\mathbf{\Gamma }(\mathbf{n}+\frac{1}{2})}=\frac{\mathbf{n}\mathbf{\Gamma }\left(\mathbf{n}\right)\sqrt{\mathit{\pi }}}{\mathbf{\Gamma }(\mathbf{n}+\frac{1}{2})}$$

Commented by TITA last updated on 22/Dec/20

$$\mathrm{thanks}$$

Answered by Dwaipayan Shikari last updated on 23/Dec/20

$$x={sin}^{2n}\theta \Rightarrow 1=2{nsin}^{2n-1}\theta cos\theta \frac{d\theta }{dx}$$$$={\int }_0^1\frac{1}{\sqrt{1-\sqrt[n]{x}}}dx=2n{\int }_0^{\frac{\pi }{2}}\frac{{sin}^{2n-1}\theta cos\theta }{\sqrt{1-{sin}^2\theta }}d\theta$$$$=2n{\int }_0^{\frac{\pi }{2}}{sin}^{2n-1}{cos}^{2\left(\frac{1}{2}\right)-1}\theta d\theta =n\frac{\mathrm{\Gamma }\left(n\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(n+\frac{1}{2})}=n\sqrt{\pi }.\frac{\mathrm{\Gamma }\left(n\right)}{\mathrm{\Gamma }(n+\frac{1}{2})}$$$$=\frac{\mathrm{\Gamma }(n+1)\sqrt{\pi }}{\mathrm{\Gamma }(n+\frac{1}{2})}$$

Answered by mathmax by abdo last updated on 22/Dec/20

$$\mathrm{A}={\int }_0^1\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^{\frac{1}{\mathrm{n}}}}}\mathrm{changement}{\mathrm{x}}^{\frac{1}{\mathrm{n}}}=\mathrm{t}\mathrm{give}\mathrm{x}={\mathrm{t}}^{\mathrm{n}}\Rightarrow$$$$\mathrm{A}={\int }_0^1\frac{{\mathrm{nt}}^{\mathrm{n}-1}\mathrm{dt}}{\sqrt{1-\mathrm{t}}}=\mathrm{n}{\int }_0^1{\mathrm{t}}^{\mathrm{n}-1}{(1-\mathrm{t})}^{-\frac{1}{2}}\mathrm{dt}=\mathrm{n}{\int }_0^1{\mathrm{t}}^{\mathrm{n}-1}{(1-\mathrm{t})}^{\frac{1}{2}-1}\mathrm{dt}$$$$=\mathrm{nB}(\mathrm{n},\frac{1}{2})=\mathrm{n}.\frac{\mathrm{\Gamma }\left(\mathrm{n}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(\mathrm{n}+\frac{1}{2})}=\frac{\mathrm{n}(\mathrm{n}-1)!\sqrt{\pi }}{\mathrm{\Gamma }(\mathrm{n}+\frac{1}{2})}(\mathrm{n}⩾1)$$

Answered by Olaf last updated on 22/Dec/20

$$...$$$$\mathrm{\Omega }=\frac{n\mathrm{\Gamma }\left(n\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(n+\frac{1}{2})}=\frac{\sqrt{\pi }.n!}{\frac{\left(2n\right)!}{2^{2n}n!}\sqrt{\pi }}=\frac{2^{2n}n{!}^2}{\left(2n\right)!}$$

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