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Question Number 161818 by HongKing last updated on 22/Dec/21

$$\mathrm{Show}\mathrm{that}:$$$$\mathrm{\Phi }=\underset{0}{\stackrel{1}{\int }}\sqrt{\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}}\mathrm{dx}=\frac{\sqrt{\pi }}{4}(\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{3}{4}\right)}-4\frac{\mathrm{\Gamma }\left(\frac{3}{4}\right)}{\mathrm{\Gamma }\left(\frac{1}{4}\right)})$$$$\mathrm{where}:\mathrm{\Gamma }-\mathrm{Gamma}\mathrm{function}$$

Answered by Lordose last updated on 22/Dec/21

$$\mathrm{\Phi }={\int }_0^1\sqrt{\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}}\mathrm{dx}={\int }_0^1\frac{1-{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^4}}\mathrm{dx}$$$$\mathrm{\Phi }\stackrel{\mathrm{x}={\mathrm{u}}^{\frac{1}{4}}}{=}\frac{1}{4}{\int }_0^1\frac{{\mathrm{u}}^{\frac{1}{4}-1}-{\mathrm{u}}^{\frac{1}{2}+\frac{1}{4}-1}}{\sqrt{1-\mathrm{u}}}\mathrm{du}$$$$\mathrm{\Phi }=\frac{1}{4}({\int }_0^1{\mathrm{u}}^{\frac{1}{4}-1}{(1-\mathrm{u})}^{\frac{1}{2}-1}-{\int }_0^1{\mathrm{u}}^{\frac{3}{4}-1}{(1-\mathrm{u})}^{\frac{1}{2}-1}\mathrm{du})$$$$\mathrm{\Phi }=\frac{1}{4}(\frac{\mathbf{\Gamma }\left(\frac{1}{4}\right)\mathbf{\Gamma }\left(\frac{1}{2}\right)}{\mathbf{\Gamma }\left(\frac{3}{4}\right)}-\frac{\mathbf{\Gamma }\left(\frac{3}{4}\right)\mathbf{\Gamma }\left(\frac{1}{2}\right)}{\mathbf{\Gamma }\left(\frac{5}{4}\right)})$$$$\mathbf{\Gamma }(1+\mathrm{x})=\mathrm{x}\mathbf{\Gamma }\left(\mathrm{x}\right)$$$$\mathrm{\Phi }=\frac{\sqrt{\mathit{\pi }}}{4}(\frac{\mathbf{\Gamma }\left(\frac{1}{4}\right)}{\mathbf{\Gamma }\left(\frac{3}{4}\right)}-4\frac{\mathbf{\Gamma }\left(\frac{3}{4}\right)}{\mathbf{\Gamma }\left(\frac{1}{4}\right)})$$

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