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Question Number 127952 by rs4089 last updated on 03/Jan/21

Answered by Ar Brandon last updated on 03/Jan/21

$$\mathrm{\Psi }={\int }_0^1{\mathrm{x}}^2\sqrt{\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}}\mathrm{dx}={\int }_0^1\{\frac{{\mathrm{x}}^2}{\sqrt{1-{\mathrm{x}}^4}}-\frac{{\mathrm{x}}^4}{\sqrt{1-{\mathrm{x}}^4}}\}\mathrm{dx}$$$${\mathrm{x}}^4=\mathrm{u}\Rightarrow {4\mathrm{x}}^3\mathrm{dx}=\mathrm{du}\Rightarrow \mathrm{dx}=\frac{\mathrm{du}}{{4\mathrm{u}}^{\frac{3}{4}}}$$$$\mathrm{\Psi }={\int }_0^1\{\frac{{\mathrm{u}}^{\frac{1}{2}}}{\sqrt{1-\mathrm{u}}}-\frac{\mathrm{u}}{\sqrt{1-\mathrm{u}}}\}\frac{\mathrm{du}}{{4\mathrm{u}}^{\frac{3}{4}}}=\frac{1}{4}{\int }_0^1\{{\mathrm{u}}^{-\frac{1}{4}}{(1-\mathrm{u})}^{-\frac{1}{2}}-{\mathrm{u}}^{\frac{1}{4}}{(1-\mathrm{u})}^{-\frac{1}{2}}\}\mathrm{du}$$$$=\frac{1}{4}\{\beta (\frac{3}{4},\frac{1}{2})-\beta (\frac{5}{4},\frac{1}{2})\}=\frac{1}{4}\{\frac{\mathrm{\Gamma }\left(\frac{3}{4}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{5}{4}\right)}-\frac{\mathrm{\Gamma }\left(\frac{5}{4}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{7}{4}\right)}\}$$

Commented by Ar Brandon last updated on 03/Jan/21

$$=\frac{1}{4}\{\frac{\sqrt{\pi }\mathrm{\Gamma }\left(\frac{3}{4}\right)}{\frac{1}{4}\mathrm{\Gamma }\left(\frac{1}{4}\right)}-\frac{\sqrt{\pi }\frac{1}{4}\mathrm{\Gamma }\left(\frac{1}{4}\right)}{\frac{3}{4}\mathrm{\Gamma }\left(\frac{3}{4}\right)}\}=\sqrt{\pi }\left[\frac{3{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right)-\frac{1}{4}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{3\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{3}{4}\right)}\right]$$$$=\sqrt{\pi }\left[\frac{3{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right)-\frac{1}{4}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{3\frac{\pi }{\sin \left(\frac{\pi }{4}\right)}}\right]=\frac{1}{3\sqrt{2\pi }}[3{\mathrm{\Gamma }}^2\left(\frac{3}{4}\right)-\frac{1}{4}{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)]$$

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