Question Number 161818 by HongKing last updated on 22/Dec/21
$$\mathrm{Show}\:\mathrm{that}: \\ $$$$\Phi\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\sqrt{\frac{\mathrm{1}\:-\:\mathrm{x}^{\mathrm{2}} }{\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} }}\:\mathrm{dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{4}}\:\left(\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:-\:\mathrm{4}\:\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\right) \\ $$$$\mathrm{where}:\:\Gamma-\mathrm{Gamma}\:\mathrm{function} \\ $$
Answered by Lordose last updated on 22/Dec/21
$$\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\mathrm{dx} \\ $$$$\Phi\:\overset{\mathrm{x}=\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}} } {=}\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} −\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\:\sqrt{\mathrm{1}−\mathrm{u}}}\mathrm{du} \\ $$$$\Phi\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−\mathrm{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:−\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{u}^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−\mathrm{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \mathrm{du}\right) \\ $$$$\Phi\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:−\:\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}\right) \\ $$$$\boldsymbol{\Gamma}\left(\mathrm{1}+\mathrm{x}\right)\:=\:\mathrm{x}\boldsymbol{\Gamma}\left(\mathrm{x}\right) \\ $$$$\Phi\:=\:\frac{\sqrt{\boldsymbol{\pi}}}{\mathrm{4}}\left(\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:−\:\mathrm{4}\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\right) \\ $$