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

Show that: Φ =∫_( 0) ^( 1) (√((1 - x^2 )/(1 + x^2 ))) dx = ((√π)/4) (((Γ((1/4)))/(Γ((3/4)))) - 4 ((Γ((3/4)))/(Γ((1/4))))) where: Γ-Gamma function

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

Answered by Lordose last updated on 22/Dec/21

Φ = ∫_0 ^( 1) (√((1−x^2 )/(1+x^2 )))dx = ∫_0 ^( 1) ((1−x^2 )/( (√(1−x^4 ))))dx Φ =^(x=u^(1/4) ) (1/4)∫_0 ^( 1) ((u^((1/4)−1) −u^((1/2)+(1/4)−1) )/( (√(1−u))))du Φ = (1/4)(∫_0 ^( 1) u^((1/4)−1) (1−u)^((1/2)−1) − ∫_0 ^( 1) u^((3/4)−1) (1−u)^((1/2)−1) du) Φ = (1/4)(((𝚪((1/4))𝚪((1/2)))/(𝚪((3/4)))) − ((𝚪((3/4))𝚪((1/2)))/(𝚪((5/4))))) 𝚪(1+x) = x𝚪(x) Φ = ((√𝛑)/4)(((𝚪((1/4)))/(𝚪((3/4)))) − 4((𝚪((3/4)))/(𝚪((1/4)))))

$$\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \sqrt{\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\mathrm{dx} \\ $$$$\Phi\:\overset{\mathrm{x}=\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}} } {=}\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} −\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\:\sqrt{\mathrm{1}−\mathrm{u}}}\mathrm{du} \\ $$$$\Phi\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−\mathrm{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:−\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{u}^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−\mathrm{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \mathrm{du}\right) \\ $$$$\Phi\:=\:\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:−\:\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}\right) \\ $$$$\boldsymbol{\Gamma}\left(\mathrm{1}+\mathrm{x}\right)\:=\:\mathrm{x}\boldsymbol{\Gamma}\left(\mathrm{x}\right) \\ $$$$\Phi\:=\:\frac{\sqrt{\boldsymbol{\pi}}}{\mathrm{4}}\left(\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:−\:\mathrm{4}\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}\right) \\ $$

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