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Prove-0-1-K-x-3-x-dx-1-96pi-3-1-24-3-24-7-24-11-24-

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Question Number 216819 by MrGaster last updated on 22/Feb/25
Prove:∫_(0 ) ^1 ((K(x))/( (√(3−x))))dx=(1/(96π(√3)))×Γ((1/(24)))Γ((3/(24)))Γ((7/(24)))Γ(((11)/(24)))
$$\mathrm{Prove}:\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{K}}\left({x}\right)}{\:\sqrt{\mathrm{3}−{x}}}{dx}=\frac{\mathrm{1}}{\mathrm{96}\pi\sqrt{\mathrm{3}}}×\Gamma\left(\frac{\mathrm{1}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{11}}{\mathrm{24}}\right) \\ $$
Answered by MrGaster last updated on 24/May/25
K(x)=∫_0 ^(π/2) (dθ/( (√(1−x sin^2 θ)))) ∫_0 ^1 ((K(x))/( (√(3−x))))dx=∫_0 ^1 ∫_0 ^(π/2) (1/( (√(1−x sin^2 θ)))) ((dx dθ)/( (√(3−x)))) =∫_0 ^(π/2) ∫_0 ^1 (1/( (√((1−x sin^2 θ)(3−x)))))dx dθ 1−x sin^2 θ_(A) ×3−x_(B) =(3−x)(1−x sin^2 θ) x=3−(2/(1+t^2 ))⇒dx=((4t)/((1+t^2 )^2 ))dt ∫_0 ^1 (dx/( (√((3−x)(1−x sin^2 θ)))))=∫_0 ^∞ ((4t dt)/((1+t^2 )(√((2/(1+t^2 ))(1−(3−(2/(1+t^2 )))sin^2 θ))))) =(√2)∫_0 ^∞ ((t dt)/((1+t^2 )^(3/2) (√(1−3 sin^2 θ+((2 sin^2 θ)/(1+t^2 )))))) Γ((1/(24)))Γ((3/(24)))Γ((7/(24)))Γ(((11)/(24)))=2^(14) π^2 (√3) ∫_0 ^(π/2) (dθ/( (√(1−(1/4)sin^2 θ))))=((Γ^2 ((1/4)))/(4(√π))) Π_(k=1) ^3 Γ((k/(24)))=2^7 π^(3/2) 3^(1/4) ∫_(0 ) ^1 ((K(x))/( (√(3−x))))dx=((√3)/(96))×((Γ((1/(24)))Γ((3/(24)))Γ((7/(24)))Γ(((11)/(24))))/(2^(14) π^2 (√3))) ∫_(0 ) ^1 ((K(x))/( (√(3−x))))dx=(1/(96π(√3)))×Γ((1/(24)))Γ((3/(24)))Γ((7/(24)))Γ(((11)/(24)))
$$\boldsymbol{\mathrm{K}}\left({x}\right)=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{d}\theta}{\:\sqrt{\mathrm{1}−{x}\:\mathrm{sin}^{\mathrm{2}} \theta}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{K}}\left({x}\right)}{\:\sqrt{\mathrm{3}−{x}}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}\:\mathrm{sin}^{\mathrm{2}} \theta}}\:\frac{{dx}\:{d}\theta}{\:\sqrt{\mathrm{3}−{x}}} \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\left(\mathrm{1}−{x}\:\mathrm{sin}^{\mathrm{2}} \theta\right)\left(\mathrm{3}−{x}\right)}}{dx}\:{d}\theta \\ $$$$\underset{{A}} {\underbrace{\mathrm{1}−{x}\:\mathrm{sin}^{\mathrm{2}} \theta}}\:×\underset{{B}} {\underbrace{\mathrm{3}−{x}}}\:=\left(\mathrm{3}−{x}\right)\left(\mathrm{1}−{x}\:\mathrm{sin}^{\mathrm{2}} \theta\right) \\ $$$${x}=\mathrm{3}−\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }\Rightarrow{dx}=\frac{\mathrm{4}{t}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\:\sqrt{\left(\mathrm{3}−{x}\right)\left(\mathrm{1}−{x}\:\mathrm{sin}^{\mathrm{2}} \theta\right)}}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{4}{t}\:{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\sqrt{\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }\left(\mathrm{1}−\left(\mathrm{3}−\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)\mathrm{sin}^{\mathrm{2}} \theta\right)}} \\ $$$$=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{t}\:{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \theta+\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \theta}{\mathrm{1}+{t}^{\mathrm{2}} }}} \\ $$$$\Gamma\left(\frac{\mathrm{1}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{11}}{\mathrm{24}}\right)=\mathrm{2}^{\mathrm{14}} \pi^{\mathrm{2}} \sqrt{\mathrm{3}} \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{d}\theta}{\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}^{\mathrm{2}} \theta}}=\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{4}\sqrt{\pi}} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\prod}}\Gamma\left(\frac{{k}}{\mathrm{24}}\right)=\mathrm{2}^{\mathrm{7}} \pi^{\mathrm{3}/\mathrm{2}} \mathrm{3}^{\mathrm{1}/\mathrm{4}} \\ $$$$\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{K}}\left({x}\right)}{\:\sqrt{\mathrm{3}−{x}}}{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{96}}×\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{11}}{\mathrm{24}}\right)}{\mathrm{2}^{\mathrm{14}} \pi^{\mathrm{2}} \sqrt{\mathrm{3}}} \\ $$$$\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{\boldsymbol{\mathrm{K}}\left({x}\right)}{\:\sqrt{\mathrm{3}−{x}}}{dx}=\frac{\mathrm{1}}{\mathrm{96}\pi\sqrt{\mathrm{3}}}×\Gamma\left(\frac{\mathrm{1}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{7}}{\mathrm{24}}\right)\Gamma\left(\frac{\mathrm{11}}{\mathrm{24}}\right) \\ $$

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