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Question Number 126803 by john_santu last updated on 24/Dec/20

  B((7/3),(2/3)) =?  B = betha function

B(73,23)=?B=bethafunction

Answered by Dwaipayan Shikari last updated on 24/Dec/20

B((7/3),(2/3))=((Γ((7/3))Γ((2/3)))/(Γ(3)))=((Γ((1/3))Γ((2/3)).(4/3).(1/3))/(2!))=(2/9).(π/(sin(π/3)))=((4π)/(9(√3)))

B(73,23)=Γ(73)Γ(23)Γ(3)=Γ(13)Γ(23).43.132!=29.πsinπ3=4π93

Answered by liberty last updated on 24/Dec/20

B((7/3),(2/3)) = ((Γ((7/3)).Γ((2/3)))/(Γ((7/3)+(2/3)))) = ((Γ((7/3)).Γ((2/3)))/(Γ(3)))  [ Γ(3)=(3−1)! = 2 ]   [ Γ(x+1) = x Γ(x) ]    ⇔ = (((4/3). (1/3).Γ((1/3)).Γ((2/3)))/2) = (2/9).Γ((1/3)).Γ((2/3))  [ Γ(x).Γ(1−x) = (π/(sin (πx))) ; x∉Z ]   ⇔ = (2/9). (π/(sin ((π/3)))) = ((4π)/(9(√3))) = ((4(√3) π)/(27))

B(73,23)=Γ(73).Γ(23)Γ(73+23)=Γ(73).Γ(23)Γ(3)[Γ(3)=(31)!=2][Γ(x+1)=xΓ(x)]=43.13.Γ(13).Γ(23)2=29.Γ(13).Γ(23)[Γ(x).Γ(1x)=πsin(πx);xZ]=29.πsin(π3)=4π93=43π27

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