Question Number 126803 by john_santu last updated on 24/Dec/20
$$\:\:{B}\left(\frac{\mathrm{7}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)\:=? \\ $$$${B}\:=\:{betha}\:{function}\: \\ $$
Answered by Dwaipayan Shikari last updated on 24/Dec/20
$${B}\left(\frac{\mathrm{7}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left(\mathrm{3}\right)}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right).\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{2}!}=\frac{\mathrm{2}}{\mathrm{9}}.\frac{\pi}{{sin}\frac{\pi}{\mathrm{3}}}=\frac{\mathrm{4}\pi}{\mathrm{9}\sqrt{\mathrm{3}}} \\ $$
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))](Q126805.png)
$${B}\left(\frac{\mathrm{7}}{\mathrm{3}},\frac{\mathrm{2}}{\mathrm{3}}\right)\:=\:\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{3}}\right).\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left(\frac{\mathrm{7}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}\right)}\:=\:\frac{\Gamma\left(\frac{\mathrm{7}}{\mathrm{3}}\right).\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\Gamma\left(\mathrm{3}\right)} \\ $$$$\left[\:\Gamma\left(\mathrm{3}\right)=\left(\mathrm{3}−\mathrm{1}\right)!\:=\:\mathrm{2}\:\right]\: \\ $$$$\left[\:\Gamma\left({x}+\mathrm{1}\right)\:=\:{x}\:\Gamma\left({x}\right)\:\right] \\ $$$$ \\ $$$$\Leftrightarrow\:=\:\frac{\frac{\mathrm{4}}{\mathrm{3}}.\:\frac{\mathrm{1}}{\mathrm{3}}.\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right).\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right)}{\mathrm{2}}\:=\:\frac{\mathrm{2}}{\mathrm{9}}.\Gamma\left(\frac{\mathrm{1}}{\mathrm{3}}\right).\Gamma\left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\left[\:\Gamma\left({x}\right).\Gamma\left(\mathrm{1}−{x}\right)\:=\:\frac{\pi}{\mathrm{sin}\:\left(\pi{x}\right)}\:;\:{x}\notin\mathbb{Z}\:\right] \\ $$$$\:\Leftrightarrow\:=\:\frac{\mathrm{2}}{\mathrm{9}}.\:\frac{\pi}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}\right)}\:=\:\frac{\mathrm{4}\pi}{\mathrm{9}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}\:\pi}{\mathrm{27}}\: \\ $$