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

$$B(\frac{7}{3},\frac{2}{3})=?$$$$B=bethafunction$$

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

$$B(\frac{7}{3},\frac{2}{3})=\frac{\mathrm{\Gamma }\left(\frac{7}{3}\right)\mathrm{\Gamma }\left(\frac{2}{3}\right)}{\mathrm{\Gamma }\left(3\right)}=\frac{\mathrm{\Gamma }\left(\frac{1}{3}\right)\mathrm{\Gamma }\left(\frac{2}{3}\right).\frac{4}{3}.\frac{1}{3}}{2!}=\frac{2}{9}.\frac{\pi }{sin\frac{\pi }{3}}=\frac{4\pi }{9\sqrt{3}}$$

Answered by liberty last updated on 24/Dec/20

$$B(\frac{7}{3},\frac{2}{3})=\frac{\mathrm{\Gamma }\left(\frac{7}{3}\right).\mathrm{\Gamma }\left(\frac{2}{3}\right)}{\mathrm{\Gamma }(\frac{7}{3}+\frac{2}{3})}=\frac{\mathrm{\Gamma }\left(\frac{7}{3}\right).\mathrm{\Gamma }\left(\frac{2}{3}\right)}{\mathrm{\Gamma }\left(3\right)}$$$$[\mathrm{\Gamma }\left(3\right)=(3-1)!=2]$$$$[\mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }\left(x\right)]$$$$$$$$\iff =\frac{\frac{4}{3}.\frac{1}{3}.\mathrm{\Gamma }\left(\frac{1}{3}\right).\mathrm{\Gamma }\left(\frac{2}{3}\right)}{2}=\frac{2}{9}.\mathrm{\Gamma }\left(\frac{1}{3}\right).\mathrm{\Gamma }\left(\frac{2}{3}\right)$$$$[\mathrm{\Gamma }\left(x\right).\mathrm{\Gamma }(1-x)=\frac{\pi }{\sin \left(\pi x\right)};x\notin \mathbb{Z}]$$$$\iff =\frac{2}{9}.\frac{\pi }{\sin \left(\frac{\pi }{3}\right)}=\frac{4\pi }{9\sqrt{3}}=\frac{4\sqrt{3}\pi }{27}$$

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