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Question Number 113600 by eric last updated on 14/Sep/20

$$Prouverque$$$$\beta (a,b)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}={\int }_0^1{x}^{a-1}{(1-x)}^{b-1}dx$$

Answered by Dwaipayan Shikari last updated on 14/Sep/20

$$\beta (a,b)={\int }_0^1{x}^{a-1}{(1-x)}^{b-1}dx$$$$\mathrm{\Gamma }\left(a\right)={\int }_0^{\mathrm{\infty }}{x}^{a-1}{e}^{-x}dx$$$$\mathrm{\Gamma }\left(b\right)={\int }_0^{\mathrm{\infty }}{y}^{b-1}{e}^{-y}dy$$$$\mathrm{\Gamma }(a+b)={\int }_0^{\mathrm{\infty }}{x}^{a+b-1}{e}^{-x}dx$$$$\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{x}^{a-1}.y^{b-1}{e}^{-(x+y)}dydx$$$$x=vt$$$$y=v(1-t)$$$$\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)={\int }_0^{\mathrm{\infty }}{v}^{a+b-1}{e}^{-v}{\int }_0^1{t}^{a-1}{(1-t)}^{b-1}dt$$$$\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)=\mathrm{\Gamma }(a+b)\beta (a,b)$$$$\beta (a,b)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }(a+b)}$$

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