show that
$\displaystyle B(x,y)= B(x+y,3) $
i tried to solve the problem as follows
$\displaystyle \frac{\Gamma{x+y} \Gamma{3}}{\Gamma{x+y+3}} $
can of confuse here on how to go forward, pls i need help here.
thanks
B(x, y) is the beta function, [tex]B(x, y)= \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+ y)}, right?
So $\displaystyle B(x+ y, 3)= \frac{\Gamma(x+ y)\Gamma(3)}{\Gamma(x+ y+ 3)}$
(You need some parentheses to clarify what you are writing!)
That is what you have. Now, can you use the fact $\displaystyle \Gamma(x)= \int_0^\infty t^{x- 1}e^{-xt}dt$, with some substitutions,
to show that?