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

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- May 13th 2013, 06:37 AMlawochekelintegral function!
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 - May 13th 2013, 09:27 AMHallsofIvyRe: integral function!
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?