How to evaluate $\int_0^z(x(z-x))^ndx$

Question

How to evaluate $\int_0^z(x(z-x))^ndx$

I was trying to find conditional distribution involving two Gamma Distributions when this integral came up. Accoring to https://www.integral-calculator.com/ the answer is $B(n+1,n+1)z^{2n+1}$ I'm not sure how though. $n$ need not be interger so I cant use binomial expansion as well. I'm aware of the identity $$B(x,y)\Gamma(x+y)=\Gamma(x)\Gamma(y)$$

@random wouldn't we get the same thing back? Not sure how that would help — Anvit, Apr 06 '19 at 12:28
@random I think you misread the question. Both $x$ and $z-x$ are raised to power $n$ — Anvit, Apr 06 '19 at 13:20

score 1 · Accepted Answer · answered Apr 06 '19 at 12:13

1

With $x=zy$ the integral becomes $z^{2n+1}\int_0^1 y^n(1-y)^n dy$, as required.

answered Apr 06 '19 at 12:13

J.G.

115,835

How to evaluate $\int_0^z(x(z-x))^ndx$

1 Answers1