# Thread: Help with this integral

1. ## Help with this integral

Define, for $x \geq y \geq 0$
$\Lambda(x,y)=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma( x-y+1)}$
Then proof for all $n>0$
$\int^n_{0}\Lambda(n,x)dx=2^n$
And also prove for $0
$\int^ \infty_{-\infty}p^x(1-p)^{n-x}\Lambda(n,x)dx=1$
I once asked someone this question and they said it looks like the Beta Function, thus maybe that will help.

If you are curious how I came to such a conjecture is because I was trying to describe a mathematical formula for the Normal Distribution and these two:
$\Lambda(n,x)$
and
$p^x(1-p)^{n-x}\Lambda(n,x)$ where $p$ is the probability.
Graph curves which look like the Normal Distribution, furthermore the second one is a density curve this is why its Area is one.

2. Originally Posted by ThePerfectHacker
Define, for $x \geq y \geq 0$
$\Lambda(x,y)=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma( x-y+1)}$
Then proof for all $n>0$
$\int^n_{0}\Lambda(n,x)dx=2^n$
If this were true then:

$I(1)=\int_0^1 \frac{\Gamma(2)}{\Gamma (x+1) \Gamma(2-x)} dx =2$.

But it seems to me that $I(1)\approx 1.2$.

RonL

3. CaptainBlack, Graph it and take its integral after that. Maybe, I am making a mistake in the way I posed the problem. Or perhaps, I made a mistake in the domain of the function. Anyways, would you confirm that these two functions look like the normal distribution curve?

I would check it but I "borrowed" someone else's computer and it does not have a graphing program.

4. Originally Posted by ThePerfectHacker
CaptainBlack, Graph it and take its integral after that. Maybe, I am making a mistake in the way I posed the problem. Or perhaps, I made a mistake in the domain of the function. Anyways, would you confirm that these two functions look like the normal distribution curve?

I would check it but I "borrowed" someone else's computer and it does not have a graphing program.
IIRC

$
p^x(1-p)^{n-x}\Lambda(n,x)
$

with $x$ and $n$ positive integers is the probability of $x$ successes in $n$
trials, where $p$ is the probability of success in a single trial.. This is the binomial
distribution and is well know to approximate the normal distribution (in a
rather peculiar sense of approximate) for large $n$

RonL