# Help with this integral

• Dec 31st 2005, 08:27 AM
ThePerfectHacker
Help with this integral
Define, for $\displaystyle x \geq y \geq 0$
$\displaystyle \Lambda(x,y)=\frac{\Gamma(x+1)}{\Gamma(y+1)\Gamma( x-y+1)}$
Then proof for all $\displaystyle n>0$
$\displaystyle \int^n_{0}\Lambda(n,x)dx=2^n$
And also prove for $\displaystyle 0<p<1$
$\displaystyle \int^ \infty_{-\infty}p^x(1-p)^{n-x}\Lambda(n,x)dx=1$
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:
$\displaystyle \Lambda(n,x)$
and
$\displaystyle p^x(1-p)^{n-x}\Lambda(n,x)$ where $\displaystyle p$ 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.
• Dec 31st 2005, 10:43 AM
CaptainBlack
Quote:

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

If this were true then:

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

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

RonL
• Dec 31st 2005, 10:59 AM
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. :)
• Dec 31st 2005, 12:06 PM
CaptainBlack
Quote:

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

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

with $\displaystyle x$ and $\displaystyle n$ positive integers is the probability of $\displaystyle x$ successes in $\displaystyle n$
trials, where $\displaystyle p$ 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 $\displaystyle n$

RonL