Show that $1=\int\int\int x^{l-1}y^{m-1}z^{n-1}(1-x-y-z)^{p-1}dxdydz; l,m,n,p\ge 1$ taken over the tetrahedron bounded by the planes $x=0,y=0,z=0,x+y+z=1$
I have a little idea about Gaussian integrals but not much.Also I know Leibnitz Integral concept but how to apply this here. Anyone please explain as to what should be my approach .This is not a HW question. I am practising such questions as I have never solved more than one or two problems on this theory.
I know that this problem can be simplified as a multivariate beta distribution much of which I did not know except the formula on Wikipedia.
Here is what I think can be done but I need further explanations:
$1=\int\int\int x^{l-1}y^{m-1}z^{n-1}(1-x-y-z)^{p-1}dxdydz$
=$\int_{0}^{1}\int_{0}^{1-x}\int_{0}^{1-(x+y)} x^{l-1}y^{m-1}z^{n-1}(1-x-y-z)^{p-1}dxdydz$
$=B(l,m,n,p)=\dfrac{\Gamma(l)\Gamma(m)\Gamma(n)\Gamma(p)}{\Gamma(l+m+n+p)}$(Is this correct?)
Now if $l=m=n=p=1,\dfrac{\Gamma(l)\Gamma(m)\Gamma(n)\Gamma(p)}{\Gamma(l+m+n+p)}=1$(But how?)
I am solving 2D case ie,
$$I=\int\int x^{l-1}y^{m-1}(1-x-y)^{p-1}dxdydz; l,m,p\ge 1$$ $$\implies I= \int_0^1x^{l-1}(\int_0^{1-x}y^{m-1}(1-x-y)^{p-1}dy)dx$$
Put $y = (1-x)z$, then the inner integral can be transformed as (using $dy= (1-x)dz$ as $x$ is constant inside the second integral) $$I= \int_0^1x^{l-1}(\int_0^{1}(1-x)^{m-1}z^{m-1}(1-x)^{p-1}(1-z)^{p-1}(1-x)dz)dx$$ $$\implies I= \int_0^1x^{l-1}(1-x)^{m+p-1}(\int_0^{1}z^{m-1}(1-z)^{p-1}dz)dx$$ $$\implies I= \int_0^1x^{l-1}(1-x)^{m+p-1}dx.\int_0^1z^{m-1}(1-z)^{p-1}dz$$ $$\implies I= B(l, m+p).B(m, p)$$ $$\implies I = \frac{\Gamma(l)\Gamma(m+p)}{\Gamma(l+m+p)}. \frac{\Gamma(m)\Gamma(p)}{\Gamma(m+p)}$$ $$\implies I= \frac{\Gamma(l)\Gamma(m)\Gamma(p)}{\Gamma(l+m+p)}$$
Simailarly, it can be proved for three dimensional case.