Double integral involving beta functions I want to prove the following result: $$\int_0^1 \int_0^1 {f(xy)(1-x)^{p-1}y^p(1-y)^{q-1}} \mathrm{d}x \, \mathrm{d}y=\frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)} \int_0^1 {f(t)(1-t)^{p+q-1}} \mathrm{d}t.$$
I have tried using the substitution $t=xy$, but am not sure what to use for the other variable. We were given no information about the function $f$.
Any help would be appreciated!
 A: By definition of the so-called Beta Integral
$$B(p, q)=\int_{0}^{1}t^{p-1}(1-t)^{q-1}dt$$
It is easy to show that for $\Re(u) > 0, \Im(v) > 0$
$$B(p, q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$$
In order to show this result, using the standard definition of the Gamma function coupled with (your correct intuition into a substitution) $t=xy$ and $s=x(1-y)$ we find that
\begin{align}
\Gamma(p)\Gamma(q) &= \int_{0}^{\infty}e^{-t}t^{p-1}dt\int_{0}^{\infty}e^{-s}s^{q-1}ds \\
                   &= \int_{0}^{\infty}\int_{0}^{\infty}e^{-(t+s)}t^{p-1}s^{q-1}dt ds
\end{align}
Applying now that change of variables and computing the Jacobian, you should find that
\begin{align}
\Gamma(p)\Gamma(q) &= \int_{0}^{1}\int_{0}^{\infty}e^{-x}x^{p-1}y^{p-1}x^{q-1}(1-y)^{q-1}x dx dy \\
                   &= \int_{0}^{\infty}e^{-x}x^{p+q-1}dx \int_{0}^{1}y^{p-1}(1-y)^{q-1}dy \\
                   &= \Gamma(p+q)B(p, q)
\end{align}
You may think I haven't answered your query but what I have shown is that your integral is a corollary in the proof of the Beta integral. I leave it top you to play around with the integral you have to show that the above computation for the Beta integral is sufficient for the evaluation of yours.
Best of luck,
Bacon
