= Solution
Multiply the two defining <Gamma integrals>. For $p,q>0$ the nonnegative product integrand allows <Tonelli's theorem>:
$$
\Gamma(p)\Gamma(q)=\int_0^\infty\int_0^\infty x^{p-1}y^{q-1}e^{-(x+y)}dx\,dy.
$$
Use $r=x+y$, $u=x/(x+y)$, so $x=ru$, $y=r(1-u)$ and the positive quadrant becomes $r>0$, $0<u<1$. The <Jacobian determinant> has absolute value
$$
\left|\det\begin{pmatrix}u&r\\1-u&-r\end{pmatrix}\right|=r.
$$
After including this factor, the transformed integral separates:
$$
\left[\int_0^\infty r^{p+q-1}e^{-r}dr\right]
\left[\int_0^1u^{p-1}(1-u)^{q-1}du\right]
=\Gamma(p+q)B(p,q).
$$
This proves the <beta-gamma identity>,
$$
\boxed{\Gamma(p)\Gamma(q)=B(p,q)\Gamma(p+q).}
$$
All changes of variables are made inside the open positive domains; their boundaries do not contribute to the integrals.
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