= Beta-gamma identity
{title2=$B(p,q)=\Gamma(p)\Gamma(q)/\Gamma(p+q)$}
For $p,q>0$, express the product of the two <Gamma integrals> as the nonnegative <double integral>
$$
\Gamma(p)\Gamma(q)=\int_0^\infty\int_0^\infty x^{p-1}y^{q-1}e^{-(x+y)}dx\,dy.
$$
Use $r=x+y$ and $u=x/(x+y)$, with inverse $x=ru$, $y=r(1-u)$ and <Jacobian determinant> of absolute value $r$. The new <domain of integration> is $r>0$, $0<u<1$, and the integrand with this factor becomes $r^{p+q-1}e^{-r}u^{p-1}(1-u)^{q-1}$. <Tonelli's theorem> and separation of the two integrals therefore give $\Gamma(p+q)B(p,q)$. The positive real parameter hypothesis ensures that all the original integrals are finite.
For complex parameters with positive real parts, the same proof first gives the undivided identity $\Gamma(p+q)B(p,q)=\Gamma(p)\Gamma(q)$. Absolute integrability follows by taking real parts in the exponents, so <Fubini's theorem> applies. The <nonvanishing of the Gamma function> then permits division by $\Gamma(p+q)$.
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