= Solution
The two <Gamma function> integrals are absolutely convergent when $\operatorname{Re}\alpha,\operatorname{Re}\beta>0$. Consequently <Fubini theorem> gives
$$
\Gamma(\alpha)\Gamma(\beta)
=\int_0^\infty\int_0^\infty x^{\alpha-1}y^{\beta-1}e^{-(x+y)}\,dx\,dy.
$$
Make the <change of variables> $u=x+y$, $t=x/(x+y)$, so $x=ut$, $y=u(1-t)$, $u>0$, $0<t<1$, and the Jacobian is $u$. Powers of positive real variables use the real logarithm, so this transformation is also valid for complex exponents. The integral factors as
$$
\left[\int_0^\infty u^{\alpha+\beta-1}e^{-u}\,du\right]
\left[\int_0^1t^{\alpha-1}(1-t)^{\beta-1}\,dt\right].
$$
Both factors converge under the same real-part assumptions. The first is $\Gamma(\alpha+\beta)$, proving the <Beta function> identity
$$
\boxed{\Gamma(\alpha)\Gamma(\beta)=\Gamma(\alpha+\beta)\int_0^1t^{\alpha-1}(1-t)^{\beta-1}\,dt.}
$$
Back to article page
