In the middle of a proof involving the product of two gamma functions, I have the following expressions:
$$\Gamma(a) \Gamma(b) = \int^{\infty}_{0}\exp(-x)x^{(a-1)}dx~ \int_{0}^{\infty}\exp(-y)y^{(b-1)}dy$$
This leads to the next expression via a variable change of y=t-x which i have no problem following:
$$\Gamma(a)\Gamma(b) = \int_{0}^{\infty} x^{a-1} \bigg(\int_{x}^{\infty} \exp(-t)(t-x)^{b-1}dt\bigg) ~dx$$
Then, the Textbook says, change the order of x and t in the integrations taking care with limits of integration, and then make a change of variables: $$x=t~p$$
And somehow, we are suppose to arrive at this result:
$$\Gamma(a)\Gamma(b)=\int^{\infty}_{0} \exp(-t) t^{(a-1)} t^{(b-1)}~t~dt \int^{1}_{0} p^{(a-1)} (1-p)^{(b-1)}~dp$$
$$=\Gamma(a+b) = \int_{0}^{1} p^{(a-1)}~(1-p)^{(b-1)}~dp$$
I was curious about the steps in-between to get from the first integral to the rearrangement and variables substitution to obtain the second set of integrals.
Switching $x$ and $y$, you have \begin{eqnarray} \Gamma(a)\Gamma(b) &=& \int_{0}^{\infty} x^{a-1} \bigg(\int_{x}^{\infty} exp(-t)(t-x)^{b-1}dt\bigg) ~dx\\ &=&\int_{0}^{\infty}\bigg(\int_{x}^{\infty} x^{a-1} exp(-t)(t-x)^{b-1}dt\bigg) ~dx\\ &=&\int_{0}^{\infty} \bigg(\int_{0}^{t} x^{a-1} exp(-t)(t-x)^{b-1}dx\bigg) ~dt\\ &=&\int_{0}^{\infty}exp(-t) \bigg(\int_{0}^{t} x^{a-1} (t-x)^{b-1}dx\bigg) ~dt. \end{eqnarray} Now you can perform the change of variable $x=tp$ to obtain the result.