# Showing that $\Gamma(x)\Gamma(y) = \Gamma(x+y)\beta(x,y)$ via change of variable

From $$\Gamma(x)\Gamma(y)=\int_0^\infty e^{-t}t^{x-1} \left( \int_0^\infty e^{-t} s^{y-1} ds \right) dt,$$ use a change of variable $s=ut$ to show $$\Gamma(x)\Gamma(y)=\Gamma(x+y)\beta(x,y).$$

Let $s=ut$ so $ds = udt + tdu$ and then

\begin{align*} \Gamma(x)\Gamma(y) &=\int_0^\infty e^{-t}t^{x-1} \left( \int_0^\infty e^{-s} s^{y-1} ds \right) dt \\ &= \int_0^\infty e^{-t} t^{x-1} \left( \int_0^{\infty} e^{-ut} (ut)^{y-1}(udt+tdu)\right)dt \\ &= \int_0^\infty e^{-t} t^{x-1} \left( \int_0^{\infty} (e^{-ut} u^y t^{y-1})dt+ \int_0^{\infty} (e^{-ut}u^{y-1}t^y)du\right)dt \\ \end{align*} My next thought was to use integration by parts, but that didn't pan out. Any suggestions?

Note that Gamma Function for $x>0$ we define: $$\Gamma(x):=\int_0^\infty e^{-t}t^{x-1}dt.$$

And for the Beta Function for $x>0$, $y>0$, we define $$\beta(x,y):=\int_0^1 t^{x-1}(1-t)^{y-1}dt.$$

Hint: Try the substitution $s+t=\alpha$ and $t=\alpha \gamma$ so $s=\alpha(1-\gamma)$ and $ds \, dt = \alpha d \alpha \, d \gamma$.
• Maybe $\beta$ isn't the best symbol to use here. – Shaun Jun 3 '18 at 21:49