Proving the identity between Beta and Gamma functions using semi-group property of the Gamma.
Solution 1:
From Wikipedia:
With $f(u) = e^{-u} u^{x-1} 1_{u > 0}$ and $g(u) = e^{-u} u^{y-1} 1_{u > 0}$, we have $$\Gamma(x) \Gamma(y) = \int_{-\infty}^\infty f(u) \, du \cdot \int_{-\infty}^\infty g(u) \, du = \int_{-\infty}^\infty (f * g)(u) \, du = B(x,y) \Gamma(x+y).$$
To show the last equality, note that $$(f*g)(u) = \int_{-\infty}^\infty f(t) g(u-t) \, dt = \int_0^u e^{-u} t^{x-1} (u-t)^{y-1} \, dt = e^{-u} u^{x+y-1} \int_0^1 s^{x-1} (1-s)^{y-1} \, ds$$ where the last step comes from the substitution $t=us$.