How to prove the following identity regarding Laplace transforms?

Question

I tried solving it by integrating by parts but i was unsuccessful.

$${\cal L}\left[\int_0^xf(x-t)g(t)\ dt\right]=F(p)G(p)$$

Answer 1

\begin{align} F(p) G(p) &= \int_0^\infty e^{-pu}f(u)\ du\int_0^\infty e^{-pv}g(v)\ dv \\ &= \int_0^\infty\int_0^\infty e^{-p(u+v)}f(u)g(v)\ du\ dv \,\,\, , \,\,\, \text{u+v=t} \,\, , \,\, \text{v=x}\\ &= \int_0^\infty\int_x^\infty e^{-pt}f(t-x)g(x)\ dt\ dx \,\, , \,\, \text{changing order of integration}\\ &= \int_0^\infty e^{-pt}\Big[\int_0^tf(t-x)g(x)\ dx \Big]\ dt \\ &= {\cal L}(f*g)(t) \end{align}