Theorem. If $$\mathcal{L}\{f_1(t)\} \,=\, F_1(s) \quad\mbox{and}\quad \mathcal{L}\{f_2(t)\} \,=\, F_2(s),$$ then $$\mathcal{L}\left\{\int_0^tf_1(\tau)f_2(t-\tau)\,d\tau\right\} \;=\; F_1(s)F_2(s).$$

Proof. According to the definition of Laplace transform, one has $$\mathcal{L}\left\{\int_0^tf_1(\tau)f_2(t-\tau)\,d\tau\right\} \;=\; \int_0^\infty e^{-st}\left(\int_0^t f_1(\tau)f_2(t-\tau)\,d\tau\right)dt,$$ where the right hand side is a double integral over the angular region bounded by the lines $\tau = 0$ and $\tau = t$ in the first quadrant of the $t\tau$ -plane. Changing the order of integration, we write $$\mathcal{L}\left\{\int_0^tf_1(\tau)f_2(t-\tau)\,d\tau\right\} \;=\; \int_0^\infty\left(f_1(\tau)\int_\tau^\infty e^{-st}f_2(t-\tau)\,dt\right)d\tau.$$ Making in the inner integral the substitution $t-\tau := u$ , we obtain $$\int_\tau^\infty e^{-st}f_2(t-\tau)\,dt \,=\, \int_0^\infty e^{-(u+\tau)s}f_2(u)\,du = e^{-\tau s}\int_0^\infty e^{-su}f_2(u)\,du = e^{-\tau s}F_2(s),$$ whence $$\mathcal{L}\left\{\int_0^tf_1(\tau)f_2(t-\tau)\,d\tau\right\} \;=\; \int_0^\infty f_1(\tau)e^{-\tau s}F_2(s)\,d\tau \,=\, F_2(s)\int_0^\infty f_1(\tau)e^{-\tau s}\,d\tau = F_1(s)F_2(s),$$ Q.E.D.