Laplace transform of convolution

Theorem.  If

$$ℒ\{f_1(t)\}=F_1(s)\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}ℒ\{f_2(t)\}=F_2(s),$$

then

$$ℒ\left\{{\int }_0^t{f}_1(\tau )f_2(t-\tau )𝑑\tau \right\}=F_1(s)F_2(s).$$

Proof.  According to the definition of Laplace transform, one has

$$ℒ\left\{{\int }_0^t{f}_1(\tau )f_2(t-\tau )𝑑\tau \right\}={\int }_0^{\mathrm{\infty }}{e}^{-st}\left({\int }_0^t{f}_1(\tau )f_2(t-\tau )𝑑\tau \right)𝑑t,$$

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 of integration, we write

$$ℒ\left\{{\int }_0^t{f}_1(\tau )f_2(t-\tau )𝑑\tau \right\}={\int }_0^{\mathrm{\infty }}\left(f_1(\tau ){\int }_{\tau }^{\mathrm{\infty }}{e}^{-st}{f}_2(t-\tau )𝑑t\right)𝑑\tau .$$

Making in the inner integral the substitution  $t-\tau :=u$,  we obtain

$${\int }_{\tau }^{\mathrm{\infty }}{e}^{-st}{f}_2(t-\tau )𝑑t={\int }_0^{\mathrm{\infty }}{e}^{-(u+\tau )s}{f}_2(u)𝑑u=e^{-\tau s}{\int }_0^{\mathrm{\infty }}{e}^{-su}{f}_2(u)𝑑u=e^{-\tau s}{F}_2(s),$$

whence

$$ℒ\left\{{\int }_0^t{f}_1(\tau )f_2(t-\tau )𝑑\tau \right\}={\int }_0^{\mathrm{\infty }}{f}_1(\tau )e^{-\tau s}{F}_2(s)𝑑\tau =F_2(s){\int }_0^{\mathrm{\infty }}{f}_1(\tau )e^{-\tau s}𝑑\tau =F_1(s)F_2(s),$$

Q.E.D.