83 Proof of convolution formula

Also, when we change order of integration, then let $u=x-y$,

$$ℒ(f\ast g)(s)={\int }_0^{\mathrm{\infty }}{e}^{-sx}f\ast g(x)𝑑x$$

$$={\int }_0^{\mathrm{\infty }}{e}^{-sx}{\int }_0^{x}f(x-y)g(y)𝑑y𝑑x$$

$$={\int }_0^{\mathrm{\infty }}{e}^{-sy}g(y)\left({\int }_y^{\mathrm{\infty }}{e}^{-s(x-y)}f(x-y)𝑑x\right)𝑑y$$

$$={\int }_0^{\mathrm{\infty }}{e}^{-sy}g(y)𝑑y{\int }_0^{\mathrm{\infty }}{e}^{-su}f(u)𝑑u$$

$$=ℒ(f)(s)ℒ(g)(s).$$

The Laplace transform converts convolution to multiplication.