# Laplace transform of convolution

Theorem.  If

 $\mathcal{L}\{f_{1}(t)\}\,=\,F_{1}(s)\quad\mbox{and}\quad\mathcal{L}\{f_{2}(t)% \}\,=\,F_{2}(s),$

then

 $\mathcal{L}\left\{\int_{0}^{t}f_{1}(\tau)f_{2}(t-\tau)\,d\tau\right\}\;=\;F_{1% }(s)F_{2}(s).$

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

 $\mathcal{L}\left\{\int_{0}^{t}f_{1}(\tau)f_{2}(t-\tau)\,d\tau\right\}\;=\;\int% _{0}^{\infty}e^{-st}\left(\int_{0}^{t}f_{1}(\tau)f_{2}(t-\tau)\,d\tau\right)dt,$

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 of integration, we write

 $\mathcal{L}\left\{\int_{0}^{t}f_{1}(\tau)f_{2}(t-\tau)\,d\tau\right\}\;=\;\int% _{0}^{\infty}\left(f_{1}(\tau)\int_{\tau}^{\infty}e^{-st}f_{2}(t-\tau)\,dt% \right)d\tau.$

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

 $\mathcal{L}\left\{\int_{0}^{t}f_{1}(\tau)f_{2}(t-\tau)\,d\tau\right\}\;=\;\int% _{0}^{\infty}f_{1}(\tau)e^{-\tau s}F_{2}(s)\,d\tau\,=\,F_{2}(s)\int_{0}^{% \infty}f_{1}(\tau)e^{-\tau s}\,d\tau=F_{1}(s)F_{2}(s),$

Q.E.D.

Title Laplace transform of convolution LaplaceTransformOfConvolution 2013-03-22 18:24:04 2013-03-22 18:24:04 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 26A42 msc 44A10 convolution property of Laplace transform Convolution