Let \(f(t),g(t)\) be continuous functions for \(t\geq 0\). The convolution of \(f\) and \(g\) is the function

\[\displaystyle (f\ast g)(t)=\int _{0}^{t}f(x)g(t-x)\,dx.\]

Let’s compute the convolution of \(t\) and \(e^{-t}\):

\begin{eqnarray*} t\ast e^{-t} &=& \displaystyle \int _{0}^{t}xe^{-(t-x)}\,dx \\ &=& \displaystyle e^{-t}\int _{0}^{t}xe^{x}\,dx \\ &=& \displaystyle e^{-t}\left .(x-1)e^{x}\right |_{0}^{t} \\ &=& \displaystyle e^{-t}\left [(t-1)e^{t}+1\right ] \\ &=& \displaystyle t-1+e^{-t}. \end{eqnarray*}

Convolution Theorem: Let \(f(t)\) and \(g(t)\) be continuous functions for \(t\geq 0\). Then \(\mathcal {L}(f\ast g)=\mathcal {L}(f)\mathcal {L}(g)\).

(The proof is left as an exercise for the reader :)

This gives us another way to compute the Convolution of two functions: Take the Laplace transform, use a PFD, and take the inverse Laplace transform.

\(\displaystyle \mathcal {L}(t\ast e^{-t})=\mathcal {L}(t) \mathcal {L}(e^{-t})=\frac {1}{s^2} \, \frac {1}{s+1}=\frac {1}{s^2(s+1)}=\frac {A}{s}+\frac {1}{s^2}+\frac {1}{s+1}\). Clearing fractions, \(1=As(s+1)+(s+1)+s^2\). At \(s=1\) this is \(1=2A+3\) implying \(A=-1\).

So \(\displaystyle \mathcal {L}(t\ast e^{-t})=\frac {-1}{s}+\frac {1}{s^2}+\frac {1}{s+1}\). And the inverse transform is \(t\ast e^{-t}=t-1+e^{-t}\), matching our direct calculation above.

Suppose \(y(t)\) is a function satisfying \(\displaystyle y(t)+\int _0^t y(x)\,dx=\frac {1}{2}t^2\) (an \(``\)integral equation").

A key observation is that \(\displaystyle y\ast 1=\int _0^t y(x)\,dx\). So we can rewrite this integral equation as \(y\,+\,y\ast 1=\frac {1}{2}t^2\). Then by taking the Laplace transform of both sides, with \(Y=\mathcal {L}(y)\), we get \(Y\,+\, Y\,\frac {1}{s}=\frac {1}{s^3}\), or equivalently, \(\displaystyle \left (\frac {s+1}{s}\right ) Y=\frac {1}{s^3}\). So \(\displaystyle Y=\frac {1}{s^2(s+1)}\), which by the above we know that the solution is \(y=t-1+e^{-t}\).