Remark 6.4.1.
The Laplace transform of \(\delta(t-a)\) would be the Laplace transform of the derivative of the Heaviside function \(u(t-a)\text{,}\) if the Heaviside function had a derivative. First,
\begin{equation*}
{\mathcal{L}} \bigl\{ u(t-a) \bigr\} = \frac{e^{-as}}{s}.
\end{equation*}
To obtain what the Laplace transform of the derivative would be, we multiply by \(s\text{,}\) to obtain \(e^{-as}\text{,}\) which is the Laplace transform of \(\delta(t-a)\text{.}\) We see the same thing using integration,
\begin{equation*}
\int_0^t \delta(s-a)\,ds = u(t-a) .
\end{equation*}
So in a certain sense
\begin{equation*}
\text{"}
\quad \frac{d}{dt} \Bigl[ u(t-a) \Bigr] = \delta(t-a) . \quad
\text{"}
\end{equation*}
This line of reasoning allows us to talk about derivatives of functions with jump discontinuities. We can think of the derivative of the Heaviside function \(u(t-a)\) as being somehow infinite at \(a\text{,}\) which is precisely our intuitive understanding of the delta function.