Even though the Dirac delta function is not a piecewise continuous, exponentially bounded function, we can define its Laplace transform as the limit of the Laplace transform of \(d_\tau(t)\) as \(\tau \to 0\text{.}\) More specifically, assume that \(t_0 \gt 0\) and
\begin{equation*}
{\mathcal L}(\delta(t - t_0)) = \lim_{\tau \to 0} {\mathcal L}(d_\tau(t - t_0)).
\end{equation*}
Assuming that \(t_0 - \tau \gt 0\text{,}\) the Laplace transform of \(d_\tau(t - t_0)\) is
\begin{align*}
{\mathcal L}(d_\tau(t - t_0)) & = \int_0^\infty e^{-st} d_\tau(t - t_0) \, dt\\
& = \int_{t_0 - \tau}^{t_0 + \tau} e^{-st} d_\tau(t - t_0) \, dt\\
& = \frac{1}{2 \tau} \int_{t_0 - \tau}^{t_0 + \tau} e^{-st} \, dt\\
& = \frac{1}{2 s \tau} e^{-st} \bigg|_{t = t_0 - \tau}^{t = t_0 + \tau}\\
& = \frac{1}{2 s \tau} e^{-st_0} (e^{s\tau} - e^{-s \tau})\\
& = \frac{\sinh s \tau}{s \tau} e^{-st_0}.
\end{align*}
We can use l’Hospital’s rule to evaluate \((\sinh s \tau)/ s \tau\) as \(\tau \to 0\text{,}\)
\begin{equation*}
\lim_{\tau \to 0} \frac{\sinh s \tau}{s \tau} = \lim_{\tau \to 0} \frac{s \cosh s \tau}{s} = 1.
\end{equation*}
Thus,
\begin{equation*}
{\mathcal L}(\delta(t - t_0)) = e^{-st_0}.
\end{equation*}
We can extend this result to allow \(t_0 = 0\text{,}\) by
\begin{equation*}
\lim_{t_0 \to 0} {\mathcal L}(\delta(t - t_0)) =\lim_{t_0 \to 0} e^{-st_0} = 1.
\end{equation*}
It is important to notice that we are using the Dirac delta function like an ordinary function. This requires some rigorous mathematics to justify that we can actually do this.
