Answer :

Solution :

Dirac Delta Function

$\delta \left(t-a\right)=\underset{ϵ\to 0}{lim}\{\begin{array}{c}0,-\mathrm{\infty }<t<a\\ \frac{1}{ϵ},a\le t\le a+ϵ\\ 0,a+ϵ<t<ϵ\end{array}$

L[δ(t – a)] = e^-as

^{${\because \; \int }_0^{\mathrm{\infty }}f\left(t\right)\delta \left(t-a\right)dt=f\left(a\right)$}

^{$\therefore L\left[\delta \left(t-a\right)\right]={\int }_0^{\mathrm{\infty }}{e}^{-st}\delta \left(t-a\right)dt=e^{-as}$}