The Dirac-delta function (δ(t - t0)) for t, t0 ∈ R, has the following property

${\displaystyle {\int }_a^b\varphi (t)\delta (t-t_0)dt=\{\begin{array}{cc}\varphi (t_0)& a<t_0<b\\ 0& \text{otherwise}\end{array}}$

The Laplace transform of the Dirac-delta function δ(t - a) for a > 0; $\mathcal{L}(\delta (t-a))=F(s)$ is

The correct option is e^-sa.

To find the Laplace transform of the Dirac-delta function $\delta (t-a)$ for $a>0$, we start by using the definition of the Laplace transform.

The Laplace transform of any function $f\left(t\right)$ is defined as:
$$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$$

Substituting $f\left(t\right)=\delta (t-a)$ into this definition, we obtain:
$$\mathcal{L}\left\{\delta \right(t-a\left)\right\}={\int }_0^{\infty }{e}^{-st}\delta (t-a)dt$$

Now we apply the sifting property of the Dirac-delta function given in the problem description:
$${\int }_a^b\varphi \left(t\right)\delta (t-t_0)dt=\varphi \left(t_0\right)$$
provided that the impulse location $t_0$ lies strictly within the integration interval $(a,b)$.

Comparing our Laplace integral to this property:
1. The function $\varphi \left(t\right)$ corresponds to $e^{-st}$.
2. The parameter $t_0$ is equal to $a$.
3. The integration interval is $[0,\infty )$. Since we are given that $a>0$, the value $t=a$ lies strictly inside the range of integration $(0,\infty )$.

Therefore, applying the sifting property allows us to evaluate the integral by simply evaluating the integrand's function $e^{-st}$ at $t=a$:
$${\int }_0^{\infty }{e}^{-st}\delta (t-a)dt=e^{-s·a}=e^{-sa}$$
Thus, the Laplace transform of the Dirac-delta function is $F\left(s\right)=e^{-sa}$.