Laplace transform of Dirac delta

The Dirac delta (http://planetmath.org/DiracDeltaFunction) $\delta$ can be interpreted as a linear functional, i.e. a linear mapping from a function space, consisting e.g. of certain real functions, to $ℝ$ (or $ℂ$), having the property

$$\delta [f]=f(0).$$

One may think this as the inner product

$$⟨f,\delta ⟩={\int }_0^{\mathrm{\infty }}f(t)\delta (t)𝑑t$$

of a function $f$ and another “function” $\delta$, when the well-known

$${\int }_0^{\mathrm{\infty }}f(t)\delta (t)𝑑t=f(0)$$

is true.  Applying this to  $f(t):=e^{-st}$,  one gets

$${\int }_0^{\mathrm{\infty }}{e}^{-st}\delta (t)𝑑t=e^{-0},$$

i.e. the Laplace transform

$ℒ\{\delta (t)\}=\mathrm{\hspace{0.33em}1}.$(1)

By the delay theorem, this result may be generalised to

$$ℒ\{\delta (t-a))\}=e{}^{-as}.$$

When introducing some “nascent Dirac delta function”, for example

${\eta }_{\epsilon }(t):=\{\begin{array}{cc}\frac{1}{\epsilon }\hspace{1em}\text{for}\mathrm{\hspace{0.33em}\hspace{0.33em}0}\le t\le \epsilon ,\hfill & \\ 0\hspace{1em}\text{for}\hspace{1em}\hspace{1em}t>\epsilon ,\hfill & \end{array}$

as an “approximation” of Dirac delta, we obtain the Laplace transform

$$ℒ\{{\eta }_{\epsilon }(t)\}={\int }_0^{\mathrm{\infty }}{e}^{-st}{\eta }_{\epsilon }(t)𝑑t={\int }_0^{\epsilon }\frac{e^{-st}}{\epsilon }𝑑t+{\int }_{\epsilon }^{\mathrm{\infty }}{e}^{-st}\cdot 0𝑑t=\frac{1}{\epsilon }{\int }_0^{\epsilon }{e}^{-st}𝑑t=\frac{1-e^{-\epsilon s}}{\epsilon s}.$$

As the Taylor expansion shows, we then have

$$\underset{\epsilon \to 0+}{lim}ℒ\{{\eta }_{\epsilon }(t)\}=\mathrm{\hspace{0.33em}1},$$

being in accordance with (1).