Looking at the Dirac Delta, we can see some interesting properties:

\begin{align*}
\int_{-\infty}^{\infty}\delta(t-\tau)f(t)dt &= f(\tau) \\
\int_{\tau^-}^{\tau^+}\delta(t-\tau)f(t)dt &= f(\tau) \\
\int_{\tau-\epsilon}^{\tau+\epsilon}\delta(t-\tau)f(t)dt &= f(\tau) \\
\int_{\tau-\epsilon}^{\tau+\epsilon}\delta(t-\tau)dt &= 1 \\
\end{align*}

What happens if we write an integral like this?
$$\int_{-\infty}^{t}\delta(\tau)d\tau$$

In turns out that this is a very important element in characterizing continuous-time LTI systems, known as the \emph{Continuous-Time Unit Step}, which can be defined as

$$ u(t) = \int_{-\infty}^{t}\delta(\tau)d\tau = \begin{cases} 0 & t<0 \\ 1 & t \geq 0 \\ \end{cases}$$

Note, however, that there is a discontinuity at $t=0$, but by convention, we shall include it for $t \geq 0$. The height of the discontinuity happens to be the coefficient in front of $\delta(t)$. We just showed how the CT unit step can be defined in terms of the integral of the Dirac Delta. Notice that then the derivative the the unit step will spawn a DIrac Delta, thus, we can define the Dirac Delta as:

$$ \delta(t) = \dot{u}(t) = \frac{d}{dt}u(t) = \frac{d}{dt}\int_{-\infty}^{t}\delta(\tau)d\tau = \delta(t) $$

If we define $u_\Delta(t)$ as
$$
u_\Delta(t)=
\begin{cases}
0 & t<-\frac{\Delta}{2} \\
\frac{1}{\Delta}t+\frac{1}{2} & -\frac{\Delta}{2}\leq t \leq\frac{\Delta}{2} \\
1 & t>\frac{\Delta}{2} \\
\end{cases}
$$

% \begin{figure}[htp]
% \centering
% \includegraphics{images/dirac/delta-unit-step.ps}
% \caption{ The unit step with the discontinuity from $-\frac{\Delta}{2}$ to $\frac{\Delta}{2}$ }
% \label{ fig:delta-unit-step }
% \end{figure}

\begin{center}
\begin{pspicture}(-3.2,-3.2)(6,6)
%\psaxes[labels=none,Dx=1]{->}(-3,0)(3.25,0)
\pscustom[fillstyle=solid,fillcolor=blue!40,linestyle=none,algebraic]{
\psplot{-1}{1}{ 1/2*x+1/2}
\psline(1,0)(-1,0)

\psplot{1}{3}{ 1 }
\psline(3,0)(1,0) %  flipped for some reason
}
\psplot[algebraic,linewidth=2pt,algebraic]{-1}{1}{1/2*x + 1/2 }
\psplot[algebraic,linewidth=2pt,algebraic]{1}{3}{1}


\rput(-1.25,-0.25){$-\frac{\Delta}{2}$}
\rput(0,-0.25){0}
\rput(1,-0.25){$\frac{\Delta}{2}$}

\rput(3.25,-0.25){$t$}

\psline[linestyle=dashed](1,0)(1,1)

\psline{<->}(2.35,0)(2.35,1)
\rput(2.65,0.5){$1$}


\end{pspicture}
\end{center}


then we can take the derivative and we find that  $\dot{u}_\Delta(t) = \delta_\Delta(t)$, where $\delta_\Delta(t)$ represents the unit area given by 
$$
\dot{u}_\Delta(t) = \delta_\Delta(t)=
\begin{cases}
0 & t<-\frac{\Delta}{2} \\
\frac{1}{\Delta} & -\frac{\Delta}{2}\leq t \leq\frac{\Delta}{2} \\
0 & t>\frac{\Delta}{2} \\
\end{cases}
$$

% \begin{figure}[htp]
% \centering
% \includegraphics{images/dirac/dirac-unitarea.ps}
% \caption{ The unit area }
% \label{ fig:dirac-unit }
% \end{figure}


\begin{center}
\begin{pspicture}(-3.2,-3.2)(6,6)
%\psaxes[labels=none, Dx=1]{->}(-3,0)(3,0)
% \pscustom[fillstyle=solid,fillcolor=blue!40,linestyle=none,algebraic]{
\psplot{-1}{1}{ 1 }
\psline(1,0)(-1,0)



% }
\psplot[algebraic,linewidth=2pt,algebraic]{-1}{1}{1}
\psline[linewidth=2pt](1,0)(1,1)
\psline[linewidth=2pt](-1,0)(-1,1)


\rput(-1.175,-0.35){$-\frac{\Delta}{2}$}
\rput(0,-0.35){0}
\rput(1,-0.35){$\frac{\Delta}{2}$}

\rput(3.25,-0.25){$t$}

\psline{<->}(1.5,0)(1.5,1)
\rput(1.725,0.5){$\frac{1}{\Delta}$}


\end{pspicture}
\end{center}

If we take the limit, 
$$ \lim_{\Delta \to 0} \dot{u}_\Delta(t) = \delta(t) = \dot{u}(t) $$
we can clearly see that the derivative relates the CT unit step and the Dirac Delta. What about the discrete-time counter parts? We can write the Kronecker Delta as $\delta(n) = u(n) - u(n-1)$, the difference of a unit step and a shifted version of a unit step. We can write the unit step in discrete-time in two ways:
\begin{align*}
u(n) &= \sum \limits_{k=0}^{\infty}\delta(n-k) \\
u(n) &= \sum \limits_{\ell=-\infty}^{n}\delta(\ell) \quad \mbox{substitute } \ell = n-k
\end{align*}
We can verify that this works because when $n<0$ we get $u(n)=0$, and for $n \geq 0$, we have $u(n)=1$.


Dirac Delta and Unit Step