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Section 16.5 Relationship between Delta and Step Functions
The Dirac Delta and Step Functions are a Derivative/Integral Pair.
We can relate the delta function to the step function in the following way. Consider the integral
\begin{equation}
\int_{-\infty}^x \delta(u-a)\,du\tag{16.5.1}
\end{equation}
Notice the variable \(x\) in the upper limit of the integral. The value of this integral is \(0\) if we stop integrating before we reach the peak of the delta function, i.e. for \(x\lt a\text{.}\) If we integrate through the peak, the value of the integral is \(1\text{,}\) i.e. for \(x>a\text{.}\) Thus, we have argued that the value of the integral, thought of as a function of \(x\text{,}\) is just the step function
\begin{equation}
\Theta(x-a) = \int_{-\infty}^x \delta(u-a)\,du\tag{16.5.2}
\end{equation}
(Recall that we don’t really care about the choice of \(\Theta(0)\text{,}\) so we don’t need to worry about the value of this function if we stop integrating exactly at \(x=a\text{.}\) )
If the step function is the integral of the delta function, then the delta function must be the derivative of the step function.
\begin{equation}
\frac{d}{dx}\Theta(x-a) = \delta(x-a)\tag{16.5.3}
\end{equation}
You should be able to persuade yourself that this statement is reasonable geometrically if you think of the derivative of a function as representing its slope.
Sensemaking 16.2 . The derivative of the step function.
Prove that the derivative of the step function is the delta function, i.e. prove
Equation (16.5.3) .
Solution .
We need to show that
\begin{equation}
\Int_{b}^{c} f(x)\,\delta(x-a) \,dx = f(a)\tag{16.5.4}
\end{equation}
for \(b\lt a\lt c\text{,}\) where
\begin{equation}
\delta(x-a)=\frac{d}{dx}\,\Theta(x-a)\tag{16.5.5}
\end{equation}
The main strategy is to use integration by parts, paying strict attention to the limits of integration.
\begin{align*}
\Int_{b}^{c} f(x)\,\delta(x-a) \,dx
\amp= \Int_{b}^{c} f(x)\,\frac{d}{dx}\, \Theta(x-a) \,dx\\
\amp= \left. f(x)\, \Theta(x-a)\right|_b^c
- \Int_{b}^{c} \frac{d}{dx}(f(x))\,\Theta(x-a) \,dx\\
\amp = \left\{ f( c )-0\right\}
- \Int_{a}^{c} \frac{d}{dx}(f(x))\,dx\\
\amp = f( c )-\left\{\left. f(x)\right|_a^c\right\}\\
\amp = f( c )-\left\{f( c )-f(a)\right\}\\
\amp = f(a)
\end{align*}
