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Section 11.2 The Dirac Delta Function

The Dirac delta function \(\delta(x)\) is not really a “function”. It is a mathematical entity called a distribution which is well defined only when it appears under an integral sign. It has the following defining properties:

\begin{equation} \delta(x) = \cases{0, \qquad \amp if $x\not= 0$\cr \infty, \qquad \amp if $x=0$\cr}\tag{11.2.1} \end{equation}
\begin{equation} \int_b^c \delta(x)\, dx = 1 \qquad\qquad b\lt 0\lt c\tag{11.2.2} \end{equation}
\begin{equation} x\,\delta(x) \equiv 0\tag{11.2.3} \end{equation}

It may be easiest to think of the delta function as the limit of a sequence of steps, each of which is higher and narrower than the previous step, such that the area under the step is always one; see Figure Figure 11.2.1.

Figure 11.2.1. The function \(\delta(x)\) can be approximated by a series of steps that get progressively thinner and higher in such a way that the area under the curve is always equal to one.

The properties of the delta function allow us to compute

\begin{align*} \int_{-\infty}^{\infty} f(x)\,\delta(x) \,dx \amp = \int_{-\infty}^{\infty} f(0)\,\delta(x) \,dx\\ \amp = f(0) \int_{-\infty}^{\infty} \delta(x) \,dx \nonumber\\ \amp = f(0) \nonumber \end{align*}

We can shift the “spike” in the delta function as usual, obtaining \(\delta(x-a)\text{.}\) This shifted delta function satisfies

\begin{equation} \int_{-\infty}^{\infty} f(x)\,\delta(x-a) \,dx = f(a)\tag{11.2.4} \end{equation}

Thus, the Dirac delta function can be used to pick out the value of a function at any desired point.

We can relate the delta function to the step function in the following way. Consider the function \(g(x)\) given by the integral

\begin{equation} g(x)=\int_{-\infty}^x \delta(u-a)\,du\tag{11.2.5} \end{equation}

Notice the variable \(x\) in the upper limit of the integral. The value of this function \(g(x)\) 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{11.2.6} \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{11.2.7} \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.

Prove that the derivative of the step function is the delta function, i.e. prove Equation (11.2.7).

Solution.

We need to show that

\begin{equation} \Int_{b}^{c} f(x)\,\delta(x-a) \,dx = f(a)\tag{11.2.8} \end{equation}

for \(b\lt a\lt c\text{,}\) where

\begin{equation} \delta(x-a)=\frac{d}{dx}\,\Theta(x-a)\tag{11.2.9} \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 \nonumber\\ \amp = \left. f(x)\, \Theta(x-a)\right|_b^c - \Int_{b}^{c} \frac{d}{dx}(f(x))\,\Theta(x-a) \,dx \nonumber\\ \amp = \left\{ f( c )-0\right\} - \Int_{a}^{c} \frac{d}{dx}(f(x))\,dx \nonumber\\ \amp = f( c )-\left\{\left. f(x)\right|_a^c\right\} \nonumber\\ \amp = f( c )-\left\{f( c )-f(a)\right\} \nonumber\\ \amp = f(a) \end{align*}