The Dirac Delta Function

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Section 8.4 The Dirac Delta Function

Definition 8.4.1. The Dirac Delta Function.

The Dirac delta function \(\delta(x)\) has has the following defining properties:

\begin{equation} \delta(x) = \begin{cases} 0\quad \amp x\not= 0\\ \infty\quad\amp x=0 \end{cases}\tag{8.4.1} \end{equation}

\begin{equation} \int_b^c \delta(x)\, dx = 1 \qquad\qquad b\lt 0\lt c\tag{8.4.2} \end{equation}

\begin{equation} x\,\delta(x) \equiv 0\text{.}\tag{8.4.3} \end{equation}

Using the Dirac Delta Function in an Integral.

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\\ \amp = f(0) \end{align*}

In the first equality, it is safe to replace the function \(f(x)\) with its value \(f(0)\) at \(x=0\) since everywhere else the integrand is zero, due to the delta function.

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{8.4.4} \end{equation}

To Remember.

The Dirac delta function can be used inside an integral to pick out the value of a function at any desired point.

Visualization of the Delta Function.

To understand why the delta function under an integral sign picks out a particular value of the integrand, it may help to think of the delta function as the limit of a sequence of steps \(\delta_{\epsilon}(x)\text{,}\) Each step is narrower (width \(2\epsilon\)) and higher (height \(1/(2\epsilon)\)) than the previous step, such that the area under each step is always one; see Figure Figure 8.4.2.

Figure 8.4.2. 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.

Then

\begin{equation} \int_{-\infty}^{\infty} f(x)\, \delta_{\epsilon}(x-a)\, dx\tag{8.4.5} \end{equation}

give the average value of \(f(x)\) on the interval determined by \(\epsilon\text{.}\) In the limit that \(\epsilon\) becomes infinitesimally small, then the peak becomes infinitely narrow and infinitely high in just the right way to pick out the value of the function at \(x=a\text{.}\)