# Dirac's Delta as Weight Function

The (notorious) Dirac delta-function is usually introduced as a function with the following properties: $\delta(x) = \begin{cases} 0 & \qquad x \neq 0 \\ \text{“\infty”} & \qquad x = 0 \end{cases}$

while $\newcommand{\d}{\,\mathrm{d}} % \int \! \delta(x) \d x = 1 \qquad \text{for each interval that includes 0.}$

But this “intuitive” characterization easily leads users astray, because it does not motivate a correct understanding of the behavior of expressions involving the delta-function. For instance, it does not prepare users for the scaling property of the delta-function: $\int \! \delta(a x) \d x = \frac{1}{a}$ Surely, $\delta(a x) = 0$ for all $x \neq 0\,$? And, at the same time, surely $a x = 0$ for $x = 0\,$? So, why is the value of the integral not 1?

A better way to think intuitively about $\delta(x)$ is not as a stand-alone function, but as a weight function or integration kernel. Properly understood, it only has meaning inside an integral. With this interpretation, $\int \! \d x \, \delta(x) \dots$ is an integral operator with the property that, when applied to a function $f(x)$: $\int \! \d x \, \delta(x) f(x) = f(0)$

When introduced like this, the familiar properties of the delta-function follow easily. The important thing is that the operator is defined exactly as written, with $\delta(x) \d x$, where the argument of the delta-function equals the dummy variable of integration given by $\d x$. It is not defined for any other forms of the kernel; if they arise, they must be reduced to the definition through variable substitutions: $\int \! \d x \, \delta(x+a) f(x) = \int \! \d u \, \delta(u) f(u-a) = f(-a)$ with $u = x+a \quad \Rightarrow \quad x = u-a \quad \Rightarrow \quad \d x = \frac{\d x}{\d u} \d u = 1 \cdot \d u = \d u$ and, in particular: $\int \! \d x \, \delta(a x) f(x) = \int \! \frac{1}{a} \d u \, \delta(u) f\left( \frac{u}{a} \right) = \frac{1}{a} f(0)$ with $u = ax \quad \Rightarrow \quad x = u/a \quad \Rightarrow \quad \d x = \frac{\d x}{\d u} \d u = \frac{1}{a} \d u$

This interpretation of $\delta(x)$ also neatly reproduces the central property of the delta-function, namely that $\int \! \delta(x) f(x) \d x = f(0)$, along the following lines: \begin{align*} \int \! f(x) \delta(x) \d x = & \int \! \left( f(0) + x f^\prime(0) + \frac{1}{2} x^2 f^{\prime \prime}(0) + \dots \right) \delta(x) \d x \\ = & f(0) \int \! \delta(x) \d x \\ = & f(0) \end{align*} All other terms vanish, because the integral operator is applied to a power of $x$, which (by the definition of the operator) yields the value of its argument at 0.

In a similar manner, and using integration by parts, the result $\int \! \d x \, x \, \delta^\prime(x) = - \int \! \d x \, \delta(x)$ can be obtained.

To repeat: the central point of my argument is that it is never proper to consider $\delta(x)$ as an entity itself. It only makes sense under an integral sign. Put differently: don’t think of $\delta(x)$ as a bare expression, only consider the integral operator $\int \! \d x \, \delta(x) \dots$.

I also want to be clear that the interpretation put forth above is intended as an “intuitive” way of reasoning about the properties of expressions involving $\delta(x)$, not a mathematically rigorous or precise definition, although it is very much in the spirit of treating $\delta(x)$ as a “distribution” in the sense of a “generalized function”.

Another good way to arrive at reliable insights about the behavior of the delta-function is to regard it as the limit of a sequence of approximations. Three such approximations are commonly used: \begin{align*} \delta(x) & = \lim_{\epsilon \rightarrow 0} \frac{1}{\sqrt{2 \pi} \epsilon} \mathrm{e}^{- \frac{1}{2} (x/\epsilon)^2} & & \text{Gaussian} \\ \delta(x) & = \lim_{\epsilon \rightarrow 0} \frac{1}{\pi \epsilon} \frac{1}{1 + (x/\epsilon)^2} & & \text{Cauchy-Poisson} \\ \delta(x) & = \lim_{\epsilon \rightarrow 0} \begin{cases} \frac{1}{2 \epsilon} & \text{for $|x| \lt \epsilon$} \\ 0 & \text{otherwise} \end{cases} & & \text{Step} \\ \end{align*}