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---
title: "Dirac delta function"
firstLetter: "D"
publishDate: 2021-02-22
categories:
- Mathematics
- Physics
date: 2021-02-22T21:35:38+01:00
draft: false
markup: pandoc
---
# Dirac delta function
The **Dirac delta function** $\delta(x)$, often just the **delta function**,
is a function (or, more accurately, a [Schwartz distribution](/know/concept/schwartz-distribution/))
that is commonly used in physics.
It is an infinitely narrow discontinuous "spike" at $x = 0$ whose area is
defined to be 1:
$$\begin{aligned}
\boxed{
\delta(x) =
\begin{cases}
+\infty & \mathrm{if}\: x = 0 \\
0 & \mathrm{if}\: x \neq 0
\end{cases}
\quad \mathrm{and} \quad
\int_{-\varepsilon}^\varepsilon \delta(x) \dd{x} = 1
}
\end{aligned}$$
It is sometimes also called the **sampling function**, thanks to its most
important property: the so-called **sampling property**:
$$\begin{aligned}
\boxed{
\int f(x) \: \delta(x - x_0) \: dx = \int f(x) \: \delta(x_0 - x) \: dx = f(x_0)
}
\end{aligned}$$
$\delta(x)$ is thus quite an effective weapon against integrals. This may not seem very
useful due to its "unnatural" definition, but in fact it appears as the
limit of several reasonable functions:
$$\begin{aligned}
\delta(x)
= \lim_{n \to +\infty} \!\Big\{ \frac{n}{\sqrt{\pi}} \exp(- n^2 x^2) \Big\}
= \lim_{n \to +\infty} \!\Big\{ \frac{n}{\pi} \frac{1}{1 + n^2 x^2} \Big\}
= \lim_{n \to +\infty} \!\Big\{ \frac{\sin(n x)}{\pi x} \Big\}
\end{aligned}$$
The last one is especially important, since it is equivalent to the
following integral, which appears very often in the context of
[Fourier transforms](/know/concept/fourier-transform/):
$$\begin{aligned}
\boxed{
\delta(x)
%= \lim_{n \to +\infty} \!\Big\{\frac{\sin(n x)}{\pi x}\Big\}
= \frac{1}{2\pi} \int_{-\infty}^\infty \exp(i k x) \dd{k}
\:\:\propto\:\: \hat{\mathcal{F}}\{1\}
}
\end{aligned}$$
When the argument of $\delta(x)$ is scaled, the delta function is itself scaled:
$$\begin{aligned}
\boxed{
\delta(s x) = \frac{1}{|s|} \delta(x)
}
\end{aligned}$$
*__Proof.__ Because it is symmetric, $\delta(s x) = \delta(|s| x)$. Then by
substituting $\sigma = |s| x$:*
$$\begin{aligned}
\int \delta(|s| x) \dd{x}
&= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|}
\end{aligned}$$
*__Q.E.D.__*
An even more impressive property is the behaviour of the derivative of
$\delta(x)$:
$$\begin{aligned}
\boxed{
\int f(\xi) \: \delta'(x - \xi) \dd{\xi} = f'(x)
}
\end{aligned}$$
*__Proof.__ Note which variable is used for the
differentiation, and that $\delta'(x - \xi) = - \delta'(\xi - x)$:*
$$\begin{aligned}
\int f(\xi) \: \dv{\delta(x - \xi)}{x} \dd{\xi}
&= \dv{x} \int f(\xi) \: \delta(x - \xi) \dd{x}
= f'(x)
\end{aligned}$$
*__Q.E.D.__*
This property also generalizes nicely for the higher-order derivatives:
$$\begin{aligned}
\boxed{
\int f(\xi) \: \dv[n]{\delta(x - \xi)}{x} \dd{\xi} = \dv[n]{f(x)}{x}
}
\end{aligned}$$
## References
1. O. Bang,
*Applied mathematics for physicists: lecture notes*, 2019,
unpublished.
|
