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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 called the **delta function**,
+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**, due 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 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:
+
+$$\begin{aligned}
+    \boxed{
+        \int f(\xi) \: \dv[n]{\delta(x - \xi)}{x} \dd{\xi} = \dv[n]{f(x)}{x}
+    }
+\end{aligned}$$