11. Dirac’s delta

Dirac’s delta \(\delta(x)\) is a distribution, or generalized function, with the following properties

(11.1)\[\int_{D} \delta(x-x_0) \, dx = 1 \quad \text{if $x_0 \in D$}\]

(11.2)\[\int_{D} f(x) \delta(x-x_0) \, dx \quad \text{if $x_0 \in D$}\]

for \(\forall f(x)\) “regular” todo what does regular mean?

11.1. Dirac’s delta in terms of regular functions

11.1.1. Piece-wise constant

\[\begin{split}\delta(x) \sim r_{\varepsilon}(x) = \begin{cases} \frac{1}{\varepsilon} & x \in \left[-\frac{\varepsilon}{2}, \frac{\varepsilon}{2} \right] \\ 0 & \text{otherwise} \end{cases}\end{split}\]

Properties - proof.

1. Unitariety

   \[\int_{x=-\infty}^{\infty} r_{\varepsilon}(x-x_0) \, dx = \int_{x=x_0-\frac{\varepsilon}{2}}^{x_0+\frac{\varepsilon}{2}} \frac{1}{\varepsilon} \, dx = 1 \ , \]

   for \(\forall \varepsilon\);

2. Shift property, using mean-value theorem of continuous functions

   \[\int_{x=-\infty}^{\infty} r_{\varepsilon}(x-x_0) f(x) \, dx = \int_{x=x_0-\frac{\varepsilon}{2}}^{x_0+\frac{\varepsilon}{2}} \frac{1}{\varepsilon} f(x) \, dx = \frac{1}{\varepsilon} \varepsilon f(\xi) \ , \]

   with \(\xi \in \left[x_0-\frac{\varepsilon}{2}, x_0+\frac{\varepsilon}{2}\right]\), for the mean value theorem. As \(\varepsilon \rightarrow 0\), \(\xi \rightarrow x_0\), and thus

   \[\int_{x=-\infty}^{\infty} r_{\varepsilon}(x-x_0) f(x) \, dx \rightarrow f(x_0) \]

11.1.2. Piecewise-linear

\[\begin{split}\delta(x) \sim t_{\varepsilon}(x) = \begin{cases} \frac{2}{\varepsilon} \left( 1 - \frac{2 |x|}{\varepsilon} \right) & x \in \left[-\frac{\varepsilon}{2}, \frac{\varepsilon}{2} \right] \\ 0 & \text{otherwise} \end{cases}\end{split}\]

Properties - proof

1. Unitariety

   \[\int_{x=-\infty}^{\infty} t_{\varepsilon}(x-x_0) \, dx = \int_{x=x_0-\frac{\varepsilon}{2}}^{x_0+\frac{\varepsilon}{2}} \frac{2}{\varepsilon} \left( 1 - \frac{2 |x|}{\varepsilon} \right) \, dx = \frac{1}{2} \varepsilon \frac{2}{\varepsilon} = 1 \ , \]

   for \(\forall \varepsilon\);

2. Shift property, using mean-value integration scheme in \(x \in \left[x_0-\frac{\varepsilon}{2}, x_0 \right]\), \(x \in \left[x_0, x_0+\frac{\varepsilon}{2} \right]\) (todo why?)

   \[\begin{split}\begin{aligned} \int_{x=-\infty}^{\infty} t_{\varepsilon}(x-x_0) f(x) \, dx & = \int_{x=x_0-\frac{\varepsilon}{2}}^{x_0+\frac{\varepsilon}{2}} \frac{2}{\varepsilon} \left( 1 - \frac{2 |x-x_0|}{\varepsilon} \right) f(x) \, dx = \\ & = \int_{x=x_0-\frac{\varepsilon}{2}}^{x_0} \frac{2}{\varepsilon} \left( 1 - \frac{2 |x-x_0|}{\varepsilon} \right) f(x) \, dx + \int_{x=x_0}^{x_0+\frac{\varepsilon}{2}} \frac{2}{\varepsilon} \left( 1 - \frac{2 |x-x_0|}{\varepsilon} \right) f(x) \, dx = \\ & = \frac{\varepsilon}{2} \frac{2}{\varepsilon} \left( 1 - \frac{2}{\varepsilon}\frac{\varepsilon}{4} \right) f\left(x_0-\frac{\varepsilon}{4} \right) \, dx + \frac{\varepsilon}{2} \frac{2}{\varepsilon} \left( 1 - \frac{2}{\varepsilon}\frac{\varepsilon}{4} \right) f\left(x_0+\frac{\varepsilon}{4} \right) \, dx = \\ & = \frac{1}{2} f\left( x_0 - \frac{\varepsilon}{4} \right) + \frac{1}{2} f\left( x_0 + \frac{\varepsilon}{4} \right) \end{aligned}\end{split}\]

   As \(\varepsilon \rightarrow 0\)

   \[\int_{x=-\infty}^{\infty} t_{\varepsilon}(x-x_0) f(x) \, dx \rightarrow f(x_0) \]

11.1.3. Gaussian approximation

For \(\alpha \rightarrow +\infty\),

\[\varphi_{\alpha}(x) = \sqrt{\frac{\alpha}{\pi}}e^{-\alpha x^2} \sim \delta(x)\]

Properties - proof

Fourier transform of \(\varphi_{\alpha}(x)\) reads

\[\begin{split}\begin{aligned} \mathscr{F}\{ \varphi_{\alpha}(x) \}(k) & = \int_{x=-\infty}^{+\infty} \varphi_\alpha(x) e^{-ikx} \, dx = \\ & = \int_{x=-\infty}^{+\infty} \sqrt{\frac{\alpha}{\pi}} e^{-\alpha x^2} e^{-ikx} \, dx = \\ & = \sqrt{\frac{\alpha}{\pi}} \int_{x=-\infty}^{+\infty} e^{-\alpha \left( x + i \frac{k}{2 \alpha} \right)^2} \, dx \, e^{-\frac{k^2}{4 \alpha}} = \\ & = \sqrt{\frac{\alpha}{\pi}} \, \sqrt{\frac{\pi}{\alpha}} \, e^{-\frac{k^2}{4 \alpha}} = e^{-\frac{k^2}{4 \alpha}} \ ,\\ \end{aligned}\end{split}\]

for \(\alpha \rightarrow +\infty\),

\[\mathscr{F}\{ \varphi_{\alpha}(x) \}(k) \rightarrow 1\]

Fourier transform of Dirac’s delta is \(1\), as shown in (14.3), thus \(\varphi_\alpha(x) \rightarrow \delta(x)\) for \(\alpha \rightarrow +\infty\).

11.1.4. Fourier anti-transform

For \(a \rightarrow + \infty\),

(11.3)\[\delta(x) \sim \int_{y=-a}^{+a} e^{i 2 \pi y x} \, dy = \frac{1}{2 \pi} \int_{k=-2\pi a}^{2 \pi a} e^{i k x} \, dk \ ,\]

or

\[\delta \sim 2 \int_{y=0}^{a} \cos(2 \pi y x) \, dy \ .\]

Proof of the equilvanece

\[\begin{split}\begin{aligned} \delta(x) & \sim \frac{1}{2 \pi} \int_{k=-2\pi a}^{2 \pi a} e^{i k x} \, dk = \frac{1}{2 \pi} \left( \int_{k=-2\pi a}^{0} e^{i k x} \, dk + \int_{0}^{k=2\pi a} e^{i k x} \, dk \right) = \frac{1}{2 \pi} \int_{k = 0}^{2 \pi a} \left( e^{ikx} + e^{ikx} \right) \, dx = \frac{1}{\pi} \int_{x=0}^{2 \pi a} \cos(k x) \, dk \\ & = \int_{y=-a}^{+a} e^{i 2 \pi y x} \, dy = \dots = \int_{y = 0}^{a} (e^{i 2 \pi y x} + e^{i 2 \pi y x}) \, dy = 2 \int_{y=0}^{a} \cos(2 \pi y x) \, dy \ . \end{aligned}\end{split}\]

11.1.5. \(\text{sinc}(x)\) approximation

For \(a \rightarrow +\infty\)

\[\delta(x) \sim \frac{\sin(2 \pi x a)}{\pi x}\]

Proof

Directly follows from integral of the approximation (11.3)

\[\int_{y=-a}^{+a} e^{i 2 \pi y x} \, dy = \frac{1}{i 2 \pi x} \left. e^{i 2 \pi y x}\right|_{y=-a}^{+a} = \frac{1}{\pi x} \frac{e^{i 2 \pi a x} - e^{-i 2 \pi a x}}{2 i} = \frac{\sin(2 \pi x a)}{\pi x}\]

11.1.6. Fourier series

For \(x \in [-\pi, \pi]\), and \(N \rightarrow +\infty\), Fourier series of Dirac’s delta (train with period \(2\pi\)) reads

\[\delta(x) \sim \frac{1}{2\pi}\sum_{n=-N}^{N} e^{i n x} = \frac{1}{2 \pi} \frac{\sin\left(\left(N+\frac{1}{2}\right)x\right)}{\sin\left( \frac{x}{2} \right)}\]

or the \(T\)-periodic Dirac’s delta train,

\[\delta(x) \sim \frac{1}{T}\sum_{n=-N}^{N} e^{i n \frac{2\pi}{T} x} \ .\]

todo Write the proof of the last expression, using the relation between complex exponentials and cosine and sine

Proof

Coefficients of the Fourier series of Dirac’s delta (train with period \(T = 2 \pi\)) are evaluated using the expression (14.2)

\[c_n = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \delta(t) e^{-i n \frac{2\pi}{2 \pi}t} = \frac{1}{2\pi} \ ,\]

and thus the complex Fourier series (14.1) of Dirac’s delta reads

\[\delta(x) \sim \sum_{n=-\infty}^{+\infty} c_n e^{i n \frac{2\pi}{T}x} = \frac{1}{2 \pi} \sum_{n=-\infty}^{+\infty} e^{i n x}\]

Obs. here, integration interval \([-\pi,\pi]\) to “avoid troubles” with Dirac’s delta on the extreme points of the interval (it would give \(1/2\) and \(1/2\) contributions on both extremes…)

It’s possible to write the \(T\)-periodic Dirac’s delta train as

\[\delta(x) \sim \sum_{n=-\infty}^{+\infty} c_n e^{i n \frac{2\pi}{T}x} = \frac{1}{T} \sum_{n=-\infty}^{+\infty} e^{i n \frac{2\pi}{T} x}\]

Integral \(\ I = \int_{-\infty}^{+\infty} e^{-\alpha x^2} \, dx\)

\[\begin{split}\begin{aligned} I^2 & = \int_{x=-\infty}^{+\infty} e^{-\alpha x^2} \, dx \, \int_{y=-\infty}^{+\infty} e^{-\alpha y^2} \, dy = \\ & = \int_{x=-\infty}^{+\infty} \int_{y=-\infty}^{+\infty} e^{-\alpha (x^2 + y^2)} \, dx \, dy = \\ & = \int_{\theta=0}^{2\pi} \int_{r=0}^{+\infty} e^{-\alpha r^2} \, r \, dr \, d \theta = \\ & = 2 \pi \frac{1}{2 \alpha} \int_{r=0}^{+\infty} e^{-\alpha r^2} d \left(\alpha r^2 \right) = \\ & = \frac{\pi}{\alpha} \left[ - e^{\alpha r^2} \right]\bigg|_{r = 0}^{+\infty} = \frac{\pi}{\alpha} \ . \end{aligned}\end{split}\]