The Dirac-Delta Function

[Eli]

Dirac delta function, Dirac-\(\delta\) is extremely very powerful tool, classically applied in calculus, for example, in transforming complicated integral functions to make them easily integrable, and in transformations between real and Fourier spaces. Most of its applications span into many areas of physics and applied mathematics.

We can think of the Dirac delta function as a rectangle function \( R(x)\) whose full-width at half-maximum (FWHM) is \(\sigma\), in which as the value of \(\sigma\) becomes smaller, the width of the rectangle becomes smaller and smaller and the height of the rectangle becomes larger but the area under the curve always remains unity, given by
\begin{align}
\delta(x-a) = \lim_{\sigma \to 0} R_{\sigma}(x).
\end{align}
Because of this property, if we integrate \(\delta(x-a)\) multiplied by an arbitrary well-behaved function, \(f(x)\), it only picks up its value at \(f(x=a)\).

The Dirac delta function can formally be defined as
\begin{align}
\delta(x) =
\begin{cases}
0; & x\ne 0 \\
\infty; & x = 0,
\end{cases}
\end{align}
such that
\begin{align}
\int_{-\infty}^{+\infty}\delta(x){\rm d}x = 1.
\end{align}
We can also define the Dirac delta function as
\begin{align}
\int_{-\infty}^{+\infty}\delta(x){\rm d}x = H(x)|_{-\infty}^{+\infty} = 1,
\end{align}
where \(H(x)\) is the Heaviside step function or the unit step function defined as
\begin{align}
H(x) = \begin{cases}
0; & x<0 \\
1; & x \ge 0.
\end{cases}
\end{align}

Here is an example of how to evaluate integrals using the Dirac delta function:
\begin{align}
\begin{split}
\int_{-\infty}^{\infty}f(x)\delta(x-a){\rm d}x &\quad = \int_{-\infty}^{\infty}f(a)\delta(x-a){\rm d}x\\
&\quad = f(a) \int_{-\infty}^{\infty}\delta(x-a){\rm d}x\\
&\quad = f(a).
\end{split}
\end{align}
This result follows from the definition, and you must have noticed that
\begin{align}
\int_{-\infty}^{\infty}f(x)\delta(x){\rm d}x = f(0).
\end{align}

The Dirac delta function is usually represented by the two functions
\begin{align}\label{dr1}
\begin{split}
\int_{-\infty}^{+\infty} f(x)\delta(x-a){\rm d}x &\quad = f(a),\\
\delta(x-a) &\quad = 0 \ \text{for} \ x \ne a.
\end{split}
\end{align}
At \(x = a\), the Dirac delta function, \(\delta(x-a)\) is not defined, instead it has an infinite value, and we present it by a spike of unit height at that point.

We can further write the Dirac delta function in the form,
\begin{align}\label{dr3}
\delta(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{\pm ikx}{\rm d}k.
\end{align}
If we shift origin in the above integral by \(x'\), we have
\begin{align}\label{dr4}
\delta(x-x') = \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{\pm ik(x-x')}{\rm d}k.
\end{align}
These two latter equations are called the integral representation of the Dirac delta function, and are extremely very useful.

Let's use the above equation and consider a well-behaved function, \(f(x)\) and note that
\begin{align}\label{fr1}
f(x) = \int_{-\infty}^{+\infty}\delta(x-x')f(x'){\rm d}x' = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty}e^{ik(x-x')}f(x'){\rm d}x'{\rm d}k.
\end{align}
If we separate out an integral with \(x'{\rm s}\) above, we have
\begin{align}\label{fr2}
F(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x')e^{-ikx'}{\rm d}x',
\end{align}
and that \(f(x)\) becomes
\begin{align}\label{fr3}
f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}F(k)e^{ikx}{\rm d}k.
\end{align}
These two latter equations, describe what is known as Fourier Integral Theorem.

The function \(F(k)\) is the Fourier transform of \(f(x')\) and \( f(x)\) is the inverse Fourier transform.

We can remove the prime on \(x'\) (since integral with \(x'\) is definite) and write
\begin{align}\label{fr4}
F(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x)e^{-ikx}{\rm d}x,
\end{align}
\begin{align}\label{fr5}
f(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}F(k)e^{ikx}{\rm d}k.
\end{align}
Explicitly, \(F(k)\) is the Fourier transform of \(f(x)\) and \(f(x)\) is its inverse Fourier transform - the Fourier transform of \(F(k)\).

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[Eli]

Related topics:

laplace-transforms-convolution-theorem-492

fourier-transform-4730

[Eli]

Great topic I wrote a while ago!

@Rest