# Dirac delta in Fourier transforms?

1. May 29, 2007

### pivoxa15

Fourier transforms were invented before dirac delta functions but hidden in every fourier transform is a dirac delta function. But it went unnoticed until dirac came along? Then they argued for the legitamacy of the delta function but it is present in every fourier transform which is legitamate.

2. May 29, 2007

### Hurkyl

Staff Emeritus
What do you mean by that? One can define Fourier transforms, and develop their theory, without ever leaving the comfortable world of multivariable calculus.

(The dirac delta "function", which is a distribution and not a function, is not part of the comfortable world of multivariable calculus)

3. Sep 1, 2011

### csyrony

I discovered yesterday that the Dirac delta function was actually invented by Fourier 106 years before Dirac rediscovered it. It appears in his 1822 book "The analytical theory of heat" on page 434 of the 1878 English translation. I wouldn't have spotted it but it was picked up by Hawkins in his 1970 book "Lebesgue's Theory".

Fourier claimed that an arbitrary function $f(x)$ could be represented in the form:
$$f(x)=\frac{1}{2\pi}\int_a^bf(\alpha)d\alpha\int_{-\infty}^{\infty}\cos(px-p\alpha)dp$$
These days we'd write this as:
$$f(x)=\int_a^bf(\alpha)\delta(x-\alpha)d\alpha$$
where $\delta(x-\alpha)$ is the Dirac delta function. Does Fourier's claim that $$\delta(x-\alpha) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\cos(px-p\alpha)dp$$ have any validity?

The standard way to generate a Dirac delta function is to start with a symmetric function that integrates to one, say $g(t)$, and shrink the width and increase the height by a factor, say $\omega$, in such a way that the integral remains constant. It is easy to show that $\delta(t)=\omega g(\omega t)$ satisfies the conditions needed for a delta function when $\omega$ is an infinite number (for example, an arbitrary infinite number from the field of hyperreals $^*\mathbb{R}$).

Let $g(t)$ be the sinc function with integral normalised to one $$g(t)=\frac{1}{\pi}\frac{\sin(t)}{t}\mbox{. Then}\quad\delta(t)=\frac{1}{\pi}\frac{\sin(\omega t)}{t}$$

Take Fourier's integral from $-\omega$ to $\omega$ and let $t=x-\alpha$. Then:
$$\frac{1}{2\pi}\int_{-\omega}^{\omega}\cos(pt)dp = \frac{1}{2\pi}\left[\frac{\sin(pt)}{t}\right]_{-\omega}^\omega = \frac{1}{\pi}\frac{\sin(\omega t)}{t} = \delta(t)$$

Fourier concluded that since $x$ appears only in $\cos(px-p\alpha)$, which represents a differentiable function, the function $f$ "acquires in a manner by this transformation, all the properties of trigonometrical quantities; differentiations, integrations and summations of series thus apply to functions in general in the same manner as to exponential trigonometric functions."

Functions that are even discontinuous everywhere on the reals satisfy $f(x)=\int_a^bf(\alpha)\delta(x-\alpha)d\alpha$. Fourier has used the Dirac delta function to turn a function that is discontinuous everywhere on the reals to one that is continuous and continuously differentiable to all orders, on the infinitesimals.