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I am curious as to why the integral of the Delta-Dirac function is one? It would seem to follow from the way in which you define the function, but any source I have read seems build this fact into the definition of the function.

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- #1

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I am curious as to why the integral of the Delta-Dirac function is one? It would seem to follow from the way in which you define the function, but any source I have read seems build this fact into the definition of the function.

- #2

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Yes, it is by definition. A dirac delta function by definition is a "function ##\delta## such that for each smooth function with compact support ##f## holds that

I am curious as to why the integral of the Delta-Dirac function is one? It would seem to follow from the way in which you define the function, but any source I have read seems build this fact into the definition of the function.

[tex]\int_{-\infty}^{+\infty}f(x)\delta(x)dx = f(0)[/tex]

The question of course is whether this function exists. It of course does not. It's not a actual function with domain ##\mathbb{R}## and codomain ##\mathbb{R}##. However, we can make it mathematically precise as a distribution.

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- #4

Matterwave

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For intuitive reasons (non-rigorous) you can think of the delta function as "eliminating all other values of f(x)". In other words, when I multiply ##f(x)## with ##\delta(x)##, intuitively speaking, I get a function that is only non-zero at x=0. What this function actually looks like is not intuitive since this function has to be non-zero at only 1 point, but have a finite area of 1 underneath. This is why the delta function is not really a function, but a distribution.

But from this definition, if we take ##f(x)=1##, then ##f(0)=1## and so we should get:

$$\int_{-\infty}^{\infty} f(x)\delta(x) dx=f(0)=1=\int_{-\infty}^{\infty} 1\delta(x) dx$$

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Yes, for regular functions we define the function and then calculate the integral. But the delta function is not a regular function, for those, we cannot define the function values. We can only define the integral. Something like ##\delta(0)## makes little sense.

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- #7

George Jones

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Have you comes across the concept of normalized functions (actual functions, not distributions like the delta function), perhaps in special functions, in electromagnetism, or in quantum theory?

- #8

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What about the three dimensional version of Dirac delta function?

- #9

HallsofIvy

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- #10

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Rather, for any

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- #12

pwsnafu

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The right hand side looks wrong.

While it is true that Dirac is usually defined as an element of ##\mathcal{D}'##, because it is a distribution with compact support, it actually sits in ##\mathcal{E}'##, where ##\mathcal{E}## is the space of smooth functions (with or without compact support) with topology given by convergence through all derivatives.Rather, for anysmoothfunction ##f## with compact support.

Or in symbols ##\mathcal{D}(\Omega) = C^\infty_c(\Omega)## and ##\mathcal{E}(\Omega) = C^\infty(\Omega)##, and there exists an embedding of ##\mathcal{E}'## into ##\mathcal{D}'##.

Last edited:

- #13

pwsnafu

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First notation. Here ##\delta^3## simply means Dirac in three dimensions. You need context though because it

IIRC, the right hand side is what Dirac equals to when you change basis to spherical coordinates.

- #14

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How do you prove that equation?

- #15

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One may do so using the test function method.How do you prove that equation?

We observe that the Laplacian of 1/r is zero every except possibly at the origin. Then we assume a test function [itex]f[/itex] with nice properties. We then integrate, splitting the integral up into two regions - one a small sphere around the origin with radius a, and another everything else (which integrates to 0).

[itex]\int_0^{2\pi} d\phi \int_0^\pi d\theta \int_0^a \mathop{dr} r^2 \sin \theta f(\mathbf{r})\boldsymbol{\nabla}^2 \frac{1}{r}[/itex]

We shrink the radius of the sphere down to zero. We argue as the radius becomes arbitrarily small, value of [itex]\mathbf{0}[/itex], [itex]f(\mathbf{r})[/itex] is essentially constant within the sphere so we can take it outside the integral. So we are left with

[itex] f(\mathbf{0})\int_0^{2\pi} d\phi \int_0^\pi d\theta \int_0^a \mathop{dr} r^2 \sin \theta\, \boldsymbol{\nabla} \cdot \left(\boldsymbol{\nabla} \frac{1}{r}\right)[/itex].

Next we use the divergence theorem, noting [itex]\boldsymbol{\nabla} \frac{1}{r} = -\frac{\hat{\mathbf{r}}}{r^2}[/itex]. The dot product with the unit normal to the sphere is the same everywhere, [itex] -\left.\frac{\hat{\mathbf{r}} \cdot \mathbf{n}}{r^2} \right|_{r =a} = -\frac{1}{a^2}[/itex].

[itex] f(\mathbf{0})\int \int \left(\boldsymbol{\nabla} \frac{1}{r}\right)\cdot \mathbf{n} \mathop{dS} =f(\mathbf{0})\int \int \left( -\frac{1}{a^2} \right)\mathop{dS} [/itex]

Since it's the same everywhere on the sphere to obtain the answer we just multiply it by the surface area of the sphere, to obtain the final result [itex]4\pi a^2\left(-\frac{1}{a^2} \right)f(\mathbf{0}) = -4\pi f(\mathbf{0}) [/itex].

So we determined

[itex]\int \int \int dV f(\mathbf{r})\boldsymbol{\nabla}^2 \frac{1}{r} = -4\pi f(\mathbf{0}) [/itex].

- #16

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It seems that there are two ways to define a dirac delta function:

(1) Say that the area under the graph is always 1, and that the "function" is everywhere zero except at the origin.(What it looks like)

(2) Say that the "function" has the property of sifting out f(0) if integrated with any f(x). (What it does)

Which one is a better definition of dirac delta function?

- #17

pwsnafu

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They are both heuristics and don't work as definitions (doing either breaks properties of integration). As has been stated above the Dirac delta is the functional defined as ##g(x) \mapsto g(0)##, for any smooth function g.

It seems that there are two ways to define a dirac delta function:

(1) Say that the area under the graph is always 1, and that the "function" is everywhere zero except at the origin.(What it looks like)

(2) Say that the "function" has the property of sifting out f(0) if integrated with any f(x). (What it does)

Which one is a better definition of dirac delta function?

The problem is that the "integral" in ##\int_{-\infty}^\infty \delta(x) g(x) dx## is not

- #18

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This is weird. I have never seen dirac delta appear without integration.

- #19

pwsnafu

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The most explicit example I can give would be Colombeau algebra (nb the Wikipedia article is not very good). There the Dirac delta is the reflection operator, i.e. ##g(x) \mapsto g(-x)##. The standard notation we use is ##T_{[\delta]} g(x) = \langle \delta(y), g(y-x) \rangle = g(-x)##.This is weird. I have never seen dirac delta appear without integration.

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