# Integral of Dirac-Delta Fuction

1. Jun 20, 2014

### Bashyboy

Hello,

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. Jun 20, 2014

### micromass

Staff Emeritus
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

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

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.

3. Jun 20, 2014

### Bashyboy

To me, that seems odd that one would define the value of a functions integral. I thought the value of an integral followed from the way in which a function is defined.

4. Jun 20, 2014

### Matterwave

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:

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

5. Jun 20, 2014

### micromass

Staff Emeritus
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.

6. Jun 20, 2014

### homeomorphic

It may help to think of the delta function as a limit of something that can only be achieved by ordinary functions in the limit. You can get ordinary functions that approximate the behavior of the delta function by taking a function that is 0 everywhere except in a small interval where you get a little triangle (with corners smoothed out, if you like) that has an area of 1 underneath it. Then you make that interval shrink down to zero. It's slightly tricky because the ordinary limit of these functions is not quite the delta function. But with the right notion of limit, you will get the delta function (you just have to make the definition in terms of those integrals).

7. Jun 20, 2014

### George Jones

Staff Emeritus
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. Jun 28, 2014

### TSC

What about the three dimensional version of Dirac delta function?

9. Jun 28, 2014

### HallsofIvy

Staff Emeritus
Pretty much the same thing. "A distribution (or "generalized function") such that $$\int_0^\infty\int_0^\infty\int_0^\infty \delta(x,y,z)f(x,y,z)dxdydz= 1$$" for any integrable function f with compact support.

10. Jun 28, 2014

### micromass

Staff Emeritus
Rather, for any smooth function $f$ with compact support.

11. Jun 28, 2014

### TSC

Why is it sometimes we see this ${\delta ^3}\left( {\bf{r}} \right) = - \frac{1}{{4\pi }}{\nabla ^2}\left( {\frac{1}{r}} \right)$ ?

12. Jun 28, 2014

### pwsnafu

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.

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: Jun 28, 2014
13. Jun 28, 2014

### pwsnafu

Edit Cross post.

First notation. Here $\delta^3$ simply means Dirac in three dimensions. You need context though because it could mean Dirac cubed. In this case it is fine because it is clear we are working in vector calculus. Some authors write $\delta(x,y,z)$ others $\delta^3(x,y,z)$ and others $\delta^3(\mathbf{x})$. They mean the same.

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

14. Jun 29, 2014

### TSC

How do you prove that equation?

15. Jun 29, 2014

### MisterX

One may do so using the test function method.

We observe that the Laplacian of 1/r is zero every except possibly at the origin. Then we assume a test function $f$ 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).

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

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

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

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

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

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 $4\pi a^2\left(-\frac{1}{a^2} \right)f(\mathbf{0}) = -4\pi f(\mathbf{0})$.

So we determined

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

16. Jun 29, 2014

### TSC

Haha, this is rather neat. Thanks a lot. It seems that the same technique can also be used in the one-dimensional case.
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. Jun 29, 2014

### pwsnafu

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.

The problem is that the "integral" in $\int_{-\infty}^\infty \delta(x) g(x) dx$ is not actually an integral, in the same way as $\frac{dy}{dx}$ is not actually a fraction. We are abusing notation because it is convenient. This is why textbooks on generalized functions avoid the integral notation and use $\langle \delta, g \rangle$ or $\delta[g]$. Books that don't specialize in this topic (especially physics books) will use the integral notation without explanation confusing students.

18. Jun 29, 2014

### TSC

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

19. Jun 30, 2014

### pwsnafu

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 $T_{[\delta]} g(x) = \langle \delta(y), g(y-x) \rangle = g(-x)$.