# The math of the Dirac delta function?

1. Jan 29, 2009

### SW VandeCarr

I'm posting this here because I'm asking about the mathematical properties of the Dirac delta function, delta(x) which is zero for all non-zero real values of x and infinite when x is zero. The integral (-inf to +inf) of this function is said to be 1. How is this derived?

2. Jan 29, 2009

### tiny-tim

distributions

Hi SW VandeCarr!

The Dirac delta function, despite its name, isn't a function …

it's a distribution (a generalised function), used only in combination with other functions (usually to help in difficult integrations) …

see http://en.wikipedia.org/wiki/Distribution_(mathematics)

3. Jan 30, 2009

### SW VandeCarr

Re: distributions

Thanks tiny-tim. I thought the modern definition of a function is any mapping from one mathematical space to another. I guess that's what you mean by a generalized function. Nevertheless the integral of the Dirac delta function is given as unity in number of sources and I suppose it wouldn't work if this weren't true. It's possible to integrate, for example, the Gaussian distribution (whose integral is also unity). Is there no analytic way to integrate this "function."? If not, where does this result come from?

4. Jan 30, 2009

### arildno

5. Jan 30, 2009

### Tac-Tics

Re: distributions

Let me put more precisely what tiny-tim said. The Dirac Delta is not R->R. Meaning, it is not a map from the real numbers into the real numbers. It IS however, a plain old function in the set-theoretic sense.

There are a few formalisms of the Dirac Delta. The branch which studies it is called Distribution Theory or something like that. A "generalized function" or a distribution is actually a function (R->R) ->R, which maps real functions to real numbers. So the Dirac Delta would be defined as $$D(f) = f(0)$$. It takes an R->R function as it's input, f, and returns f evaluated at 0. We can define all sorts of distributions this way, often using integrals. Another common one would be $$G(f) = \int_{-\infty}^\infty f(x) e^{-x^2} dx$$, which takes a function f and returns the infinite integral of f times the gaussian function.

The nice thing about Distribution Theory is it gives a very clean way of handling Fourier Transforms without sweeping infinities "under the rug."

6. Jan 30, 2009

### Ben Niehoff

Re: distributions

It's not a result. It is the definition of the delta function. Really, the only necessary defining property of the delta function is that

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

for all test functions f(x) which are continuous at x=0.

7. Jan 31, 2009

### Preno

Distributions are basically "integral factors" of other functions. It works kinda like this:

You choose a suitable space of ("test") functions - for example, the space of smooth real functions R^n -> R with compact support or smooth complex functions R^n -> C which decrease fast enough near infinity (any derivative of a test function times any polynomial converges to zero). The latter are called "tempered distributions". Then you define distributions as continuous linear functionals on that space of test functions. ("Continuous" in the sense that, roughly, the derivatives of f_n [times any polynomial] uniformly approach the derivatives of f [times the polynomial].)

Clearly, the functional $\delta_a$ that assigns to a test function f its value f(a) at some point a is continuous and linear, so is the functional generated by any locally integrable function f that grows slowly near infinity (i.e. at most polynomially, along with its derivatives):

$\varphi \rightarrow \int_{R^n} f(x) \varphi(x) \textrm{d}V$

So we can look at "well-behaved" functions as both regular functions and distributions. One can then define for example the derivative and the Fourier/Laplace transform of a distribution so that the derivative/transform of a distribution generated by f is the distribution generated by the derivative/transform of f. In particular, <d',f> := - <d,f'> for any distribution d and test function f (<d,f> meaning d applied to f). You can try proving for yourself that the Dirac delta distribution is the derivative of the Heaviside step function.

ETA: note that the integral in

$\int_R \delta(x) \varphi(x) dx$

is just a notation for $\delta(\varphi)$, i.e. the application of the distribution delta to the test function phi, not an actual Lebesgue integral of some function.

Last edited: Jan 31, 2009
8. Jan 31, 2009

### SW VandeCarr

Thanks all. I will need to do more research based on your responses. If I can prove that the Dirac delta distribution is a derivative of the Heaviside step function, will that allow for an analytic solution such that the integral (-inf,+inf) of the Dirac delta distribution is unity, or is Ben Neihoff correct when he states ( if I understand correctly) that this integral of the Dirac delta is simply defined this way and not derived analytically?

As an aside, the function F(x)=log(x) (for all positive real bases) is infinite at x=0 and arguably the sum for all non-zero positive real test values 'a' of F(x) is zero. I don't know if this has anything to do with the Dirac delta, but I thought of it when trying to think of any common continuous symmetrical distributions that are infinite at x=0.

Last edited: Jan 31, 2009
9. Feb 1, 2009

### arildno

You won't be able to do that, since the Heaviside function is discontinuous, and hence not differentiable in the strict meaning of that term.
No.

As has been said, the Dirac delta "function" is not a "normal" function at all, it is what we call a "generalized function", or distribution, which can be regarded as a proper function on a function space.

The legitimacy of the abuse of notation of the integral representation of the effect of the Dirac delta "function", comes from the provable fact that we may form a sequence of integrals with respect to a class of test functions, so that in the limit of this integral sequence, the end result is to get out f(0), i.e, the "output" of the Dirac delta distribution.

I advise you to read my tutorial on this, or some other work.
I have no idea of what you are trying to say here, it sounds like complete nonsense.

10. Feb 1, 2009

### Preno

No, it's not an integral in any regular sense (Riemann, Lebesgue, whatever), it's just an intuitive notation for the application of the continous linear functional delta to a function. In particular,

$\int_{-\infty}^\infty \delta(x) \textrm{d}x = \delta(1) = 1(0) = 1,$

where 1 is the function f: f(x) = 1 for all x.
It's ok to think of the Dirac delta as being "infinite x=0", but be aware that this is just a metaphor, distributions don't actually have "values" at different points.

Anyway, "symmetric" distributions "infinite" at x=0 (other than multiple of delta and delta + a symmetric distribution) are for example the even derivatives of delta. An odd distribution (odd distributions are naturally defined by d(f(-x))=-d(f(x)) ) "infinite at x=0" (or rather "zero but behaving weirdly at x=0") is for example:

$\left(\textrm{v.p.} \frac{1}{x}\right)(\varphi) = \lim_{\epsilon\rightarrow 0+} \int_{|x|\geq \epsilon} \frac{\varphi(x)}{x} \textrm{d}x$
We're talking about distributional derivatives. Every distribution is infinitely differentiable (because the differentiation gets moved to the test function, which is by definition smooth), and it's easy to prove (a simple application of the definition of a distributional derivative and Newton's formula) that the derivative of the Heaviside step function is indeed the delta distribution.

Last edited: Feb 1, 2009
11. Feb 1, 2009

### arildno

I am perfectly well aware of both "weak derivatives" and distributional derivatives, thank you, and that, in a distributive sense, the derivative of the Heaviside function is the Delta function.

That is, however, quite clearly not the meaning by which the OP imbued the word "derivative", that is why it is correct to answer his question in the negative on that point.
Please try to include the OP next time you utilize the pronoun in "we were talking about.."-sentences.

12. Feb 1, 2009

### Preno

The OP was responding to my suggestion that he test his understanding of the definition of the distributional derivative by proving that the Dirac delta is the derivative of the Heaviside function. Talking about distributional derivatives was the context.

13. Feb 1, 2009

### arildno

Really? I thought the folllowing sentrence should make it clear that the OP is rather uncertain about what a "distribution" is:
Just because OP utilizes the word "distribution", doesn't mean he has properly learnt its import.
This is not to denigrate either you or the OP, but that the OP first and foremost needs to understand how a distribution differs from a normal function.
Speaking then of "distributional derivatives" is way too early.

14. Feb 1, 2009

### Preno

True.
Possibly. I don't have didactic experience with explaining distribution theory to people.

15. Feb 1, 2009

### arildno

Anyhow, just to make something clear: