# The math of the Dirac delta function?

• SW VandeCarr
In summary, the Dirac delta function is a distribution, not a function, and is used in combination with other functions to help with difficult integrations. It can be defined as a continuous linear functional on a space of "test" functions, and its defining property is that its integral over the real numbers is equal to the value of the function being tested at the origin. The Dirac delta distribution can be seen as the derivative of the Heaviside step function, and the integral of the Dirac delta distribution is defined to be 1, rather than being derived analytically.
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?

distributions

SW VandeCarr said:
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?

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)

tiny-tim said:
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)

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?

SW VandeCarr said:
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.

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."

SW VandeCarr said:
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?

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.

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:
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:
SW VandeCarr said:
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,
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.
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?
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.
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.

I have no idea of what you are trying to say here, it sounds like complete nonsense.

SW VandeCarr said:
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?
No, it's not an integral in any regular sense (Riemann, Lebesgue, whatever), it's just an intuitive notation for the application of the continuous 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.
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.
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:

$\left(\textrm{v.p.} \frac{1}{x}\right)(\varphi) = \lim_{\epsilon\rightarrow 0+} \int_{|x|\geq \epsilon} \frac{\varphi(x)}{x} \textrm{d}x$
arildno said:
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.
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:
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.

arildno said:
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.
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.

Preno said:
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. That was the context, and I think I know what I meant when I said that.
Really? I thought the folllowing sentrence should make it clear that the OP is rather uncertain about what a "distribution" is:
will that allow for an analytic solution such that the integral (-inf,+inf) of the Dirac delta distribution is unity,

Just because OP utilizes the word "distribution", doesn't mean he has properly learned 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.

arildno said:
Just because OP utilizes the word "distribution", doesn't mean he has properly learned its import.
True.
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.
Possibly. I don't have didactic experience with explaining distribution theory to people.

Anyhow, just to make something clear:

arildno said:
I have no idea of what you are trying to say here, it sounds like complete nonsense.

It was poorly stated. All I was really trying to say is that the logarithms for the totality of positive real numbers sum to zero and that the logarithm of zero is negative infinity (which Preno addressed) . Thanks to all for your help. I need to work on this.

## 1. What is the Dirac delta function?

The Dirac delta function, denoted as δ(x), is a mathematical function that is defined as 0 for all values of x except at x = 0 where it is infinite. It is commonly used in physics and engineering to represent a point mass or impulse at a specific location.

## 2. How is the Dirac delta function related to the Kronecker delta?

The Kronecker delta, denoted as δij, is a discrete version of the Dirac delta function. It is defined as 1 if i = j and 0 if i ≠ j. The Dirac delta function can be seen as the continuous limit of the Kronecker delta as the spacing between the discrete values approaches 0.

## 3. What is the integral of the Dirac delta function?

The integral of the Dirac delta function is 1. This can be seen from the definition of the function, where it is infinite at x = 0 and 0 everywhere else. When integrated over the entire real line, the function "picks out" the value at x = 0, which is 1.

## 4. How is the Dirac delta function used in Fourier analysis?

The Dirac delta function is commonly used in Fourier analysis to represent a periodic function as a sum of sine and cosine waves. By taking the inverse Fourier transform of a function's Fourier series, the Dirac delta function can be used to reconstruct the original function.

## 5. What are some applications of the Dirac delta function in real-world problems?

The Dirac delta function has many applications in physics and engineering, such as in electrical circuits, signal processing, and quantum mechanics. It is also used in probability and statistics to model point events. Additionally, the Dirac delta function has been used in image processing and computer graphics to simulate sharp edges and boundaries.

• Calculus
Replies
25
Views
2K
• Calculus
Replies
2
Views
2K
• Calculus
Replies
32
Views
4K
• Calculus
Replies
2
Views
968
• Calculus
Replies
18
Views
2K
• Calculus
Replies
1
Views
1K
• Calculus
Replies
4
Views
1K
• Calculus
Replies
5
Views
3K
• Calculus
Replies
3
Views
3K
• Calculus
Replies
24
Views
2K