Quick Question on the Dirac Delta Function

In summary, the Dirac delta function is not a function, but rather a distribution that is defined by the property that when integrated with a function, it returns the value of that function at the origin. However, when used in integrals with different bounds, it can lead to different results depending on the chosen representation for the delta function. This suggests that there may not be a simple answer to the question posed, and that the use of limits and delta functions in integrals should be approached with caution.
  • #1
G01
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The Dirac delta function, [tex]\delta (x)[/tex] has the property that:

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

Will this same effect happen for the following bounds on the integral:

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

My intuition tells me that it should, but the fact that the peak of the delta function lies on one of the bounds makes me think I should double check my reasoning. So, does anyone know if (2) above is correct?

Thanks for any advice you can offer.
 
Last edited:
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  • #2
According to mathematica integral from 0 to infinity and the integral from negative infinity to 0 are equal to f(0)/2.
 
  • #3
I think I've seen sometimes integration limits like this

[tex]
\lim_{\epsilon\to 0^+}\int\limits_{0-\epsilon}^{\infty}
[/tex]

to avoid this problem, which suggests that there probably is not very simple answer to the question. For example

[tex]
\lim_{n\to\infty} \int\limits_0^{\infty} f(x) \sqrt{\frac{n}{\pi}}e^{-nx^2} dx = \frac{1}{2}f(0)
[/tex]

but suppose you modify the kernel like this

[tex]
\sqrt{\frac{n}{\pi}} e^{-n(x-x_0(n))^2}
[/tex]

where [itex]x_0:\mathbb{N}\to\mathbb{R}[/itex] is some sequence such that [itex]x_0(n)\to 0[/itex] as [itex]n\to\infty[/itex]. That still behaves as a delta function if the zero is contained in an open integration interval, but if the zero is on the boundary of the integration interval, the answer could be anything, depending on the chosen [itex]x_0[/itex].
 
  • #4
[tex]\delta (x)[/tex] isn't really a function.

It only has meaning inside [tex]\int_{-\infty}^{+\infty} f(x) \delta (x) dx[/tex].

When it is on its own as part of a calculation, that is because it will be put inside such an integral in the next step of the calculation.

This is because it is a symbol which is defined by the property [tex]\int_{-\infty}^{+\infty} f(x) \delta (x) dx = f(0)[/tex]

Inside [tex]\int_{0}^{+\infty} f(x) \delta (x) dx[/tex] , it's a symbol without any meaning.

(Though of course, you can write [tex]\int_{-\infty}^{+\infty} \theta (x) f(x) \delta (x) dx[/tex], which looks pretty similar.)
 
  • #5
The answer comes by joining the posts of jostpuur and tiny-tim. :smile:

The second integral is
[tex]I=\int_{0}^{+\infty} f(x) \delta (x) dx = \int_{-\infty}^{+\infty} \theta (x) f(x) \delta (x) dx[/tex]
where [itex]\theta(x)[/itex] is the step function. Thus [itex]I[/itex] equals

[tex]I=\theta(0)\,f(0)[/itex]

and the question now is what's the definition of the step function. The one which is most commonly used is that [itex]\theta(0)=\frac{1}{2}[/itex], thus

[tex]I=\frac{1}{2}\,f(0)[/tex]
 
  • #6
Rainbow Child said:
The answer comes by joining the posts of jostpuur and tiny-tim. :smile:

The second integral is
[tex]I=\int_{0}^{+\infty} f(x) \delta (x) dx = \int_{-\infty}^{+\infty} \theta (x) f(x) \delta (x) dx[/tex]
where [itex]\theta(x)[/itex] is the step function. Thus [itex]I[/itex] equals

[tex]I=\theta(0)\,f(0)[/itex]

and the question now is what's the definition of the step function. The one which is most commonly used is that [itex]\theta(0)=\frac{1}{2}[/itex], thus

[tex]I=\frac{1}{2}\,f(0)[/tex]

This doesn't fully make sense. If f is a function, then

[tex]
\int\limits_0^{\infty} f(x) dx = \int\limits_{-\infty}^{\infty} \theta(x)f(x) dx
[/tex]

is true with arbitrary value [itex]\theta(0)[/itex]. Why should the [itex]\theta(0)=1/2[/itex] be chosen when there is [itex]f(x)\delta(x)[/itex] in the integrand instead?

I think the original question is of similar nature as the question about what 0/0 is supposed to be. As I showed in my first response, if you try to approach this integral with some limits, you can get different results with different ways of taking the limit. Thus, we say that the limits cannot be used to give proper meaning for the this integral.
 
  • #7
jostpuur said:
This doesn't fully make sense. If f is a function, then

[tex]
\int\limits_0^{\infty} f(x) dx = \int\limits_{-\infty}^{\infty} \theta(x)f(x) dx
[/tex]

is true with arbitrary value [itex]\theta(0)[/itex]. Why should the [itex]\theta(0)=1/2[/itex] be chosen when there is [itex]f(x)\delta(x)[/itex] in the integrand instead?

Because when the integrand contains a Dirac delta, it is not a function, and so the condition "if f is a function" does not apply.

jostpuur said:
I think the original question is of similar nature as the question about what 0/0 is supposed to be. As I showed in my first response, if you try to approach this integral with some limits, you can get different results with different ways of taking the limit. Thus, we say that the limits cannot be used to give proper meaning for the this integral.

Well, it's more like asking what half of infinity is. But, yeah, to make Dirac deltas at all consistent (let alone rigorous), you need to enforce a definition of them as a limit of some sequence of well-defined functions. As you demonstrated, which exact sequence you choose will affect the answers you get if you ask questions like the one in this thread. That said, most people who are that concerned about consistency forgo the Dirac delta entirely and instead use distributions. The rest of us just live with the fact that certain operations (again, like the OP) aren't defined.
 
  • #8
I'll be more explicit!

Consider the sequence of following functions

[tex]
\delta_{n,1}(x) = \sqrt{\frac{n}{\pi}} \exp\big(-n(x-n^{-1/4})^2\big) = \sqrt{\frac{n}{\pi}} \exp\big(-(\sqrt{n}x -n^{1/4})^2\big)
[/tex]

It is easy to believe, that the distributions, represented by these functions, converge towards the delta distribution, because this is just the usual Gaussian peak representation, but displaced slightly. Let a and b be such that [itex]a\leq 0 < b[/itex]. Let us calculate the next integral.

[tex]
\lim_{n\to\infty}\int\limits_a^b f(x)\delta_{n,1}(x) dx = \lim_{n\to\infty} \sqrt{\frac{n}{\pi}} \int\limits_a^b f(x)\exp\big(-(\sqrt{n}x-n^{1/4})^2\big) dx = \lim_{n\to\infty} \frac{1}{\sqrt{\pi}} \int\limits_{\sqrt{n}a-n^{1/4}}^{\sqrt{n}b-n^{1/4}} f\big(\frac{y}{\sqrt{n}} + n^{-1/4}\big) e^{-y^2} dy
[/tex]
[tex]
= \frac{1}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty} f(0) e^{-y^2} dy = f(0)
[/tex]

However, notice that this is indeed true even with a=0.

If we instead chose the sequence

[tex]
\delta_{n,2}(x) = \sqrt{\frac{n}{\pi}} e^{-nx^2}
[/tex]

we would have

[tex]
\lim_{n\to\infty} \int\limits_a^b f(x)\delta_{n,2}(x) dx = f(0),\quad\quad a<0<b
[/tex]

and

[tex]
\lim_{n\to\infty} \int\limits_a^b f(x)\delta_{n,2}(x) dx = \frac{1}{2}f(0),\quad\quad a=0<b.
[/tex]

This means, that the distributions, represented by functions [itex]\delta_{n,1}[/itex] and [itex]\delta_{n,2}[/itex], converge towards the same delta distribution. This is so, because if you want to calculate

[tex]
\int\limits_a^b f(x)\delta(x) dx,\quad\quad a<0<b
[/tex]

you can choose either of these sequences, and they give the same result. But when you calculate

[tex]
\int\limits_0^b f(x)\delta(x) dx,
[/tex]

choosing different representations for the delta distribution, you get different numbers out of the integral. Thus, the delta distribution does not contain enough information to calculate this integral.
 
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  • #9
quadraphonics said:
Because when the integrand contains a Dirac delta, it is not a function, and so the condition "if f is a function" does not apply.

I was not applying any function assumption to any calculation with distributions. I merely noted, that with integrals of functions, the value of [itex]\theta(0)[/itex] does not matter, and then asked that why should some specific value for it be chosen when we integrate distributions.

But, yeah, to make Dirac deltas at all consistent (let alone rigorous), you need to enforce a definition of them as a limit of some sequence of well-defined functions.

Not precisely true. Distributions can be defined without sequences of functions, but the sequences of functions still remain as an important way to handle distributions.

As you demonstrated, which exact sequence you choose will affect the answers you get if you ask questions like the one in this thread.

Yes. This was my point. I explained it even more explicitly in my last post #8.
 
  • #10
The endearing quality of distributions

jostpuur said:
{snip} I merely noted, that with integrals of functions, the value of [itex]\theta(0)[/itex] does not matter, and then asked that why should some specific value for it be chosen when we integrate distributions. {snip}

Because that's how distributions are defined, and what they were created for! :smile:

It is distributions' most endearing (and, in my opinion, only endearing :frown:) quality.
 
  • #11
But, yeah, to make Dirac deltas at all consistent (let alone rigorous), you need to enforce a definition of them as a limit of some sequence of well-defined functions.
Not at all.

Let F be some function space over R, I some open interval in R, and define the functional D as follows:
[tex]D(f,I)=f(0), 0\in{I},f\in{F},D(f,I)=0,0\notin{I},f\in{F}[/tex]
This is the delta functional, and it can be shown to be linear, i.e, a distribution.
This is how you rigorously define the Dirac Delta "function".
 
  • #12
jostpuur said:
This doesn't fully make sense. If f is a function, then

[tex]
\int\limits_0^{\infty} f(x) dx = \int\limits_{-\infty}^{\infty} \theta(x)f(x) dx
[/tex]

is true with arbitrary value [itex]\theta(0)[/itex].
Why should the [itex]\theta(0)=1/2[/itex] be chosen when there is [itex]f(x)\delta(x)[/itex] in the integrand instead?

But that's the whole point! [itex]\delta(x)[/itex] is a distribution thus it does not behaves like ordinary functions. Your point of view is that the integral does not exist, because it can take multiple values, but I say that it can be defined if you choose the value of [itex]\theta(0)[/itex] yielding to

[tex]I=\int\limits_0^{\infty} f(x)\,\delta(x) dx = \theta(0)\,f(x)[/tex]
 
  • #13
Rainbow Child said:
Your point of view is that the integral does not exist, because it can take multiple values, but I say that it can be defined if you choose the value of [itex]\theta(0)[/itex] yielding to

[tex]I=\int\limits_0^{\infty} f(x)\,\delta(x) dx = \theta(0)\,f(x)[/tex]

When you choose the value for theta(0), in effect you are choosing the value of the entire integral. The number, that is supposed to come out of the integral, is not something that should be chosen. It should come out from some calculation. Right now the calculations don't give a unique number, and I don't feel like choosing some unique number by force would be a very good way to deal with this.
 

Related to Quick Question on the Dirac Delta Function

1. What is the Dirac Delta Function and what does it represent?

The Dirac Delta Function, also known as the impulse function, is a mathematical function used to represent a point mass or point charge in physics. It is characterized by being zero everywhere except at the origin, where it is infinite. It is often used to model discontinuous phenomena such as electrical impulses or point masses in mechanics.

2. How is the Dirac Delta Function defined mathematically?

The Dirac Delta Function is defined as follows:

&delta(x) = 0 for x ≠ 0
&delta(x) = ∞ for x = 0

And it satisfies the property:

∫ &delta(x-a)dx = 1 for all a

3. What is the physical interpretation of the Dirac Delta Function?

The Dirac Delta Function represents an idealized point mass or point charge that has infinite density or infinite charge. It is often used in physics to model the behavior of particles or forces that are concentrated at a single point.

4. How is the Dirac Delta Function used in engineering and mathematics?

The Dirac Delta Function is used in engineering and mathematics as a tool for simplifying calculations and modeling real-world phenomena. It is particularly useful in integral calculus, where it can be used to represent discontinuous functions and solve differential equations. It is also used in signal processing and control systems to model impulses and shocks.

5. Is the Dirac Delta Function a real function or just a theoretical concept?

The Dirac Delta Function is a theoretical concept that has many real-world applications. While it is not a function in the traditional sense, it can be defined and manipulated mathematically to solve problems in physics and engineering. It is often referred to as a distribution or generalized function, and is a valuable tool in mathematical modeling and analysis.

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