# What Are Common Misconceptions About the Dirac Delta Function?

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• gionole
gionole
I have read lots but still, there're some really unproductive explanations of dirac delta function. So hopefully, you can explain it by following my arguments and not formal definition because I've read it all.

It's shown to be as ##\delta (x) = 0## when ##x \neq 0## and ##\delta (x) = \infty## when ##x=0##. I get that it's not a traditional sense function, because if I try to take ##\int \delta(x) dx##, we got ##0## times ##\infty## which is undefined.

Then some people say that this is not a dirac definition and it's ##\int f(x) \delta(x) dx = f(0)##. Then some say that it's a "distribution". But somehow, in the end, I understood nothing even though I read a lot. I think all these different representations make me confused more.

Could we engage in this forum back-and-forth to help me understand this somehow ? As an example, where do I start from ? I tried starting it from point particle, but I really couldn't understand why ##\int f(x) \delta(x) dx = f(0)## is correct or how the integral is 1.

For me the best heuristic (non-rigorous) conception is to consider it a rectangular function with and area = 1, so the height is defined by 1/width. Then the δ function is created by making the width arbitrarily small, approaching zero. So the integral is part of the function definition and is always equal to one.

There's much more a mathematician can say about this, but this is a good practical model to keep in mind. The real function that is squeezed is a continuous one (shown here), but its the same idea.

SammyS and Dale
DaveE said:
consider it a rectangular function with and area = 1
I also like this approach. You can take any convenient family of functions with area equals 1 and variable width. As you take the limit as the width goes to zero you get the right answer rather independently of the exact shape of the peak

I would suggest you consider your questions:
• δ(x)=0 when x≠0​
• ∫δ(x)dx=1​
• ∫f(x)δ(x-a)dx=f(a)​
more like definitions.

I would just avoid the question about δ(0)=∞. Infinities are unnecessarily confusing. Think of it as an arbitrarily large value. From an engineer's perspective, δ(0) by itself isn't useful; you'll always have it inside an integral in practice.

Office_Shredder
Thanks for the answer.

To be honest, I also was considering this but here is 2 things that prevent me from understanding your conception:

Question 1: Even if you don't treat ##\delta (x)## to be infinity when ##x = 0##, but some other value, there's still a problem. Here is why: imagine the rectangle of ##-\epsilon##, ##+\epsilon## width and height as ##\frac{1}{2\epsilon}##. If you narrow the width, ##|\epsilon|## is more and more approaching 0, but the value of function is already some value even before ##\epsilon## becomes the exact 0. So what I emphasize by it is that with this example, ##\delta (x)## ends up having a non-zero value even at ##x \neq 0## and it contradicts the definition that it's 0 everywhere except at ##x=0## only.

Question 2: My question 1 seems an obstacle to me, but I prefer to ask this as well anyways. You choose such a rectangle that its area is 1. Now, when you meet integral in real use case problems such as ##\int f(x)\delta (x) dx## where, say, ##f(x) = 2x+5##, then you end up getting ##f(0) = 5##, because ##\delta (x) dx## is 1 per definition of your rectangle perception. How do you know that answer you now got is correct in your problem ? If you could define ##\int \delta (x) dx = 2##, then the answer you would get is ##2f(0) = 10##. So your definition changes the actual answers of the actual problem which makes me ask: how are we sure that defining the integral to be 1 is 100% correct and it won't make our problem solutions wrong ?. Hope that makes sense what I'm asking.

gionole said:
δ(x) ends up having a value even at x≠0.
Are you familiar with ε-δ proofs?

At δ(a), any a≠0, if you propose any value δ(a)≡b≠0, I can prove that |δ(a)|<|b|. This is what is meant by "arbitrarily close to zero". This is the sort of process you often need when working with Limits.

You are correct, for the exponential function usually used in the Dirac-Delta definition, it always has a value ≠ 0. But it's value is "arbitrarily close to zero".

gionole said:
If you could define ∫δ(x)dx=2, then the answer you would get is 2f(0)=10. So your definition changes the actual answers of the actual problem.
Of course. That is how the world works. Changing definitions often changes results.

gionole said:
your definition changes the actual answers of the actual problem
That is always true, with almost any definition.

You certainly could define a ##\delta_2## such that ##\int_{-\infty}^{\infty}f(x)\delta_2(x)\ dx=2\ f(0)##. It could still work just fine, you would just have lots of factors of ##1/2## in your formulas.

This is like the difference between decay rates and half lives. If you are using half lives you will see lots of factors of 0.693 in equations without those factors for decay rates.

DaveE
Point 1: Consider that ##b## is approaching to 0 pretty close.

Let's say ##x=1##, then ##\delta## ends up ##\lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} 0 = 0##. If ##x## is super small, since ##b## approaches to 0, it will be even less, so we would get the same value for ##\delta## function to be 0. It's as if whatever ##x## you come up with, ##b## can be found even smaller and making the end result 0. It's only for ##x=0##, when ##\delta(0) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} 1 = \infty##.

So I think by this ##\lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}## exponential function definition, I actually match the definition such as: "whose value is zero everywhere except at zero", but this holds true when ##b## is approaching to 0 very close and I think this is correct the definition(using limits and making ##b## approach to 0 pretty close - i.e inserting ##b## values such as 1, or 0.5 or 0.003 doesn't work as they will yield non-zero values for ##\delta## for ##x \neq 0## and this contradicts the definition.

Point 2: If you got an rectangle such as its height is 1/width, then sure, ##\int \delta(x) dx## yields 1, but note that such rectangle doesn't represent the same thing as ##\delta_b(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}##, because on the image on wikipedia, it's clear that this doesn't look like rectangle and its area is not 1.

DaveE
gionole said:
inserting b values such as 1, or 0.5 or 0.003 doesn't work as they will yield non-zero values for δ for x≠0 and this contradicts the definition.
Yes, but which definition? If you mean the ones I posted above, then that's because I was sloppy and those are technically wrong. We (incorrectly) use the shorthand δ(a)=0, a≠0, when we should say δ(a)→0. You'll find this lack of rigor commonplace in calculus, since we all learned it in school and just want to get on with communicating something else related to the definitions.

This is why my original reply said that mathematicians would have much more to say; for example, that wikipedia page.

gionole
Got it. Thanks so much. Could you also answer my Point 2 ? copying it here:

Point 2: If you got an rectangle such as its height is 1/width, then sure, ##\int \delta(x) dx## yields 1, but note that such rectangle doesn't represent the same thing as ##\delta_b(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}##, because on the image on wikipedia, it's clear that this doesn't look like rectangle and its area is not 1.

gionole said:
Got it. Thanks so much. Could you also answer my Point 2 ? copying it here:

Point 2: If you got an rectangle such as its height is 1/width, then sure, ##\int \delta(x) dx## yields 1, but note that such rectangle doesn't represent the same thing as ##\delta_b(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}##, because on the image on wikipedia, it's clear that this doesn't look like rectangle and its area is not 1.
a) you're right, exponentials and rectangle functions are different.

b) rectangles aren't actually used in good definitions, mostly because they aren't continuous. But is is common in introducing the concept that the area is constant even though the width approaches 0.

My main point was that you can only say ##\int \delta(x)f(x)dx = f(0)## when ##\delta## is such as its area is 1 and I said that true definition of ##\delta## which is ##\delta_b(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}##, doesn't represent area of 1, i.e ##\int \delta(x)f(x)dx## can't be ##f(0)##, but it still is. hahaha

gionole said:
it's clear that this doesn't look like rectangle and its area is not 1
That is why I said:
Dale said:
You can take any convenient family of functions with area equals 1 and variable width. As you take the limit as the width goes to zero you get the right answer rather independently of the exact shape of the peak
Also, the area is 1

@Dale

1. Can you give one example of such function with its limit as well ?

2. since the real, true definition of ##\delta## is the following(##\delta_b(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}##), are you saying that if I do: ##\int \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2} dx ##, I will get 1 ?

gionole said:
1. Can you give one example of such function with its limit as well ?
Sure. Consider an isosceles triangle with base from ##-a## to ##a## and height ##1/a##. That will work also.

gionole said:
2. since the real, true definition of ##\delta## is the following
I disagree with calling that the real true definition. The real true definition is that it is equal to zero everywhere except at zero, and that its integral is equal to one.

The following is not a real true definition but merely one of many functions that meet the definition.

gionole said:
##\delta_b(x) = \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2}##), are you saying that if I do: ##\int \lim_{b \to 0} \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2} dx ##, I will get 1 ?
If you calculate ##\int \frac{1}{|b| \sqrt{\pi}} e^{-(\frac{x}{b})^2} dx ## you will get ##1## for any ##b##

gionole
Ah, I think I start to understand the usecase of this whole thing.

So basically, if I am working on some problem, whether it's in electromagnetism or anywhere else, and somehow, I realize that the followings are true:
1. some property or some function exactly behaves the same way as ##\delta (x)## (i.e it's zero everywhere except at one point where it goes to infinity)
2. and I somehow , who knows why, end up with integral such as ##\int f(x) \delta (x) dx## where ##\delta (x)## is how my problem behaves:

then I can just immediatelly say the answer is ##f(0)##..

It seems to me that there was no need to come up with dirac function at all. If I had a problem which acts as its effect is 0 everywhere except at one point, and I had to do integral, I could just solve it by myself - it seems like dirac just said: "hey, if your problem is like that, just remember to not think about too much anymore and quickly remind of the rule that it's ##f(0)##".

Yes. That is reasonable.

The only thing that is tricky is trying to construct an actual function that behaves that way. The issue is
gionole said:
where it goes to infinity
because infinity is not a real number. So they have to do lots of “limit gymnastics” to make it well defined.

Of course, you can take a hyperreal function and directly define a Dirac delta as a legitimate function. Because the hyperreals include infinite and infinitesimal numbers this doesn’t cause any difficulties in the hyperreals like it does in the reals.

Thanks so, so much <3

I am burnt out now, so will sleep and tomorrow, I will sum things up and hopefully, all will be okay. Have a good day/night both of you

Dale
gionole said:
It seems to me that there was no need to come up with dirac function at all
It turns out to be extremely useful. You'll see it later in your studies. It's a mainstay in modeling linear systems. It's often use to model an impulse of something (energy usually). A finite amount of stuff delivered at a specific instant. You'll also see it in Fourier transforms and analyzing the frequency content of waveforms.

Dale
Take two seconds to truly digest the dirac delta. It says if you have a value at a point, it gives you one. Any other point, you have no value AKA 0.

Now, apply this to integrals, what changes? Nothing. If you still don't have that value, you'll get zero. And if you have that value, guess what? The delta goes away, and you're FORCED to use that point, because that's where you have a value! In other words, if you have ##f(3)## and ##\delta(3) = 0## then does it matter what ##f(3)## is? No, because ##f(3)*\delta(3) = 0##, so that's why you're forced to get ##f(0)##, because for any other point, you'll get 0. Seems kind of silly, sure, but let's see if we can see something "physical" that has these structures.

Since it seems like you're doing E+M, let's go ahead and imagine point charges. Don't they give a similar vibe? At this place in space, I have a value, the charge. Any other place in space I don't, so my charge returns no value. With the above knowledge, that "screams" dirac delta, and of course, smarter people than I put the pieces together, and soon you will too (look at discrete charge distributions if you need a starting point for textbooks). All of this is in a more basic language, but this is truly how I view this function.

My subjective intuitions on this topic aside, I found this blog post interesting (https://galileo-unbound.blog/2022/12/17/paul-diracs-delta-function/) which led me to these lecture notes which give some historical perspectives in the first few pages (and I haven't finished reading it all, but caught my attention enough to read the first few pages): https://www.reed.edu/physics/facult...th/Delta Functions/Simplified Dirac Delta.pdf (and notice how he uses point masses for motivation, how powerful mathematics can truly be!)

Hi guys. I've somehow ended up in a rabbit hole because I wanted to learn more about this and now, I'm at a big confusing state. I'd appreciate so much if you could answer these below questions one by one.

I learned that dirac delta is a distribution(i.e mapping) from function to real numbers. This was all clear until I looked into the "point mass" example and density.

Imagine, there's a point mass of 1kg at the origin. For sure, if you wanna figure out the density(forget dirac for now), you get infinity at ##x=0## and 0 everywhere else. Though, if you integrate this function, you don't get mass of it which is 1. So we got a problem.

So we call dirac distribution to be mapping whose input is ##\varphi(x)## and value is ##\lim_{\varepsilon \to 0}\int f_\epsilon (x)\varphi(x) dx## which ends up to be ##\varphi(0)##.

Attaching the image from the book.

Question 1: The book proceeds to mention the following: "It is this functional that we take for our sought-for density ##\delta(x)##. And this is the famous delta-function of Dirac." - If ##\delta## is a distribution, does the author(when he mentions ##\delta(x)## mean that argument (which is ##x##) is assumed to be a function ?

Question 2: Also, how does ##\lim_{\varepsilon \to 0}\int f_\epsilon (x)\varphi(x) dx## really represent the density ? Where does this show that at ##x=0##, the density is infinite ? I'm saying all this because author says: ""It is this functional that we take for our sought-for density ##\delta(x)##""

Question 3: The author of the book restores the mass by inputting ##\varphi(x) = 1## to the distribution which for sure, gives back 1 again, but why does he input 1 ? It should have a logical justification why such ##\varphi(x) = 1## is inserted - I mean, justification for our mass density situation/scenario.

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Are you interested in hyperreals, or do you want to stick only with real numbers? If you want to stick purely with real numbers then this statement is incoherent:
gionole said:
you get infinity at x=0 and 0 everywhere else

True, but I'm trying to make sense of the author's analysis. I'm also attaching the "before explanation".

Could you check my 3 questions after you take a look at what author is saying ?

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gionole said:
True, but I'm trying to make sense of the author's analysis. I'm also attaching the "before explanation".

Could you check my 3 questions after you take a look at what author is saying ?
Sorry, but that didn’t answer the question I asked you above. I cannot proceed without you answering that question because it dramatically changes the answer.

Well, density when you got a point mass is definitely infinite. So let's proceed with hyperreals(whatever includes infinity).

Dale
gionole said:
Well, density when you got a point mass is definitely infinite. So let's proceed with hyperreals(whatever includes infinity).
So in the hyperreals the discussion is not nearly so complicated. First, we define the hyperreal function: $$\delta(x) = \begin{cases} 1/\epsilon & |x|\leq \epsilon/2 \\ 0 & \text{otherwise} \end{cases}$$ where ##\epsilon## is any positive infinitesimal hyperreal number. There is no need to call this a distribution rather than a function, it is a legitimate function taking any hyperreal argument ##x## and returning a hyperreal number, either ##0## or the infinite hyperreal number ##1/\epsilon##. There is also no need to take limits as ##\epsilon## approaches 0. Any infinitesimal ##\epsilon## is already small enough. This function has the desired properties, for any non-zero real ##x## its value is 0, its value is infinite at ##x=0##, and its integral (over any real region containing 0) is 1.

With that then instead of defining the mass as a point mass, we define it as an infinitesimal sphere of uniform density with a total mass ##m##. From the perspective of the real numbers that is a point mass, but when we zoom in infinitesimally it is a small sphere. We can then define the density straightforwardly as $$\rho(r)=m\frac{6}{\pi \epsilon^2} \delta(r)$$ Note that the density is infinite inside ##r\leq \epsilon/2## and zero outside. When we integrate the density we get $$\int_0^\infty \rho(r) \ 4 \pi r^2 \ dr = m$$

gionole said:
Question 1: The book proceeds to mention the following: "It is this functional that we take for our sought-for density δ(x). And this is the famous delta-function of Dirac." - If δ is a distribution, does the author(when he mentions δ(x) mean that argument (which is x) is assumed to be a function ?
There is no need to make it a functional. It is a standard hyperreal function.

gionole said:
Question 2: Also, how does limε→0∫fϵ(x)φ(x)dx really represent the density ? Where does this show that at x=0, the density is infinite ? I'm saying all this because author says: ""It is this functional that we take for our sought-for density δ(x)""
There is no need to take the limits at all. We can simply write the density explicitly in terms of our hyperreal ##\delta(x)## function.

gionole said:
Question 3: The author of the book restores the mass by inputting φ(x)=1 to the distribution which for sure, gives back 1 again, but why does he input 1 ? It should have a logical justification why such φ(x)=1 is inserted - I mean, justification for our mass density situation/scenario.
I chose the number ##m \ 6/\pi\epsilon^2## in the definition of ##\rho## specifically so that the integral would come out right given my definition of ##\delta##. If I had changed my definition of ##\delta## then I would have needed a different constant in ##\rho##

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