This is a follow up from a previous question asked by me.

I know that $$\delta(x) = \begin{cases} 0 & \space \mathrm{for} \space x \ne 0 \\\infty&\ \mathrm{for} \space x = 0 \end{cases} $$ and that $$\int_{-\infty}^{\infty} \delta(x) \mathrm{d}x = 1$$

I also know that the product $$f(x) \delta(x)= 0\space\forall \space x\ne 0$$

I can summarize my lack of understanding with basically two questions:

$\color{red}{\mathrm{Question} \space1:}$ If I'm allowed to write $f(x) \delta(x) = f(0)\delta(x)$ then why not $f(x) \delta(x) = f(3)\delta(x)$ since $3 \ne 0$?

$\color{blue}{\mathrm{Question} \space2:}$ Also, why do we just substitute $x=0$ into the function $f(x)$ and not $\delta(x)$? In other words why don't we write $f(0)\delta(0)$ or $f(3)\delta(3)$ instead of $f(0)\delta(x)$ and $f(3)\delta(x)$ respectively. I know that $f(0)\delta(0)$ is undefined, but the point is that the $\delta$ still takes $x$ as its argument as well as $f$.

(As ever, apologies for the abuse of notation; this Dirac measure is all very new to me, hence all the questions about it)

  • 3
    $\begingroup$ To get $f(3)$ you may use $f(x) \delta(x-3) = f(3)\delta(x-3)$ (because the only value of interest is where the parameter of $\delta$ becomes $0$). $\endgroup$ – Raymond Manzoni Oct 6 '15 at 22:36


In This Answer and This Answer, I provided more detailed primers on the Dirac Delta. Herein, we condense the content of those answers.

The Dirac is not a function, but rather a Generalized Function also known as a Distribution.

The symbol $\int_{-\infty}^{\infty}\delta (x)f(x)\,dx$ is ,in fact, not an integral. It is a Functional that maps a test function $f$ into the number given by $f(0)$. (Note that whereas a function is a mapping, or "rule" that assigns to a number in a domain, a number in the range, a functional is a "rule" that assigns to functions in a vector space domain, a number.

We write

$$\int_{-\infty}^{\infty}\delta (x-a)f(x)\,dx=f(a)$$

but alternatively, and more compactly, we can write

$$\langle \delta_a,f \rangle=f(a)$$

For $a=0$, we have

$$\int_{-\infty}^{\infty}\delta(x)f(x)\,dx=\langle \delta_0,f \rangle=f(0)$$

Now, for this specific question, we have

$$\begin{align} \int_{\infty}^{\infty}f(x)\delta(x)\,dx&=f(0)\\\\ &=\int_{-\infty}^{\infty}f(0)\delta(x)\,dx\\\\ &\ne f(3)\\\\ &=\int_{-\infty}^{\infty}f(3)\delta(x)\,dx\end{align}$$

Therefore, $f(x)\delta(x)=f(0)\delta(x)\ne f(3)\delta(x)$ as was to be shown.

  • 2
    $\begingroup$ I would have written more here, but am at the gym and typing from a "not so smart phone." When I get back home, I'll include some references to answers I've posted to questions on the Dirac Delta herein. $\endgroup$ – Mark Viola Oct 6 '15 at 22:09
  • 3
    $\begingroup$ Well, no it doesn't have any meaning. You see, the Dirac Delta is not a function. It is a Generalized Function, also known as a Distribution. And the integral symbol used is NOT an integral. Rather, it is a Functional and the functional with the Dirac Delta assigns to any test function $f$ a number equal to $f(0)$. There is no meaning to $\delta(3)$ as a number nor to $\delta(x)$ as a function. $\endgroup$ – Mark Viola Oct 6 '15 at 22:22
  • 2
    $\begingroup$ But why can I write $\displaystyle\int f(3)\delta(x)\,\mathrm dx = f(3)$? this is not equal to $f(0)$. What is a "Functional"? $\endgroup$ – BLAZE Oct 6 '15 at 22:29
  • 1
    $\begingroup$ @BLAZE $f(3)$ is a constant. So, the Dirac Delta assigns to that constant, itself. I've added a Primer with references to other answers I've posted. $\endgroup$ – Mark Viola Oct 6 '15 at 23:19
  • 1
    $\begingroup$ As always, please let me know how I can improve my answer further as I really want to help and give you the best answer I can. $\endgroup$ – Mark Viola Oct 6 '15 at 23:19

Note that the "definition" of $\delta(x)$ that you cite is more of an informal description, to aid with intuition. The actual definition is by the relation $$\int f(x)\delta(x)\,\mathrm dx = f(0)$$

Question 1:

We have, by definition, $$\int f(x)\delta(x)\,\mathrm dx = f(0)$$ We also obviously have $$\int f(0)\delta(x)\,\mathrm dx = f(0)$$ but $$\int f(3)\delta(x)\,\mathrm dx = f(3) \ne f(0)$$

Question 2:

Basically you already answered your question yourself: That replacement would be undefined, and thus especially not give the same value.

  • $\begingroup$ Thanks for your answer, but for question 2 what about writing $f(3)\delta(3)$ is that 'allowed'? $\endgroup$ – BLAZE Oct 6 '15 at 22:05
  • 1
    $\begingroup$ Well $\int f(3)\delta(3)\,\mathrm dx=0\ne f(0)$ in general. It's always the same: Plug it in the integral and look if it is (a) defined and (b) gives the same result. If the answer to both is yes, then the replacement is allowed, otherwise it isn't. $\endgroup$ – celtschk Oct 6 '15 at 22:35

It can be true that $f(x)\delta(x) = f(0)\delta(x)$, depending on your initial definition of $\delta(x)$. But it is not the case with your definition.

Really, $\delta(x)$ is not a function. This seems to be causing you a bit of trouble. It is something termed a "distribution", which morally means it is defined only via how it acts with other functions when integrated against. The defining characteristic of $\delta(x)$ is that $$ \int_\mathbb{R} f(x) \delta(x) dx = f(0).$$

I can imagine difficulties arising from courses or books trying to use the Dirac delta distribution without really trying to understand it (since it's useful in differential equations, for instance, even if just treated as a black box).

It is true however that $$ \int_\mathbb{R} f(x) \delta(x) dx = \int_\mathbb{R} f(0) \delta(x)dx,$$ and neither of these are equal to $$ \int_\mathbb{R} f(3) \delta(x) dx.$$

  • $\begingroup$ Thanks for your answer, my first question has been answered, but for question 2 what about writing $f(3)\delta(3)$ is that 'allowed'? $\endgroup$ – BLAZE Oct 6 '15 at 22:14
  • 1
    $\begingroup$ It depends on how you define $\delta(x)$. From a slightly sophisticated point of view, $\delta(x)$ is not a function, and so you shouldn't try to substitute values into it at all. $\endgroup$ – davidlowryduda Oct 6 '15 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.