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I know Dirac delta can be defined rigorously with distributions, but is the following definition absurd?

Definition

I think so since the integral should always be 0 in the sense of Lebesgue (it takes only one positive value, but on a singleton).

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    $\begingroup$ The Dirac delta is not a function in the usual sense. It is what's called a distribution. $\endgroup$ Jul 3 at 5:13
  • $\begingroup$ This is not a formal definition but rather some intuitive explanation. $\endgroup$
    – PC1
    Jul 3 at 5:26
  • $\begingroup$ See math.stackexchange.com/a/1413826/960197 $\endgroup$
    – PC1
    Jul 3 at 5:27
  • $\begingroup$ The definition is absurd, yes, although it might provide some intuition or some motivation for the rigorous definition. $\endgroup$
    – littleO
    Jul 3 at 5:27
  • $\begingroup$ Please consider writing up the excerpt in your question, not pasting a screenshot of text, as using a screenshot that makes the key part of your question unsearchable and impossible for text-to-voice devices (used by the visually impaired) to understand. $\endgroup$
    – KCd
    Jul 3 at 5:49

2 Answers 2

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In the way it is written, it is absurd. But it can be understood in the following way. If you define the Dirac measure $\delta$ as $$ \delta(E)=\begin{cases}1,&\ x_0\in E\\[0.3cm] 0,&\ x_0\not\in E\end{cases} $$ then it is easy to check that the Lebesgue integral for any $f$ is $$\tag1 \int_{\mathbb R}f(x)\,\delta(x)=f(x_0). $$ The notation that physicists use comes from the Radon-Nikodym Theorem: if $\mu$ is absolutely continuous with respect to Lebesgue measure, then there exists a function $g$ such that $$\tag2 \int_{\mathbb R}f(x)\,d\mu(x)=\int_{\mathbb R}f(x)\,g(x)\,dx. $$ Of course the Dirac measure is not absolutely continuous with respect to Lebesgue measure. But physicists only use the Dirac delta in the equality $(1)$, never as a standalone function, so they don't run into contradictions.

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    $\begingroup$ I would say (from my perspective as a physicist) that physicists often run into contradictions with the Dirac delta, usually when they're not being sufficiently careful with their integrals. :-) $\endgroup$ Jul 3 at 13:42
  • $\begingroup$ +1. Comments: a usual notation is $\delta_{x_0}$ for the shifted Dirac delta measure you define ;) As I explained in my answer here math.stackexchange.com/questions/3801916/…, a better notation for integrals with another measure is $\int f(x)\,\mu(\mathrm d x)$. With these notations, one could write indeed $$ \int f(x) \,\delta_{x_0}(\mathrm d x) = f(x_0) $$ $\endgroup$
    – LL 3.14
    Jul 6 at 10:26
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Within the theory of integration the definition doesn't make sense. You can even call it absurd.

But it common, especially among physicists, to abuse or extend the integral notation (call it whatever you want) to cover the application of a distribution on a test function (which is a kind of generalization of an integral), and as such it makes sense.

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