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I can not undestand this mathematical formula:

$$ \large \int_{a-\epsilon}^{a+\epsilon}f(x)\delta(x-a)dx=f(a) $$

I understand that it is the derivative of an integral evaluated in a, but still can not explain in mathematically.

How do you get $f(a)$ from the integral and what are the assumptions you have to do? Any help?

I found a great answer here (pags 2-3)

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closed as off-topic by AccidentalFourierTransform, Emilio Pisanty, Kyle Kanos, John Rennie, Yashas Apr 30 '17 at 9:40

  • This question does not appear to be about physics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I'm not sure how this is a physics rather than a Mathematics question. $\endgroup$ – ACuriousMind Apr 29 '17 at 18:49
  • $\begingroup$ I think the delta function has a history as a convenient tool in applied contexts that was only justified rigorously later. So I think the question would be OK here or at MSE. $\endgroup$ – daniel Apr 29 '17 at 19:09
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    $\begingroup$ What about it don't you understand? How can you reasonably expect anyone to answer a question if you give them no details about it. $\endgroup$ – Kyle Kanos Apr 30 '17 at 2:59
  • $\begingroup$ The answer to the question, "How do you get $f(a)$ out" is simply follow the definition, so it makes little sense to even ask. Have you bothered to even look up the extremely detailed Wikipedia entry on the delta function? $\endgroup$ – Kyle Kanos May 1 '17 at 0:07
  • $\begingroup$ Your "explanation" also doesn't seem to follow or even closely correct (what derivative do you see in the integral?), which suggests you probably need a refresher on calculus. $\endgroup$ – Kyle Kanos May 1 '17 at 0:11
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Unless we use the concept of 'distribution', that formula cannot be rigorous since $\delta$ is not a function, in fact. Giving up a rigorousness and assuming that $f$ is continuous, you can understand as follows: $$\int_{a-\epsilon}^{a+\epsilon}f(x)\delta(x-a)~\mathrm dx \approx f(a)\int_{a-\epsilon}^{a+\epsilon}\delta(x-a)~\mathrm dx = f(a) \cdot 1$$ by definition of integrating $\delta$ if $\epsilon$ is small enough.

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  • $\begingroup$ @santimirandarp Assuming $f$ is continuous, $|f(x) - f(a)|$ is small enough by definition if $|x-a|<\epsilon$ is small whence we can assume that is $f(a) + \alpha$ for another small $\alpha.$ $\endgroup$ – Kanu Kim Apr 29 '17 at 19:09
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Informally we get $f(a)$ from the integral as follows.

Consider the curve $f(x)$ weighted (multiplied by) the delta function in a small vicinity of the point $a.$ However we define the delta function, it is essentially an impulse (a narrow rectangle or triangle or gaussian) of unit area. As the rectangle gets very narrow it grows taller, preserving unit area. In a sense it ulimately picks out the valus of $f$ in a very small interval about $a,$ $[a+\epsilon,a-\epsilon],$ multiplying that value by one.

Since the delta function is zero outside this narrow range, there is no need to worry about the function elsewhere.


Note: In Dirac's 'Quantum Mechanics' p. 58 ff. the author goes to some trouble to explain the delta function and show its utility in QM. This I think supports that the question is properly in the ambit of physics as well as mathematics.

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I'm not totally sure what you're asking, but the standard rigorous way to think of the delta function is as the limit of a normalized Gaussian as the width goes to 0.

$$\delta(x-a) = \lim \limits_{w \to 0} \frac{1}{\sqrt{2 \pi} w} e^{-\frac{(x-a)^2}{2w}}$$

[Here's a set of plots](https://www.wolframalpha.com/input/?i=graph+1%2Fsqrt(2piw%5E2)+e%5E(-((x-2)%5E2)%2F(2*w))+for+w+in+%5B0.5,0.2,0.01,0.001%5D) that show the convergence.

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  • $\begingroup$ the question is how to get f(a) from the integral $\endgroup$ – santimirandarp Apr 29 '17 at 17:44
  • $\begingroup$ You plug a into the thing that multiplies the delta function... $\endgroup$ – Bobak Hashemi Apr 29 '17 at 17:46
  • $\begingroup$ And why in f and not in $\delta$? $\endgroup$ – santimirandarp Apr 29 '17 at 17:47
  • $\begingroup$ This is literally the definition of the $\delta$ function... $\endgroup$ – Noiralef Apr 29 '17 at 17:59
  • $\begingroup$ It's not the only definition. Physicists often use the equation in the OP to define the delta function as well, but then it's not a 'well defined' function. $\endgroup$ – Bobak Hashemi Apr 29 '17 at 19:09
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Most introductory texts in either Quantum Mechanics or Electromagnetism define the Dirac delta function and explain a few of its properties. I recommend taking a look in either of those places for some good discussion.

Bobak is correct that most rigorous expositions define it as the limit of a Gaussian. Informally, one thinks of the $\delta$-function as a function given by $$\begin{align} \delta(x) = 0 &\text{ if } x \neq 0 \\ \delta(x) = \infty &\text{ if } x = 0 \end{align} $$ It has the additional property that $$\int_{-\infty}^{\infty} \delta(x)dx = 1$$ or in fact $$\int_{a}^{b} \delta(x)dx = 1$$ as long as $0 \in (a,b)$. We can "shift" where the peak is by translating right or left. If we take $\delta(x-a)$, then when we plug in $a$ we get $\delta(0) = \infty$. Now if we integrate over some interval containing $a$, we get 1. You can use this to essentially "pick out" values of an arbitrary function $f(x)$. We do this by forming the integral $$ \int_{a-\varepsilon}^{a+\varepsilon}f(x)\delta(x-a)dx = f(a) $$

Again, this is all informal. But you can think of the $\delta$ function as "killing" the function everywhere but at $a$. This is kind of analogous to a situation where you have $g(x)=1$ on some interval $[a, b]$ and zero elsewhere. What can you then say about $$ \int_{-\infty}^{\infty} f(x)g(x)dx $$ ?

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  • $\begingroup$ I think that maybe you are right, but i do not like this form of reasoning. Anyway, thanks for helping.. $\endgroup$ – santimirandarp Apr 29 '17 at 19:20
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    $\begingroup$ Yes, it is very informal and therefore limited. I wasn't sure what level of machinery you were comfortable with. Cheers. $\endgroup$ – gabe Apr 29 '17 at 19:24
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Actually, that property you stated for the δ-function is more precisely defined as $$ \int\ f(x)δ(x-a)=f(a) \ $$ when the limits extend from -oo to +oo. Thus, your expression must be corrected as in the following $$ (1/2ε)*(your\ integral) $$ and since δ-function gives 1 (one) only for x=α and zero elsewere, you will get

$$ (1/2ε)*f(α)*\int_$dx\ $$ and the last integral with limits from (α-ε) to (α+ε) yields $$ \int = (α+ε)-(α-ε)=2ε $$ so you end up with $$ (1/2ε)*f(α)*2ε $$ Just mention that f(α) represents a certain value that is it is a number so it can go out of the Integral.

Hope these help.

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