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We see the Dirac delta representation as follows,

$$\delta(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega x} d\omega$$

I want to know, how does this satisfy the following? ($k > 0$)

$$\int_{-k}^k \delta(x) dx = 1$$

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    $\begingroup$ This is a Mathematics question. $\endgroup$ Sep 27, 2022 at 17:39
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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Sep 28, 2022 at 5:20
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    $\begingroup$ Dirac delta is also used in Physics. So it is a physics question. $\endgroup$
    – Abhinav
    Sep 29, 2022 at 16:33

1 Answer 1

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We see the Dirac delta representation as follows, $$\delta(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega x} d\omega$$ I want to know, how does this satisfy the following? ($k > 0$) $$\int_{-k}^k \delta(x) dx = 1$$

$$ \int_{-k}^k \delta(x) dx = \int_{-k}^k dx \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega x} d\omega = \int_{-\infty}^{\infty} d\omega \left[\frac{1}{2\pi}\int_{-k}^k dxe^{i\omega x}\right] = \int_{-\infty}^{\infty} d\omega \frac{\sin(k\omega)}{\omega\pi}\;, $$ where the latter integrand is well defined at $\omega = 0$ because $\lim_{\omega\to0} \sin(k\omega)/\omega \to k$. Because of this we can also write this integral as: $$ \int_{-\infty}^{\infty} \frac{d\omega}{\pi} \frac{\sin(k\omega)}{\omega+i\epsilon}\;, $$ where $\epsilon$ is infinitesimal. This is also equal to: $$ \int_{-\infty}^{\infty} \frac{d\omega}{2\pi i} \frac{e^{ik\omega}-e^{-ik\omega}}{\omega+i\epsilon}\;. $$ The first term in the integrand can be closed in the upper-half plane and gives zero. The second term can be closed in the lower-half plane and gives: $$ \frac{-2\pi i}{2\pi i}\left(-e^{ik(-i\epsilon)}\right) = 1\;, $$ since $\epsilon \to 0$.


On the other hand, when $k=0$: $$ \int_{-\infty}^{\infty} d\omega \frac{\sin(k\omega)}{\omega\pi} =\lim_{N\to\infty} \int_{-N}^{N} d\omega \left(0\right) = 0 $$

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    $\begingroup$ Downvoting without explanation why will not help you get a "better" answer... $\endgroup$
    – hft
    Sep 27, 2022 at 18:29
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    $\begingroup$ Downvote is probably someone who thinks the question should not have gotten an answer. The other answer currently also has a down vote. I wouldn't sweat it. Those who leave downvotes without a comment aren't going to read / reply to your comment anyway $\endgroup$ Sep 27, 2022 at 18:40
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    $\begingroup$ @hft Why is $\int_{-k}^k dx \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega x} d\omega =\int_{-\infty}^{\infty} d\omega \frac{\sin(k\omega)}{\omega\pi} $? $\endgroup$
    – jelly ears
    Sep 28, 2022 at 5:25
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    $\begingroup$ @jellyears Swap the order of integration and use the Euler formula to find $\sin$ in terms of a linear combination of (complex) exponentials. $\endgroup$
    – Jakob
    Sep 28, 2022 at 7:02
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    $\begingroup$ But swapping the integrals here is not rigorous. I gave a rigorous explanation here math.stackexchange.com/questions/4527480/… $\endgroup$
    – LL 3.14
    Sep 30, 2022 at 7:09

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