# How does the Dirac delta, written as $\delta(x)=\frac1{2\pi}\int e^{i\omega x}d\omega$, satisfy $\int_{-k}^k \delta(x)dx=1$?

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? ($$k > 0$$)

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

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

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

– hft
Sep 27, 2022 at 18:29
• 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 Sep 27, 2022 at 18:40
• @hft Why is $\int_{-k}^k dx \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega x} d\omega =\int_{-\infty}^{\infty} d\omega \frac{\sin(k\omega)}{\omega\pi}$? Sep 28, 2022 at 5:25
• @jellyears Swap the order of integration and use the Euler formula to find $\sin$ in terms of a linear combination of (complex) exponentials. Sep 28, 2022 at 7:02
• But swapping the integrals here is not rigorous. I gave a rigorous explanation here math.stackexchange.com/questions/4527480/… Sep 30, 2022 at 7:09