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

Question

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? ($k > 0$)

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

Answer 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? ($k > 0$) $$\int_{-k}^k \delta(x) dx = 1$$

$$ \int_{-k}^k \delta(x) dx = \int_{-k}^k dx \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega x} d\omega = \int_{-\infty}^{\infty} d\omega \left[\frac{1}{2\pi}\int_{-k}^k dxe^{i\omega x}\right] = \int_{-\infty}^{\infty} d\omega \frac{\sin(k\omega)}{\omega\pi}\;, $$ 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 $\epsilon \to 0$.

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