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? ($k > 0$)
$$\int_{-k}^k \delta(x) dx = 1$$
 A: 
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 $k=0$:
$$
\int_{-\infty}^{\infty} d\omega  \frac{\sin(k\omega)}{\omega\pi}
=\lim_{N\to\infty} \int_{-N}^{N} d\omega  \left(0\right) = 0
$$
