Dirac Delta and Exponential integral I am able to derive the following equation by substituting the definition of a Fourier transform into it's inverse. 
$$2\pi\delta(x-x') = \int_{-\infty}^{\infty} e^{ik(x-x')} dk$$
How do you prove that the Dirac Delta is equal to an integral of the exponential function? How do you prove the above equation is true?
 A: Let $$h_a(x)= \int_{-a}^a e^{i k x} dk =  \frac{2 \sin(a x)}{x}= a \, H'(ax), \\ H(x) = \int_{-\infty}^x \frac{2 \sin(y)}{y}dy, \qquad H(-\infty) = 0, H(+\infty) = C$$
where for some reason $C = 2\pi$
If $\phi,\phi'$ are $L^1$ then
$$\lim_{a \to \infty}\int_{-\infty}^\infty h_a(x) \phi(x) dx =  -\lim_{a \to \infty}\int_{-\infty}^\infty H(ax) \phi'(x) dx\\ = -\int_{-\infty}^\infty H(+\infty x) \phi'(x) dx =  -\int_0^\infty C  \phi'(x) dx= 2\pi \phi(0)$$
ie. in the sense of distributions $$\int_{-\infty}^\infty e^{ik x}dk \overset{def}=\lim_{a \to\infty} h_a = 2\pi \delta$$ 
Note how this proves the Fourier inversion theorem.
A: We can give a meaning to $\int_{-\infty}^{\infty} e^{ikx} \, dk$ by introducing a damping factor $e^{-\frac12\epsilon k^2}$ inside the integral and at the end let $\epsilon \to 0$:
\begin{align*}
\lim_{\epsilon\to 0}\int_{-\infty}^{\infty} e^{-\frac12\epsilon k^2} e^{ikx} \, dk
&= \lim_{\epsilon\to 0}\int_{-\infty}^{\infty} e^{-\frac12\epsilon (k-ix/\epsilon)^2} e^{-\frac12 x^2/\epsilon} \, dk \\
\\
&= \lim_{\epsilon\to 0}e^{-\frac12 x^2/\epsilon} \int_{-\infty}^{\infty} e^{-\frac12\epsilon (k-ix/\epsilon)^2} \, dk 
\\
&= \lim_{\epsilon\to 0}\sqrt{\frac{2\pi}{\epsilon}} \, e^{-\frac12 x^2/\epsilon}
\\
&= 2\pi \, \delta(x)
\end{align*}
