The Exponential Representation of the Delta Function
As discussed in § {Representations of the Dirac Delta Function}, the Dirac delta function can be written in the form \begin{equation} \delta(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}\, dk .\\ \end{equation} We outline here the derivation of this representation.
In order to evaluate the integral, we introduce a regularization factor, $e^{-k\epsilon}$, as follows: \begin{eqnarray} \int_{-\infty}^{\infty} e^{ikx}\, dk &= \int_{-\infty}^0 e^{ikx}\, dk + \int_0^\infty e^{ikx}\, dk \nonumber\\ &= \int_0^\infty e^{-ikx}\, dk + \int_0^\infty e^{ikx}\, dk \nonumber\\ &= \int_0^\infty (e^{ikx} + e^{-ikx})\, dk \nonumber\\ &= \lim_{\epsilon\to0^+}\int_0^\infty (e^{ikx} + e^{-ikx}) \,e^{-k\epsilon}\, dk \nonumber\\ &= \lim_{\epsilon\to0^+}\int_0^\infty (e^{ik(x+i\epsilon)} + e^{-ik(x-i\epsilon)}) \,dk \nonumber\\ &= \lim_{\epsilon\to0^+} \left[ \frac{e^{ik(x+i\epsilon)}}{i(x+i\epsilon)} + \frac{e^{-ik(x-i\epsilon)}}{-i(x-i\epsilon)} \right]_0^\infty \nonumber\\ &= \lim_{\epsilon\to0^+} \left( 0 + 0 - \frac1{i(x+i\epsilon)} - \frac1{-i(x-i\epsilon)} \right) \nonumber\\ &= \lim_{\epsilon\to0^+} \left( \frac{i}{x+i\epsilon} - \frac{i}{x-i\epsilon} \right) \nonumber\\ &= \lim_{\epsilon\to0^+} \frac{2\epsilon}{x^2+\epsilon^2} = \cases{0 & $x\ne0$ \cr \infty & $x=0$ \cr} \end{eqnarray} where we have used the fact that, for $\epsilon>0$, $e^{-k\epsilon}$ goes to $0$ as $k$ goes to $\infty$. 1)
It remains to show that the final expression has the correct normalization. But \begin{equation} \int_{-\infty}^\infty \frac{2\epsilon}{x^2+\epsilon^2} \,dx = 2 \arctan\left(\frac{x}{\epsilon}\right) \Bigg|_{-\infty}^\infty = 2\pi , \end{equation} which is independent of $\epsilon$. Thus, \begin{eqnarray} \frac{1}{2\pi} \int_{-\infty}^\infty f(x) \,e^{ikx}\, dk &=& \lim_{\epsilon→0^+} \frac{1}{2\pi} \int_{-\infty}^\infty \frac{2\epsilon f(x)}{x^2+\epsilon^2} \,dx \nonumber\\ &=& \lim_{\epsilon→0^+} \frac{1}{2\pi} \int_{-\infty}^\infty \frac{2\epsilon f(0)}{x^2+\epsilon^2} \,dx \nonumber\\ &=& f(0) , \end{eqnarray} which is the defining property of the delta function.