TITLE: Rigorous derivation/explanation of delta function representation?
QUESTION [14 upvotes]: I am interested in a derivation of the following representation for the Dirac delta function:
$$\delta(x-a)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i p (x-a)}dp$$
It is clear to me how the property $\delta(x-a)=\infty$ for $x=a$ arizes. What I am most interested in, is how $\delta(x-a)=0$ for $x\neq a$ comes about when evaluating the integral? Also, please explain the normalization factor.

REPLY [2 votes]: It might be too late but I thought I share my justification (non-rigorous) for it:
\begin{eqnarray}
\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{ikx}dkf(x)dx &= \lim_{K\rightarrow +\infty}\int_{-\infty}^{+\infty}\left(\frac{e^{ikx}}{ix}\right)_{-K}^K f(x)dx \\
&= \lim_{K\rightarrow +\infty}\int_{-\infty}^{+\infty}\frac{2i\sin(Kx)}{ix}f(x)dx \\
&= \lim_{K\rightarrow +\infty}\int_{-\infty}^{+\infty}\frac{2\sin(y)}{y}f(y/k)dy, y=Kx \\
&= \int_{-\infty}^{+\infty}\frac{2\sin(y)}{y}\lim_{K\rightarrow +\infty}f(y/k)dy \\
&= \int_{-\infty}^{+\infty}\frac{2\sin(y)}{y}f(0)dy \\
&= 2f(0)\int_{-\infty}^{+\infty}\frac{\sin(y)}{y}dy \\
&= 2\pi f(0)
\end{eqnarray}
where I used $e^{iKx} = \cos(iKx)+i\sin(iKx)$