The defining property of the Dirac delta function is that for any function $f(x)$: $$\int^{\infty}_{-\infty}f(x)\delta(x-a)dx=f(a)$$
The Dirac delta function can be expressed as: $$\delta(x-a) = \frac{1}{2\pi}\int^{\infty}_{-\infty}e^{ip(x-a)}dp$$
How can we show that the latter expression possesses the property defined by the first equation? I.e. is there a proof of the following:
$$\frac{1}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x)e^{ip(x-a)}dp dx=f(a)$$
First note that the Dirac is a distribibution, not a function. Suppose $f$ is an absolutely integrable and piecewise continous function, then, using the Fourier transfom $\mathcal{F}$, the inversion property :
$$f(a) = \mathcal{F}^{-1}\mathcal{F}(f)(x) \\ = \frac{1}{2\pi}\int e^{-ipa}\int e^{ips}f(s)ds dp$$
With Fubini theorem, you have that :
$$f(a) = \frac{1}{2\pi}\int \int e^{ip(s-a)}f(s)ds dp$$