Use Dirac function to prove inverse Fourier Transform

Question

I am having trouble understanding how to show the following:

Use the $ \delta $ function to show that

$$ p(x) = {1 \over{2\pi}} \int_{-\infty}^{\infty} \varphi(k) e^{-ikx}dk $$

is the inversion of the Fourier transform.

How do I start?

Answer 1

We are going to use dirac function $\delta(\tau-x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}{e^{ik(\tau-x)}\ dk}$ and the property $p(x)=\int_{-\infty}^{\infty}{p(\tau)\ \delta(\tau-x) d\tau.}$

$$ \varphi(k)=\int_{-\infty}^{\infty}\ {p(\tau)e^{ik\tau}}\ d\tau $$

$$ \begin{aligned} \frac{1}{2\pi}\int_{-\infty}^{\infty}{\varphi(k)\ e^{-ikx}\ dk} {}&= {} \frac{1}{2\pi}\int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}\ {p(\tau)e^{ik\tau}}\ d\tau\ e^{-ikx}\ dk}\\ & =\frac{1}{2\pi}\int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}\ {p(\tau)e^{ik(\tau-x)}}\ d\tau\ dk}\\ & =\frac{1}{2\pi}\int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}\ {p(\tau)e^{ik(\tau-x)}}\ dk\ d\tau}\\ & =\int_{-\infty}^{\infty}{p(\tau)\frac{1}{2\pi}\int_{-\infty}^{\infty}\ {e^{ik(\tau-x)}}\ dk\ d\tau}\\ & =\int_{-\infty}^{\infty}{p(\tau)\ \delta(\tau-x)\ d\tau}\\ & =p(x) \end{aligned} $$