I think I knew how to show it before. But suddenly just realize I can’t.
Say let’s define $latex F(w) = \frac{1}{2 \pi} \int f(t) e^{-i wt} dt$ and the inverse transform
$latex \hat{f}(t)= \int F(w) e^{i w t}dw = f(t)$. So we expect
$latex f(t)=\frac{1}{2 \pi} \int \int f(\tau) e^{-i w\tau} d\tau e^{iwt}dw$
$latex =\frac{1}{2 \pi} \int \int f(\tau) e^{i w(t-\tau)} d\tau dw$
$latex \overset{*}{=}\int f(\tau) \left[\frac{1}{2 \pi}\int e^{i w(t-\tau)} dw\right] d\tau $
So one can finish the proof as we can show that $latex \frac{1}{2 \pi}\int e^{i w(t-\tau)} dw=\delta(t-\tau)$, note that
$latex \frac{1}{2 \pi}\int_{-\infty}^{\infty} e^{i w(t-\tau)} dw=\lim_{L \rightarrow \infty} \frac{1}{2 \pi}\int_{-L}^{L} e^{i w(t-\tau)} dw$
$latex = \frac{1}{2 \pi}\frac{e^{i w(t-\tau)}}{i(t – \tau)}|_{-L}^{L}$
$latex = \frac{\sin(L(t-\tau))}{(t-\tau)\pi}=\delta(t-\tau)$ as $latex L\rightarrow \infty$.
“Intuitively”, the above does appear to be a $latex \delta$-function. A more rigorous proof is shown here. For elementary proof, once we accept that it is a $latex \delta$-function, we just want to verify that normalization factor is right, i.e., $latex \int \frac{\sin(Lx)}{x}dx=\pi$. Note that
$latex \int \frac{\sin(Lx)}{x}dx=\int \frac{\sin(Lx)}{Lx}d(Lx)=\int \frac{\sin(x’)}{x’}dx’$, where the last term is shown to be $latex \pi$ by complex calculus.
N.B. Strictly speaking, $latex *$ is wrong as the hypothesis of Fubini algorithm is not satisfied as stated here
.