Inverse Fourier Transform of the Fourier Transform The FT of a function of time $f(t)$ transforms the function from the time domain to the frequency domain, such that:
$$\mathscr{F}[f(t)]=F(\omega)=\int ^{\infty}_{-\infty}f(t)e^{-i\omega t}dt$$
And the IFT of a function $F(\omega)$ transforms it from the frequency domain back to the time domain, such that:
$$\mathscr{F}^{-1}[F(\omega)]=f(t)=\frac{1}{2\pi}\int ^{\infty}_{-\infty}F(\omega)e^{i\omega t}d\omega$$
So, logically, applying the IFT to the FT should give the original function, such that:
$$\mathscr{F}^{-1}[\mathscr{F}[f(t)]]=f(t)$$
But upon trying to solve this, it goes as:
$$\mathscr{F}^{-1}[\mathscr{F}[f(t)]]=\frac{1}{2\pi}\int ^{\infty}_{-\infty}\int ^{\infty}_{-\infty}f(t)e^{-i\omega t}e^{i\omega t}dtd\omega=\frac{1}{2\pi}\int ^{\infty}_{-\infty}\int ^{\infty}_{-\infty}f(t)dtd\omega$$
which diverges. Whys is this the case?
 A: Following the discussion in the comments, one should rather write
\begin{aligned}
\mathscr{F}^{-1}\mathscr{F} f(t) &= \frac{1}{2\pi} \int \left(\int f (\tau) e^{-\text i \omega \tau}\,\text d\tau\right) e^{\text i \omega t}\,\text d\omega \, , \\
&= \frac{1}{2\pi} \iint f (\tau) e^{\text i \omega (t-\tau)}\,\text d\tau\,\text d\omega \, .
\end{aligned}
Using the change of variable $\theta = t-\tau$, Fubini's theorem, and properties if the Dirac distribution,
\begin{aligned}
\mathscr{F}^{-1}\mathscr{F} f(t) &= \frac{1}{2\pi} \iint f (t-\theta) e^{\text i \omega \theta}\,\text d\theta\,\text d\omega \, , \\
&= \int f (t-\theta) \left(\frac{1}{2\pi}\int e^{\text i \omega \theta}\,\text d\omega \right)\text d\theta\, , \\
&= \int f (t-\theta) \delta(\theta)\,\text d\theta\, , \\
&= f (t)\, .
\end{aligned}
This is a shortened "proof" of the theorem, assuming that the inverse Fourier transform of the constant 1 is the Dirac delta (which is a bit unfair...). The full proof of the Fourier inversion theorem holds for absolutely integrable continuous functions, with absolutely integrable Fourier transforms.
