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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?

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  • $\begingroup$ $\mathcal{F}^{-1}(\mathcal{F}(f(t)) = \mathcal{F}^{-1}(\mathcal{F}(\omega)) = f(t)$. $\endgroup$ – Maxtron Aug 11 '18 at 5:00
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    $\begingroup$ I think you are getting confused by reusing your $t$'s instead of introducing a dummy variable when writing down your expression for the function composition in the inner integral. You can't bring in your "$e^{i\omega t}$" inside the integral here. $\endgroup$ – JessicaK Aug 11 '18 at 5:02
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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.

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