Proof that $FT^{-1}FT = I$ (Fourier Transform $\cdot$ Inverse Fourier Transform = Identity)

Question

I am trying to understand the proof below. I understand all but how to get to the last line, which I'm completely confused by. Can anyone help me understand it? Thanks! \begin{align} FT^{-1}FT\left(s(t)\right)=& \dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty S(\omega)e^{i\omega t}d\omega \\ =& \dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty\left(\int_{t'=-\infty}^\infty s(t')e^{-i\omega t'}dt'\right)e^{i\omega t}d\omega \\ =& \int_{t'=-\infty}^\infty s(t')\dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty e^{-i\omega t'}e^{i\omega t}d\omega dt' \\ =& \int_{t'=-\infty}^\infty s(t')\delta(t-t')dt'=s(t) \end{align}

Answer 1

Took me a while to figure out as i thought you were integrating with respect to $t$, not $t'$

Let $$f(t')=\dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty e^{-i\omega t'}e^{i\omega t}d\omega=$$

$$=\dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty e^{i\omega(t-t')}d\omega$$

By the time domain shift theorem

$$f(-t'-t)=\dfrac{1}{2\pi}\int_{\omega=-\infty}^\infty e^{i\omega(t')}d\omega=\delta(-t'+t)=...$$

To shift $f(-t'-t)$ back to $f(t)$, I'd need to multiply it with the Heaviside step function, which as far as i know is impossible. Anyone feel free to chime in and continue from here.

Sources: Inverse Fourier Transform
https://www.thefouriertransform.com/pairs/fourier.php https://en.wikipedia.org/wiki/Fourier_transform#Translation_/_time_shifting

Answer 2

A way to prove the inverse Fourier transform. This time there is no change of variable and we play with the independent variables $\omega$ and $\tilde t$ .

Thanks to this there is no confusion with dummy variables and we guess, with the magic of Fourier transformations, the $\delta(\tilde t-t)$.

$$\mathscr{F}(f):w \mapsto \hat{f}(\omega) = F(\omega)= \int_\mathbb{R} f(t)~e^{-i \omega t} dt = \int_{-\infty}^{+\infty} f(t)~e^{-i \omega t} dt $$

\begin{aligned} \mathscr{F}^{-1}~~\Biggl(\mathscr{F} \Bigl(1\Bigl)(\omega)\Biggl)~~(\tilde t) &= \int_\mathbb{R} \left(\int_\mathbb{R} 1~e^{-i \omega t} dt\right) e^{i \omega \tilde t} \frac{d\omega}{2\pi} \\ &= \int_\mathbb{R} \left(\int_\mathbb{R} e^{i \omega (\tilde t-t)} \frac{d\omega}{2\pi} \right) ~dt \\ &= \int_\mathbb{R} \left(\mathscr{F}^{-1} ~1\right)_{(\tilde t-t)} ~dt \\ &= \int_\mathbb{R} \delta(\tilde t-t) ~dt \\ &= 1 \\ \end{aligned}

Answer 3

More generally:

A way to prove the inverse Fourier transform. This time there is no change of variable and we play with the independent variables $\omega$ and $\tilde t$ .

Thanks to this there is no confusion with dummy variables and we guess, with the magic of Fourier transformations, the $\delta(\tilde t-t)$.

$$\mathscr{F}(f):w \mapsto \hat{f}(\omega) = F(\omega)= \int_\mathbb{R} f(t)~e^{-i \omega t} dt = \int_{-\infty}^{+\infty} f(t)~e^{-i \omega t} dt $$

\begin{aligned} \mathscr{F}^{-1}~~\Biggl(\mathscr{F} \Bigl(f(t)\Bigl)(\omega)\Biggl)~~(\tilde t) &= \int_\mathbb{R} \left(\int_\mathbb{R} f(t)e^{-i \omega t} dt\right) e^{i \omega \tilde t} \frac{d\omega}{2\pi} \\ &= \int_\mathbb{R} \left(\int_\mathbb{R} e^{i \omega (\tilde t-t)} \frac{d\omega}{2\pi} \right) f(t)~dt \\ &= \int_\mathbb{R} \left(\int_\mathbb{R}~1~e^{i \omega (\tilde t-t)} \frac{d\omega}{2\pi} \right) f(t)~dt \\ &= \int_\mathbb{R} \left(\mathscr{F}^{-1} ~1\right)_{(\tilde t-t)} f(t) ~dt \\ &= \int_\mathbb{R} \delta(\tilde t-t) f(t)~dt \\ &= f(\tilde t)\int_\mathbb{R} \delta(\tilde t-t) ~dt \\ &= f(\tilde t)~1 \\ &= f(\tilde t) \\ &= f(\tilde t)|_{\tilde t \rightarrow t} = f(t)\\ \end{aligned}