What are Fourier Pairs? Given Fourier's Inversion theorem:
$$f(t)=\displaystyle\frac{1}{2\pi}\int_{\omega =-\infty}^\infty e^{i \omega t} \, \mathrm{d}\omega \int_{u=-\infty}^\infty f(u) e^{-i\omega u} \,\mathrm{d}u  \tag{1}$$
and the Fourier transform of $f(t)$:
$$\displaystyle\mathcal{F}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{t=-\infty}^{\infty}\color{red}{f(t)}e^{-i\omega t}\mathrm{d}t\tag{A}$$
and the Inverse Fourier transform of $\mathcal{F}(\omega)$:
$$\displaystyle \color{red}{f(t)}=\color{blue}{\frac{1}{\sqrt{2\pi}}\int_{\omega=-\infty}^{\infty}\mathcal{F}(\omega)e^{i\omega t}\mathrm{d}\omega}\tag{B}$$
In trying to understand Fourier pairs I substitute $(\mathrm{B})$ into $(\mathrm{A})$ by replacing the part marked $\color{red}{\mathrm{red}}$ in  $(\mathrm{A})$ with the expression  $(\mathrm{B})$ which gives
$$\displaystyle\mathcal{F}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{t=-\infty}^{\infty}\color{blue}{\frac{1}{\sqrt{2\pi}}\int_{\omega=-\infty}^{\infty}\mathcal{F}(\omega)e^{i\omega t}\mathrm{d}\omega}e^{-i\omega t}\mathrm{d}t$$
$$\implies \mathcal{F}(\omega)=\color{#180}{\displaystyle\frac{1}{2\pi}\int_{t=-\infty}^{\infty} \int_{\omega=-\infty}^{\infty}\mathcal{F}(\omega)\mathrm{d}\omega \mathrm{d}t}\tag{C}$$
Now unless I'm mistaken the RHS of $(\mathrm{C})$ marked $\color{#180}{\mathrm{green}}$ should reduce to $\mathcal{F}(\omega)$ on the LHS of $(\mathrm{C})$.
What steps do I need to take to show this?
Thank you.
 A: You should be slightly more careful with your notation. We have the Fourier transform
$$
 \hat{f}(\omega) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty f(x) e^{-i\omega x} \,dx
$$
and Fourier inversion
$$
 f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \hat{f}(\xi) e^{i\xi x} \,d\xi.
$$
You want to insert the first equation into the second:
$$
  \hat{f}(\omega) =\frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \hat{f}(\xi) e^{-i\omega x} e^{i\xi x} \,d\xi\,dx = \frac{1}{2\pi} \int_{-\infty}^\infty \int_{-\infty}^\infty \hat{f}(\xi) e^{i(\xi - \omega)x} \,d\xi\,dx
$$
The thing to realize is that the $\omega$ appearing in the argument of $\hat{f}$ on the left is not the same as the one that appears in the definition of the Fourier inversion (your Equation B). By using different letters for the dummy integration variables, you see that the exponentials do not cancel, and the correct version of your equation (C) is the one I have above. At this step there is not much else that can be simplified. Rather than thinking of simplifying your green equation (or, rather, the corrected version I propose), think of it as a result.
