I have an issue. I want to prove what the Fourier transform of a Fourier Transform is. Let $f(t)$ be the analysed function and $F(\omega)$ its fourier transform. One book claims that $\mathcal{F}(\mathcal{F}(f(t))) = f(-t)$. I believe that $\mathcal{F}(\mathcal{F}(f(t)))=2\pi f(-t)$ is more likely. Starting from there:
$$\mathcal{F}(f(t))=\int_{-\infty}^{+\infty}f(t)e^{-i\omega t}d\omega$$
$$\Longleftrightarrow \mathcal{F}(\mathcal{F}(f(t)))=\int_{-\infty}^{+\infty}\left[\int_{-\infty}^{+\infty}f(T)e^{-i\omega T}dT\right]e^{-i\omega t}d\omega$$
Using that $\mathcal{F}(e^{i\omega_0t})=2\pi \delta(\omega-\omega_0)$ I get the $2\pi$ result.
Please help !
NB: I consider the non-symmetric Fourier transform, so you have a $1/2\pi$ for the inverse transform.
UPDATE: some comments say that it depends on the Fourier transform definition. I don't understand that. This a duality property. I.e., the inverse fourier transform of a fourier transform will always equal $f(t)$, regardless of the definition of the Fourier transform, right ?
In this answer I'll try to show you how the normalisation constant does modify the multiplicative constant when applying the transform twice, among other things.
Let's index all possible standard conventions for the Fourier transforms in the fashion below: $$\mathcal{F}_{a,b}[f](\xi) := a \int_\mathbb{R} f(x) e^{-ibx\xi}\,\mathrm{d}x$$ where $(a,b) \in (0, \infty) \times (\mathbb{R} \setminus \{0\})$ (some conventions do use negative $b$ for the direct Fourier transform and not the inverse one yes).
I'll assume that $f$ is for example of Schwartz class for the purpose of this post just to alleviate any issue about integrability, inverse Fourier transforms, etc... but of course you can relax that assumption.
First, let's find the expression of the inverse Fourier transform for each of these conventions. With the previous indexing, we can index inverse Fourier transforms as a Fourier transform with second parameter $-b$, thus we are looking for $c$ such that: $$\mathcal{F}_{c,-b} \circ \mathcal{F}_{a,b} = \mathrm{id}$$ Instead of redoing the proof of the Fourier inversion theorem for every convention (aka copy-pasting everything while observing the differences) however, we'll write everything with one convention for which we know for certain what the inverse Fourier transform is like. I'll choose $\mathcal{F}_{1,\pm 2\pi}$ simply because we know that $\mathcal{F}_{1,2\pi}^{-1} = \mathcal{F}_{1,-2\pi}$ and vice-versa, which simplifies things quite a bit (see Wikipedia for example).
We have: $$\begin{split} \mathcal{F}_{a,b}[f](\xi) &= a\int_\mathbb{R} f(x)e^{-ibx\xi}\,\mathrm{d}x\\ &= a\int_\mathbb{R} f(x) e^{-i2\pi x\frac{b\xi}{2\pi}}\mathrm{d}x\\ &= a\,\mathcal{F}_{1,\,2\pi}[f]\left(\frac{b\xi}{2\pi}\right) \end{split}$$ Hence, substituting the argument $\xi$ for $\zeta := \frac{b \xi}{2\pi}$ and integrating against $e^{i2\pi \zeta x}$ for some fixed $x \in \mathbb{R}$, we have: $$\frac{1}{a}\int_\mathbb{R} \mathcal{F}_{a,b}[f]\left(\frac{2\pi\zeta}{b}\right)e^{i2\pi \zeta x}\mathrm{d}\zeta = \int_\mathbb{R} \mathcal{F}_{a,b}[f](\zeta)e^{i2\pi \zeta x}\mathrm{d}\zeta = \mathcal{F}_{1,-2\pi}[\mathcal{F}_{1,2\pi}[f]](x) = f(x)$$ Yet the left hand side becomes, going back to $\xi = \frac{2\pi\zeta}{b}$, $\mathrm{d}\xi = \frac{2\pi}{b}\mathrm{d}\zeta$: $$\frac{1}{a}\int_\mathbb{R} \mathcal{F}_{a,b}[f]\left(\frac{2\pi\zeta}{b}\right)e^{i2\pi \zeta x}\mathrm{d}\zeta = \frac{|b|}{2\pi a}\int_\mathbb{R} \mathcal{F}_{a,b}[f](\xi)e^{ib\xi x}\mathrm{d}\xi$$ where the $|b|$ appears to have the correct orientation of the integral. Which means that: $$\mathcal{F}^{-1}_{a,b}[f] = \mathcal{F}_{\frac{|b|}{2\pi a},-b}[f]$$ It can then be checked that this is coherent when reapplying that operation, as we also need to have $\mathcal{F}_{\frac{|b|}{2\pi a},-b}^{-1} = \mathcal{F}_{a,b}$ too: $$\frac{|-b|}{2\pi\frac{|b|}{2\pi a}} = \frac{|b|}{|b|}\frac{2\pi}{2\pi} a = a$$ Having found the correct normalisation constant for the inverse Fourier transform, we'll now proceed to see which constant comes out when applying the Fourier transform twice in a row: $$\begin{split} \mathcal{F}_{a,b}[\mathcal{F}_{a,b}[f]](x) &= a\int_\mathbb{R} \mathcal{F}_{a,b}[f](\xi) e^{-ib\xi x} \,\mathrm{d}\xi\\ &= a \cdot \frac{2\pi a}{|b|}\cdot \frac{|b|}{2\pi a}\int_\mathbb{R} \mathcal{F}_{a,b}[f](\xi) e^{-ib\xi x} \,\mathrm{d}\xi\\ &= \frac{2\pi a^2}{|b|}\mathcal{F}_{\frac{|b|}{2\pi a},-b}[\mathcal{F}_{a,b}[f]](-x)\\ &= \frac{2\pi a^2}{|b|}\mathcal{F}_{a,b}^{-1}[\mathcal{F}_{a,b}[f]](-x)\\ &= \frac{2\pi a^2}{|b|}f(-x) \end{split}$$ which naturally is not in general $C f(-x)$ for some $C$ independent of $a$ and $b$.
This also means that in general we do not have the identity $\mathcal{F}_{a,b}^{\circ 4} = \mathrm{id}$ which is true for $(a,b) = (1, \pm 2\pi)$, but rather $\mathcal{F}_{a,b}^{\circ 4} = \frac{4\pi^2 a^4}{|b|^2}\mathrm{id}$ which is a bit messier.
Hopefully this answers your concerns.