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Let $f:\mathbb R\longrightarrow \mathbb R$ a function. The Fourier transform is given by $$\hat f(\alpha )=\int_{\mathbb R}f(x)e^{-2i\pi \alpha x}dx$$ and the inversion by $$f(x)=\int_{\mathbb R}\hat f(\alpha )e^{2i\pi \alpha x}d \alpha .$$

If I set $\alpha =-\beta $, don't we get that $$f(x)=\int_{\mathbb R}\hat f(-\beta )e^{-2i\pi \beta x}d\beta.$$ If I compare with the Fourier transform, I have that $f$ is the fourier transform of $x\longmapsto \hat f(-x)$ no ? If yes, is it interesting ?

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    Yes, this is well known. Usually written as $\large f = \hat{\tilde{\hat f}}$ – 2017-01-15
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    Yes, assuming $\hat{f}$ has a Fourier transform, this is how you get it. It is a well-known fact. – 2017-01-15
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    @AdamHughes: thank you. But is it an interesting result ? – 2017-01-15
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    @user330587 It depends on what you mean by "interesting." Certainly it is extremely useful like all of the basic results on any topic are, so in that sense, yes. In mathematics one more often hears "interesting result" used to describe something deeper than one-line basics though, so if you mean it in terms of depth or if it is anything other than obvious, then no. – 2017-01-15
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    To make this rigorous, one needs to say more about $f$ than just "a function". It would be difficult to make sense of these statements when, for example, $f(x)=e^x$. A conservative approach is to assume that both $f$ and $\hat f$ are in $L^1$, but more general forms exist which require a careful interpretation of the integrals. – 2017-01-16

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