Abstract nonsense: the Fourier inversion formula is valid on tempered distributions. Let $T\in \mathscr S'(\mathbb R^n)$ and define the Fourier transform $\hat T$ by the bracket of duality
$$
\langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}=
\langle T,\hat\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}.
$$
Let $\mathcal C$ be the map $T(x)\mapsto T(-x)$ (which makes sense weakly) and let $\mathcal F$ be the Fourier transformation. Then the Fourier transformation is an isomorphism of $\mathscr S'(\mathbb R^n)$ and we have
$$
\mathcal C \mathcal F^2=Id, \quad [\mathcal C ,\mathcal F]=0,
$$
and as a result
$
T=\mathcal F \mathcal C \hat T
$
for any tempered distribution. If $\hat T$ happens to belong to $L^1(\mathbb R^n)$, it is enough to get $T(x)=\int e^{2iπ x\cdot \xi} \hat T(\xi) d\xi$ and $T$ is a continuous function with limit 0 at infinity: note that there is something to prove, but it is easy by using test functions in the Schwartz class. To sum-up, Fourier inversion formula is always true for tempered distributions and when the integrals make sense they coincide with the weak definition. Note also that the proof of the abstract nonsense argument is trivial and follows from the inversion formula in $\mathscr S(\mathbb R^n)$ since
$$
\langle \mathcal C\mathcal F^2 T,\phi \rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}=
\langle \mathcal F^2 T,\mathcal C\phi \rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}
=
\langle T,\underbrace{\mathcal F^2\mathcal C\phi}_{=\phi} \rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}.
$$

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