Let $\alpha\in\mathbb{C}$ with $\Re(\alpha)>0$.
It is well known that the Fourier transform of $f(x)=\operatorname{sgn}(x)e^{-\alpha|x|}$ is : $$ \widehat{f}(\xi)=-\frac{4i\pi\xi}{\alpha^{2}+4\pi^{2}\xi^{2}} $$ There exists a proposition :
If $f$ is continuous and integrable and if $\widehat{f}\in L^{1}(\mathbb{R})$, then $\forall x\in\mathbb{R}$, we have that : $$ \mathscr{F}\widehat{f}(x)=f_{\sigma}(x)=f(-x) $$
I was wondering since $\widehat{f}(\xi)$ is not integrable at $\text{infinity}$ and since $f$ is not continuous at $0$ does this mean that the inverse Fourier of $\widehat{f}(\xi)$ which is $\widehat{g}(x)=\mathscr{F}\widehat{f}(\xi)$ does not exist?
Edit : I am not sure if the correct term should be "not-unique" instead of "does not exist".