$\def\R{\mathbb R}$Question.
Does anyone know what is the Fourier transform of$$ f(x)=e^{-i/x} $$on the real line? I would like to compute it explicitly, or to establish some properties to have a good feeling of “how it looks like”.
What I know
I know that $\widehat f$ is real-valued by the symmetries of the Fourier transform, since $f(-x)=\overline{f(x)}$. It is natural to consider $h(x)=f(x)-1$. It is immediate that $h$ decays like $-i/x$ for large $x$, so $\widehat h$ belongs to $L^p(\mathbb R)$ for any $p\in [2,\infty)$ by Hausdorff-Young inequality. Then $\widehat f$ is recovered just by adding a Dirac delta.
Motivation.(not necessary to understand the problem)
For some reason, I was looking at the “anti-transport” equation$$ \left\{\begin{aligned}&u_t-\partial_x^{-1}u=0,\\&u|_{t=0}=u_0.\end{aligned}\right. $$Even if it does not entirely make sense, one can consider the unitary group $e^{t\partial_x^{-1}}$, which is well-defined on $L^2$, that acts on the Fourier side as a multiplication by $e^{-it/\xi}$. The kernel of this PDE coincides with a rescaled version of the anti-Fourier transform of $e^{-i/\xi}$, in particular it holds$$ u(t)=u_0+H_t*u_0, \quad\text{with}\quad H_t(x):=t\hat h(tx) $$for any $u_0\in L^1(\R)\cap L^2(\R)$ (up to multiplicative constants). This is why I considered the above problem.
I was wondering what is the time-regularity of solutions of such PDEs, even for smooth initial data: I find a bit funny that the operator $\partial_x^{-1}$ should in principle “smoothen things out” if well-defined, but it seems that the time-derivative of a solution of the PDE is not necessarily smooth for smooth data, unless one imposes some low-frequency condition on the initial datum (see below).
A similar PDE with dispersion relation that is singular at low frequencies pops up in the linear water wave theory for the deep water case: the PDE would look something like$$ u_t-|\partial_x|^{-1/2}u=0. $$
Edit: as pointed out by Matthew Cassell, one can write the PDE as a system:$$ \left\{\begin{aligned}u_t&=v\\v_x&=u\end{aligned}\right. $$In principle, $v$ is defined from $u$ only up to an additive constant, which could depend on time in a non-trivial way.
This system gives the following information for solutions defined via the $L^2$-group, $u(t)=e^{t\partial_x^{-1}}u_0$: if the initial datum $u_0$ lies in $\partial_x L^2(\R)$, then one can see that $u\in C(\mathbb R;\partial_x L^2(\R))$, in particular the anti-derivative $v$ is well-defined and $v\in C(\mathbb R; L^2(\R))$, so since $u_t=v$, we obtain that $u-u_0\in C^1(\R; L^2(\R))$. This procedure can be iterated:
LemmaLet $u_0\in \partial_x^j L^2(\R)\cap H^k(\R)$, and $u(t):=e^{t\partial_x^{-1}}u_0$. One has $u-u_0\in C^j(\R;H^{j+k}(\R))$.
This is a way of seeing how the time regularity of the solution is related to the low frequencies of the initial datum $u_0$ vanishing. More precise information can be derived (like fractional regularity in time) by looking at the Fourier transform of the solution in space-time.