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$\def\R{\mathbb R}$ 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:

Lemma Let $u_0\in \partial_x^j L^2(\R)\cap H^k(\R)$, 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.

Lorenzo Pompili
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    Is there a reason why you wouldn't just consider $u = \partial_{x} v$? – Matthew Cassell May 25 '24 at 00:58
  • @MatthewCassell: I’m interested in your suggestion and motivation. – A rural reader May 25 '24 at 03:04
  • @MatthewCassell thank you, I have edited the post after your suggestion. Is there something in particular you were thinking that I did not write? – Lorenzo Pompili May 25 '24 at 09:05
  • @MatthewCassell oh, now I see what you mean. Thank you! This is quite interesting on its own, and it might actually give a way of computing the Fourier transform of $f$ in terms of the kernel of Klein-Gordon. – Lorenzo Pompili May 25 '24 at 11:03
  • @LorenzoPompili That was my thought as well. – Matthew Cassell May 25 '24 at 11:44
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    Mathematica gives the result $$\mathcal{F}x\lefte^{-\frac{i}{x}}\right=\frac{\frac{\pi (\omega -| \omega | ) \sqrt{| \omega | } J_1\left(2 \sqrt{| \omega | }\right)}{\omega ^2}+2 \pi \delta (\omega )}{\sqrt{2 \pi }}=\frac{2 \pi \delta (\omega )+\frac{\pi (\text{sgn}(\omega )-1) , _0\tilde{F}_1\left(;2;-\frac{\omega }{\text{sgn}(\omega )}\right)}{\text{sgn}(\omega )^2}}{\sqrt{2 \pi }}$$ assuming the Fourier transform of $f(x)$ is defined as $$\mathcal{F}_xf(x)=\frac{1}{\sqrt{2 \pi }} \int{-\infty}^{\infty} f(x), e^{i \omega x} , dx.$$ – Steven Clark May 25 '24 at 14:49
  • @StevenClark thank you!! This seems to make sense: the group velocity of the above PDE is sign-definite, and in fact the function you wrote is only supported for negative $\omega$. Using the properties of the Bessell function $J_1$, it also seems to be bounded near zero, and to decay like $x^{-3/4}$ at $-\infty$. – Lorenzo Pompili May 25 '24 at 17:52

1 Answers1

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To close the question (changing a little to use the definition of Fourier transform that I like), the Fourier transform of $$ f(x)=e^{-i/(2\pi x)} $$ using the definition $$ \hat f(\xi)=\int_{\mathbb R} e^{-2\pi i \xi x}f(x)\,dx, $$ is $$ \delta_0 -\chi_{\{\xi>0\}}(\xi)\frac{J_1(2|\xi|^{1/2})}{|\xi|^{1/2}} , $$ where $J_1$ is the Bessel function of the first kind. At least this is what Mathematica says.

It is supported on $\{\xi\geq 0\}$ as a measure/distribution. As noted in a comment, it has a well-defined limit as $\xi\to 0^+$, and decays like $|\xi|^{-3/4}$ as $\xi\to+\infty$.

In a comment I said roughly that it should be supported for negative $\xi$ because of the group velocity of the above PDE. This actually depends on how you define the Fourier transform. What I get is supported on positive $\xi$ because I use a negative sign in the exponential, and to compute the actual kernel of the equation I would have to take the anti-Fourier transform of $f$ using my definition. So the kernel of the PDE should be the reflection of $\hat f$.

Thanks to the answer given in the comments.

Lorenzo Pompili
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