Let $|x|=\sqrt{x_1^2+\cdots+x_n^2} ~$ for $~x\in\mathbb{T}^n=[-\frac{\pi}{2},\frac{\pi}{2}]^n$, $~n\geq 2$.
We can define the Fourier transform of $f(x)=\frac{1}{|x|}$ as
$$ \hat{f}(k)= \int_{\mathbb{T}^n}\frac{e^{2ik\cdot x}}{|x|} dx$$
for $k\in\mathbb{Z}^n$, since $f\in L^{1}(\mathbb{T}^n)$ if $n\geq 2$.
My question: is $\hat{f}\in l^{p}(\mathbb{Z^n})$ for $p=1$, or other possible value of $p$ ?
By Haursdorff-Young's inequality, we can partially answer this question. But I want some refined estimates.
Another related function is $g(x)=\frac{1}{w(x)}$, where
$$ w(x)=\sqrt{\sin^{2}\left({x_1}\right)+\sin^{2}\left({x_2}\right)+\cdots+\sin^{2}\left({x_n}\right)}.$$
And I have the same question to $\hat{g}$.
In my point of view, the property of $\hat{g}$ is more important.
If we can obtain the sharp upper decay of $\hat{f}$ or $\hat{g}$, or we can prove that $\hat{g}\in l^1(\mathbb{Z^n})$, that will be excellent.
The goal is to show the following assertion. Any comments are welcome.
"The Fourier series of $g$ defined above has decay order of $\mathcal{O}(|k|^{-n+1})$ as $|k|\rightarrow \infty$, $n\geq 2$."
The idea is essentially from Example2.4.9 of the book GTM249.
In $\mathbb{R}^n$, denote by $|x-y|$ the general distance of $x,y\in\mathbb{R}^n$. First notice that $g\in l^1(\mathbb{T}^{n})$. From the classical distribution theory (Thm2.4.6 of GTM249), we know
\begin{equation}\label{*} \mathcal{F}({|x|^{z}})=C_{n,z}|x|^{-n-z}~~~~(*) \end{equation}
for all $z$ with $-n<\Re z<0$, where $\mathcal{F}$ is the Fourier transform of $|x|^{z}$ as a distribution, $C_{n,z}$ is a constant dependent on $n$ and $z$. Thus if $z=-1$ then the decay at infinity behaves like $|x|^{-n+1}$. Notice that $g(x)\sim |x|^{-1}$ near the origin, so a similar result is expected.
The first step is to make the integral to the whole $\mathbb{R}^n$, we use a decomposition
$$ \sum_{j\in\mathbb{Z}^n}\eta(x+\pi j)=1$$
holds for all $x\in\mathbb{R}^n$, where $\eta\in C_{0}^{\infty}(\mathbb{R}^n)$ with appropriate support (not including other point in $\pi j$, $j\in\mathbb{N}^n\backslash\{0\}$), and equal to 1 near the origin. Then by the periodicity of $g$ and a translation, we get
$$ \hat{g}(k)=\sum_j \int_{\mathbb{T}^n}\frac{\eta(x+\pi j)}{w(x)}e^{2ik\cdot x}dx=\int_{\mathbb{R}^n}\frac{\eta(x)}{w(x)}e^{2ik\cdot x}dx.$$
Therefore, $\hat{g}(k)=\mathcal{F}{(\frac{\eta}{w})}(2k)$. The latter $\mathcal{F}$ is the Fourier transform in $\mathbb{R}^n$, here we can understand $k\in \mathbb{R}^n$.
Next we write
$$ \mathcal{F}{\left(\frac{\eta}{w}\right)}=\mathcal{F}{\left(\left(\frac{\eta(\cdot)|\cdot|}{w(\cdot)}-1\right)\frac{1}{|\cdot|}\right)}+\mathcal{F}{\left(\frac{1}{|\cdot|}\right)}=:\mathcal{F}(g_1)+\mathcal{F}(g_2). $$
Then using a smooth cut-off $\psi$ with $\psi=1$ near the origin, we get
$$ \mathcal{F}{\left(\frac{\eta}{w}\right)}=\mathcal{F}(g_1\psi)+\mathcal{F}(g_1(1-\psi))+\mathcal{F}(g_2)=:I_1+I_2+I_3. $$
We shall see that $$ I_{1}=\mathcal{O}(|k|^{-n}),~~I_{2}=\mathcal{O}(|k|^{-N}),~~I_{3}\sim |k|^{-n+1} $$ as $|k|\rightarrow \infty$ for all large $N$ and our proof is complete (note that here $\mathcal{F}{\left(\frac{\eta}{w}\right)}$ coincide with the Fourier transform as a distribution).
For $I_{3}$, we use (*). Now we analysis $I_1$, by a direct computation and an induction (it is important) we get
$$ \varphi(x):=\frac{1}{|x|}g_1(x)\psi(x) $$ has derivative estimate $\partial^{\alpha}\varphi(x)=\mathcal{O}(|x|^{-|\alpha|})$ as $|x|\rightarrow 0$ for all multi-index $\alpha$.
Therefore,
$$ \partial^{\beta}(g_1(x)\psi(x))=\partial^{\beta}(|x|\varphi(x))=\mathcal{O}(|x|^{1-|\beta|}) $$
as $|x|\rightarrow 0$ for all $\beta\in\mathbb{N}^n$. So $$\partial^{\beta}(g_1\psi)\in L^{1}(\mathbb{R}^n)$$ for $|\beta|=n$ and we obtain the desired estimate for $I_1$ by the relationship between Fourier transform and the derivative.
As for $I_2$, a similar argument shows that
$$ \partial^{N}(g_1(x)(1-\psi(x)))\in L^{1}(\mathbb{R}^n) $$
for $N\in\mathbb{N}$ large enough. Thus we have finished.