I want to calculate the Fourier transform of $$ f: \mathbb{R}^d \to \mathbb{R}, \; x \mapsto \mathrm{sgn}(x \cdot \theta), $$ where $\theta \in \mathbb{R}^d$ is fixed. As $f$ is not in $L^1$ I consider it as a tempered distribution and expect its Fourier transform to be of the same class. My idea was to generalize the calculation of the Fourier transform of the signum function, but I couldn't get everything to work out.
My approach
For the Fourier transform of the signum function $\mathrm{sgn}: \mathbb{R} \to \mathbb{R}$ I know the following argument: One can easily show that (in the distributional sense) the derivative of $\mathrm{sgn}$ is $2 \delta$ (where $\delta$ is the delta distribution). Hence $$ i \chi \mathcal{F} \mathrm{sgn} = \mathcal{F} \mathrm{sgn}' = 2 \mathcal{F} \delta = 2 $$ shows that $\mathcal{F} \mathrm{sgn} = \frac{2}{i \chi}$, where $\chi$ is the identity function of $\mathbb{R}^d$.
To transfer this argument to the case of $d>1$, I first assume that $\theta = e_1$, the vector with a $1$ in the first row and zero everywhere else. Then I obtain that the derivative with respect to the first component $\partial_1 f$ is the distribution $T$ with $$ \langle T, \varphi \rangle = 2 \int_{\mathbb{R}^{d-1}} \varphi(0, y_1, \dots, y_n) \; d(y_1, \dots, y_n)$$ for all $\varphi \in \mathcal{S}(\mathbb{R}^d)$. Now, the problem is that the Fourier transform of $T$ is harder to compute than the Fourier transform of $\delta$ in the case $d=1$.
Any help is appreciated!
My impression is that $T = 2\delta_0 \otimes {\bf 1}$, where $\delta_0$ is the Dirac distribution on $\mathbb{R}$ and ${\bf 1}$ the constant function equal to $1$ on $\mathbb{R}^{n-1}$. Up to multiplicative constants (depending on the definition chosen for the Fourier transform), you derive $$X_1\mathcal{F}f = \mathcal{F}(\partial_1 f) = {\bf 1} \otimes \delta_0,$$ but this time ${\bf 1}$ is a function on $\mathbb{R}$ and $\delta_0$ is the Dirac distribution on $\mathbb{R}^{n-1}$. I think you fill find up to multiplicative constants $$\mathcal{F}f = [PV(1/X_1)+c\delta_0] \otimes \delta_0.$$ But the constant $c$ is $0$ because the Fourier transform of an odd distribution is odd.