Hilbert Transform
Buildup
$$ f(x,y)=\dfrac{1}{2} \mathcal{B} \left\{ \mathcal{F}^{-1} \Big[ |S|\mathcal{F} (\mathcal{R}f) (S,\ \theta) \Big]> \right\} (x,y) $$
This formula is for obtaining $f$ from Radon transform $\mathcal{R}f$ of $f$. First, recall the following property of the Fourier transform.
$$ \mathcal{F} [f^{\prime} ] (\xi) = i\xi \mathcal{F}(\xi) $$
Here, if we substitute $\mathcal{R}f$ for $f$, we get the following.
$$ \begin{equation} \mathcal{F} \left( \dfrac{\partial (\mathcal{R}f)(t,\ \theta) } {\partial t} \right) (S,\ \theta) = i S \mathcal{F}(\mathcal{R}f)(S,\ \theta) \label{eq1} \end{equation} $$
And let’s represent it as $|S|=S\cdot \mathrm{sgn}(S)$. $\mathrm{sgn}$ is the sign function.
$$ \mathrm{sgn}(S):=\begin{cases} 1 & S>0 \\ 0 & S=0 \\ -1 & S<0 \end{cases} $$
Then $\eqref{eq1}$ is as follows.
$$ \mathcal{F} \left( \dfrac{\partial (\mathcal{R}f)(t,\ \theta) } {\partial t} \right) (S,\ \theta) = i \dfrac{|S|}{\mathrm{sgn}(S)} \mathcal{F}(\mathcal{R}f)(S,\ \theta) $$
If we multiply both sides by $i \cdot \mathrm{sgn}(S)$, we obtain the following.
$$ i \cdot \mathrm{sgn}(S)\mathcal{F} \left( \dfrac{\partial (\mathcal{R}f)(t,\ \theta) } {\partial t} \right) (S,\ \theta) =- |S|\mathcal{F}(\mathcal{R}f)(S,\ \theta) $$
The right-hand side of the above equation appears in the Radon inversion. Therefore, substituting it yields the following.
$$ f(x,y)=-\dfrac{1}{2} \mathcal{B} \left\{ \mathcal{F}^{-1} \Big[ i \cdot \mathrm{sgn}(S)\mathcal{F} \left( \dfrac{\partial (\mathcal{R}f)(t,\ \theta) } {\partial t} \right) (S,\ \theta) \Big]\right\} (x,y) $$
Definition
For $g: \mathbb{R} \to \mathbb{R}$, we define the Hilbert transform $\mathcal{H}g$ of $g$ as follows.
$$ \mathcal{H} g(t) :=\mathcal{F}^{-1} \big[i \cdot \mathrm{sgn} (S) \cdot \mathcal{F}g(\xi) \big] (t) $$
When representing the Radon inversion through the Hilbert transform, it is as follows.
$$ f(x,y)=-\dfrac{1}{2} \mathcal{B} \left[ \mathcal{H} \left( \dfrac{\partial (\mathcal{R}f) (t,\ \theta)}{\partial t} \right) (S,\ \theta) \right] (x,y) $$