Biomedical Imaging

The representation of sgn(t) can be proved by complex analysis and is used here to study the action of the Riesz operator with α = 1 − n on g

where we have used the definition of the Hilbert transform in the last step. The Hilbert transform has the following property

${\left(\frac{\partial }{\partial s}\right)}^{n-1}\left(\mathcal{H}g\left(z\right)\right)\left(s\right)=\left(\mathcal{H}{\left(\frac{\partial }{\partial z}\right)}^{n-1}g\left(z\right)\right)\left(s\right).\left(4.53\right)$

Plugging the last expression for (I1−ng)(s) as well as the last mentioned property of the Hilbert transform into the general inversion formula (4.40) and with s = x ⋅ nθ we find that

$\begin{array}{l}f\left(\underset{\_}{x}\right)=\frac{i}{2}{\left(2\text{π}\right)}^{-n+1}{\displaystyle \int d{\underset{\_}{n}}_{\theta }\frac{1}{\text{π}}{\displaystyle \int dz\frac{1}{s-z}{\left(\frac{\partial }{\partial s}\right)}^{n-1}\frac{\left(\mathcal{R}f\right)\left({\underset{\_}{n}}_{\theta },z\right)}{s-z}{|}_{s=\underset{\_}{x}\cdot {\underset{\_}{n}}_{\theta }}}}\\ =\frac{i}{2}{\left(2\text{π}\right)}^{-n+1}{\left(-i\right)}^{n-1}\left({\mathcal{R}}^{\#}\mathcal{H}{\left[\frac{\partial }{\partial z}\right]}^{n-1}\left(\mathcal{R}f\right)\left(z\right)\right)\left(\underset{\_}{x}\right)\end{array}$