Image Reconstruction

Let $\stackrel{\rightharpoonup }{u}=\stackrel{\rightharpoonup }{\Phi }+t\stackrel{\rightharpoonup }{x};d\stackrel{\rightharpoonup }{y}={\left|t\right|}^{3}d\stackrel{\rightharpoonup }{x}.$ The above expression becomes

$G\left(\stackrel{\rightharpoonup }{\Phi },\stackrel{\rightharpoonup }{\beta }\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{\infty }{\int }}f\left(\stackrel{\rightharpoonup }{y}\right)e^{\frac{-2\pi i}{t}\left(\stackrel{\rightharpoonup }{y}-\stackrel{\rightharpoonup }{\Phi }\right)\cdot \stackrel{\rightharpoonup }{\beta }}}}}}\frac{1}{{\left|t\right|}^3}dtd\stackrel{\rightharpoonup }{y}.$

Let $s=\frac{1}{t};ds=\frac{1}{t^2}dt.$ We then have

$\begin{array}{l}G\left(\stackrel{\rightharpoonup }{\Phi },\stackrel{\rightharpoonup }{\beta }\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(\stackrel{\rightharpoonup }{y}\right)e^{-2\pi is\left(\stackrel{\rightharpoonup }{y}-\stackrel{\rightharpoonup }{\Phi }\right)\cdot \stackrel{\rightharpoonup }{\beta }}{\left|s\right|}^3\frac{1}{s^2}d\stackrel{\rightharpoonup }{y}ds}}},\\ G\left(\stackrel{\rightharpoonup }{\Phi },\stackrel{\rightharpoonup }{\beta }\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}({\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(\stackrel{\rightharpoonup }{y}\right)e^{-2\pi i\stackrel{\rightharpoonup }{y}\cdot \left(s\stackrel{\rightharpoonup }{\beta }\right)}d\stackrel{\rightharpoonup }{y})\left|s\right|e^{2\pi is\left(\stackrel{\rightharpoonup }{\Phi }\cdot \stackrel{\rightharpoonup }{\beta }\right)}ds}}}}.\end{array}$

Recognizing that the inner triple integral is the 3D Fourier transform of f, that is,

$F\left(s\stackrel{\rightharpoonup }{\beta }\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f(\stackrel{\rightharpoonup }{y}}}}$