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The Radon transform records line integrals of a planar density. Apply the transverse Hilbert transform and a derivative to each projection, then average the filtered values over all lines through the reconstruction point. With $Hh(\rho )={\pi }^{-1}\mathrm{PV}\int h(r)/(\rho -r)dr$, the filter has Fourier multiplier $|\kappa |$. The Fourier inversion theorem, or the jump of the complexified transport Cauchy kernel, proves the displayed formula. The factor is $1/(2\pi )$ when directions are integrated over a half-circle. The growing Fourier multiplier explains amplification of measurement noise and the need for regularization of an inverse problem.

The distributional identity ${\partial }_{\bar{z}}(1/(\pi z))=\delta$ gives the Cauchy-Green operator solutionwhere $\tilde{f}$ is $f$ in the changed real coordinates. The singularity $1/(z-z^{\prime })$ is locally integrable. For Schwartz functions, splitting the integral into a region near the origin and a rapidly decaying far region proves $\mu =O(|z{|}^{-1})$. If two solutions obey this decay, their difference has zero Wirtinger derivative and is an entire holomorphic function tending to zero; the Liouville theorem proves uniqueness.

The change-of-variables Jacobian has absolute value $a_r/|b_r|$. Therefore an equivalent form of the complexified transport Cauchy kernel isIn the original complex coordinate $w=x_1+i{x}_2$, it isFor $|\lambda |<1$ the denominator can be rewritten to give the kernel $\lambda /[\pi ((w-w^{\prime })-{\lambda }^2(\bar{w}-{\bar{w}}^{\prime }))]$. For $|\lambda |>1$ it is ${\lambda }^{-1}/[\pi ((\bar{w}-{\bar{w}}^{\prime })-{\lambda }^{-2}(w-w^{\prime }))]$. Consequently the spectral solution is holomorphic separately inside and outside the unit circle, with $\mu =O(\lambda )$ at zero and $\mu =O({\lambda }^{-1})$ at infinity. These normalizations will determine its jump reconstruction.

The two boundary values have jumpThe line-tail contributions cancel. Because the complexified transport Cauchy kernel is holomorphic on both sides of the unit circle and vanishes at spectral infinity, the Cauchy integral formula reconstructs it from this jump:There is no additional entire term: its inside and outside boundary values would agree, so it would define an entire function vanishing at infinity. Expand outside the circle as $\mu ={\mu }_1/\lambda +O({\lambda }^{-2})$. ThenIn the original coordinate $w=x_1+i{x}_2$, the transport equation is $(\lambda {\partial }_w+{\lambda }^{-1}{\partial }_{\bar{w}})\mu =f$. Its constant term at infinity is ${\partial }_w{\mu }_1=f$. Sincewe obtain the inverse Radon transform in filtered backprojection form:Opposite directions represent the same unoriented line. Reversing $\theta$ by $\pi$ reverses $\rho$; the Hilbert transform changes sign under this reversal, and its derivative changes it back. Thus the integrand agrees for opposite directions and the factor becomes $1/(2\pi )$ if the angular integral is over $[0,\pi )$. With the convention $\hat{h}(\kappa )=\int e^{-i\kappa \rho }h(\rho )d\rho$, the Hilbert transform has Fourier multiplier $-i\mathrm{sgn}\kappa$; the derivative makes the filter $|\kappa |$. This also identifies the formula with the standard ramp-filter reconstruction.

Keep the coordinates and operator $D_k$ from part (a). The attenuation $f$ is known, and the source $g$ is to be reconstructed. Put $F(\rho ,\tau ,\theta )={\int }_{\tau }^{\infty }f(s\cos \theta -\rho \sin \theta ,s\sin \theta +\rho \cos \theta )ds$ and $S(\rho ,\theta )=\mathcal{R}f={\int }_{\mathbb{R}}fd\tau$. The directed forward attenuated Radon transform, for a detector at the positive end of the line, isThis weight accounts for absorption from each emission point to the detector. It is also derived by solving the real transport equation ${\partial }_{\tau }u+fu=g$ with zero incoming intensity: its outgoing limit is $A_f$. In particular reversing the line direction generally changes the measurement. The pair reconstructs $g$ from full directed data and known $f$; it does not infer both unknown functions from the same data.

We must specify the sign convention for the projectors. Use the Hilbert transform $Hv(\rho )={\pi }^{-1}\mathrm{PV}{\int }_{\mathbb{R}}v(s)/(\rho -s)ds$ and the signed Cauchy boundary operatorsThen $P^{+}-P^{-}=I$, and the positive idempotent projectors are $P^{+}$ and $-P^{-}$. This is the signed boundary-value convention in the given limits of $M$. To express everything with ordinary complementary projectors, one may instead write ${\Pi }_{+}=P^{+}$, ${\Pi }_{-}=-P^{-}$. The given limits becomeBoth $a_{\pm }$ depend on the transverse coordinate and direction but are constant along $\tau$.

The equation $D_{k}M=f$ supplies an integrating factor:where $T_k$ is the inverse of unattenuated complex transport, normalized to vanish at spatial infinity. To see the normalization and analyticity rather than assume them, let $w=x_1+i{x}_2$. The complexified transport Cauchy kernel isIn the rotated coordinates, $E_k=[2\pi |B|(\tau +iA\rho /B){]}^{-1}$. The identity ${\partial }_{\bar{z}}(1/(\pi z))=\delta$ and the Jacobian $A/|B|$ prove $D_k{E}_k=\delta$ with exactly this normalization. The denominator cannot vanish for nonzero real $w$ off the unit circle. The kernel is analytic separately inside and outside that circle, is $O(k)$ at zero and $O(k^{-1})$ at infinity. Thus $M=T_{k}f$ and the displayed $\mu$ have the corresponding spectral behavior for the smooth decaying data. In particular $\mu =O(k^{-1})$ at infinity and is bounded, indeed $O(k)$, at zero.

The limiting unattenuated inverse on any smooth decaying input $h$ isOne may verify this directly in the rotated Fourier variables: $D_k$ has multiplier $iA\xi -B\eta$, so its inverse has a pole at $\xi =-iB\eta /A$. As $B\to 0^{+}$, the positive transverse frequencies have the negative tail primitive and the negative transverse frequencies have the positive past primitive; for $B\to 0^{-}$ these allocations reverse. Splitting with the Fourier projectors ${\Pi }_{\pm }$ yields exactly the two formulas above. Applying them to $f$ reproduces the assumed limits of $M$ and checks the signs.

Now apply the same limiting inverse to $h=e^{M^{\pm }}g$. Because $a_{\pm }$ are independent of $\tau$, its line integral is $e^{a_{\pm }}{A}_f$, while its tail is $e^{a_{\pm }}{\int }_{\tau }^{\infty }{e}^{-F}gds$. HenceThe unknown incomplete line integral cancels in their difference. All remaining quantities are obtained from the measurements and known attenuation:Each projection acts in $\rho$ at fixed $\theta$, before evaluating $\rho =-x_1\sin \theta +x_2\cos \theta$ and $\tau =x_1\cos \theta +x_2\sin \theta$. Pulling a transverse-dependent exponential outside a projector would be incorrect.

The sectionally analytic $\mu$ therefore solves an additive Riemann-Hilbert problem on the counterclockwise unit circle, with the plus value inside. The normalized Cauchy integral formula isThe Sokhotski–Plemelj theorem gives its jump; boundedness at zero and decay at infinity make it unique by the Liouville theorem. Its expansion at infinity isSubstitute into the original equation. The $k{\partial }_{x_1}-ik{\partial }_{x_2}$ part contributes at order one, the reciprocal derivative and $f\mu$ vanish at that order. This derives the projection formula for inverse attenuated Radon transform:Together with the boxed forward transform this is the requested pair. The derivation uses no small-attenuation approximation. For smooth compactly supported data the operations above are justified directly; sufficient integrable decay permits the same limiting construction.

As a normalization check, set $f=0$. Then $S=F=0$, $A_f=\mathcal{R}g$ and $J=(P^{-}+P^{+})\mathcal{R}g=iH\mathcal{R}g$. Since the line transform is independent of $\tau$ and $e^{i\theta }({\partial }_{x_1}-i{\partial }_{x_2})={\partial }_{\tau }-i{\partial }_{\rho }$, the formula reduces tothe full-angle inverse Radon transform. For the radial Gaussian $g=e^{-|x{|}^2}$, $\mathcal{R}g=\sqrt{\pi }{e}^{-{\rho }^2}$ and ${\partial }_{\rho }H\mathcal{R}g(0)=2$, so the reconstructed value at the origin is $1$, checking the $1/(4\pi )$ factor.

In single-photon emission computed tomography, $g$ represents the internal emission density, $f$ the known absorption coefficient and $A_f$ the directed detector counts. Each path integral weights a source point by its survival probability to the detector. Thus the attenuated Radon transform is the forward measurement model and its derived inverse reconstructs the emission distribution, providing the mathematical basis of SPECT.