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The integral of a source over a directed line, weighted by exponential absorption between each source point and a detector. For line coordinates $x=p\mathbf{n}+s\mathbf{d}$ and a detector at the negative end, it is $\int f(p,s)\exp [-{\int }_{-\infty }^s\mu (p,r)dr]ds$. The integrating factor of a transport equation derives the weight. Reversing the detector direction changes the weight for spatially varying attenuation, so the two directed measurements need not coincide. Setting $\mu =0$ recovers the Radon transform.

Transmission measurements determine line integrals of an absorption coefficient after taking the logarithm of incident intensity divided by transmitted intensity. Under the ideal straight-ray model, these measurements are a Radon transform; filtered backprojection reconstructs a planar slice. Angular coverage, calibration and noise control matter because differentiating measured projections amplifies high frequencies. In contrast, single-photon emission computed tomography reconstructs an emitting source through a possibly attenuating medium.

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.

A mathematician whose work includes Radon's theorem, Radon measures, the Radon-Nikodym theorem and the Radon transform.

Define the Radon transform and its transverse Hilbert transform byUse the signed Cauchy boundary operators $P^{+}=(I+iH)/2$ and $P^{-}=(-I+iH)/2$. Thus $P^{+}$ and $-P^{-}$ are the complementary idempotent projections; $P^{-}$ itself carries the signed lower-boundary convention. The PDF defines the outside limit ${\lambda }^{-}=(1+\epsilon )e^{i\theta }$ but omits the definition of the inside limit in its displayed formula. We take ${\lambda }^{+}=(1-\epsilon )e^{i\theta }$, as required by that formula.

As $r\to 1$, apply the Sokhotski–Plemelj formula to the kernel from part (b):The principal value contribution is $i\mathrm{sgn}(b_r)HRf/2$. The delta contribution is the same from either side and equalsSince $b_r<0$ for the inside limit, the resulting boundary value isThe outside limit similarly has $P^{+}Rf$ in place of $-P^{-}Rf$. Naming the signed convention explicitly prevents a sign error from treating $P^{-}$ as a positive projection.

In ideal computed tomography, the unknown is an absorption coefficient $a(x)$. Along a straight ray the transmitted intensity satisfies $dI/d\tau =-aI$, soThus calibrated transmission measurements supply the Radon transform of the slice. The inversion above reconstructs the coefficient by filtered backprojection: filter each projection in its transverse variable and then accumulate it over the rays through each image point. This turns a collection of line-integral measurements into a spatial map, which is the key usefulness of the result.

The ideal derivation assumes known ray geometry, sufficiently complete angular sampling and a straight-ray attenuation model. The ramp Fourier multiplier grows at high frequency, so measurement noise is amplified. Practical reconstruction therefore uses discretization and regularization of an inverse problem, for example a window on the ramp filter. Limited angles or missing detector positions cannot be treated as if they were complete noiseless Radon transform data.

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.

Let $\mathcal{R}g(s,\omega )$ be the Radon transformTaking the large-radius limit in the Kirchhoff formula, with $\xi =t-r$ fixed for the outgoing limit and $\eta =t+r$ fixed for the incoming limit, gives the radiation fieldsIndeed, the expanding spheres converge after multiplication by $r/t$ to the planes $y\cdot \omega =-\xi$ and $y\cdot \omega =\eta$, respectively.

The Radon transforms of smooth compactly supported functions are smooth. If the data are supported in a ball of radius $R$, these transforms vanish for $|s|>R$. Hence both radiation fields are well-defined smooth functions of compact support in the null-time variable.

Tomography reconstructs an interior distribution from measurements along multiple lines or paths through it. A linear unattenuated model is the Radon transform; absorption changes it to an attenuated Radon transform.