= Solution
The two boundary values have jump
$$
J(\theta)=\mu(\lambda^+)-\mu(\lambda^-)=-iHRf(\rho,\theta).
$$
The 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:
$$
\mu(\lambda)=\frac1{2\pi i}\oint_{|\zeta|=1}\frac{J(\arg\zeta)}{\zeta-\lambda}d\zeta.
$$
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 $\mu=\mu_1/\lambda+O(\lambda^{-2})$. Then
$$
\mu_1=-\frac1{2\pi}\int_0^{2\pi}e^{i\theta}J(\theta)d\theta
=\frac{i}{2\pi}\int_0^{2\pi}e^{i\theta}HRf(\rho,\theta)d\theta.
$$
In the original coordinate $w=x_1+ix_2$, the <transport equation> is $(\lambda\partial_w+\lambda^{-1}\partial_{\bar w})\mu=f$. Its constant term at infinity is $\partial_w\mu_1=f$. Since
$$
\partial_w\rho=\frac12(\partial_{x_1}-i\partial_{x_2})\rho=-\frac i2e^{-i\theta},
$$
we obtain the \b[inverse Radon transform] in <filtered backprojection> form:
$$
\boxed{f(x_1,x_2)=\frac1{4\pi}\int_0^{2\pi}
(\partial_\rho HRf)(-x_1\sin\theta+x_2\cos\theta,\theta)d\theta.}
$$
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 $\widehat h(\kappa)=\int e^{-i\kappa\rho}h(\rho)d\rho$, the <Hilbert transform> has <Fourier multiplier> $-i\operatorname{sgn}\kappa$; the derivative makes the filter $|\kappa|$. This also identifies the formula with the standard ramp-filter reconstruction.
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