= Solution
\b[The printed line-density statement is incorrect for a phase-mixed <orbit>]. A steady <line density on a Kepler orbit> is inversely proportional to speed: for a <cross-sectional-area current> $J_\sigma$, the area per unit arc length is $d\sigma/ds=J_\sigma/v$. Equivalently, <phase mixing> gives $d\sigma=(\sigma_{\rm tot}/P)\,dt$, uniform in <mean anomaly>, where $P=2\pi\sqrt{a^3/\mu}$ is the <orbital period>.
For an <optically thin> population of <blackbodies> in <radiative equilibrium>, a fragment absorbs $\sigma L_\star/(4\pi r^2)$ and reradiates the same <luminosity>. Thus the <fractional luminosity of a phase-mixed eccentric wire> is
$$
f=\frac{\sigma_{\rm tot}}{4\pi P}\int_0^P\frac{dt}{r^2}.
$$
The <specific angular momentum> gives $dt/r^2=d\theta/h$, so the integral is $2\pi/h$. Since $Ph=2\pi a^2\sqrt{1-e^2}$,
$$
\boxed{f=\frac{\sigma_{\rm tot}}{4\pi a^2\sqrt{1-e^2}}}.
$$
This derives the intended result after explicitly correcting the density to $1/v$.
The error is consequential. If one instead imposes the literal density $d\sigma/ds\propto v$, its time weighting is $v^2dt$. Using $\langle v^2\rangle=\mu/a$, $\langle r^{-2}\rangle=[a^2\sqrt{1-e^2}]^{-1}$ and $\langle r^{-3}\rangle=[a^3(1-e^2)^{3/2}]^{-1}$ gives
$$
f_{\lambda_\ell\propto v}=\frac{\sigma_{\rm tot}}{4\pi a^2}\frac{1+e^2}{(1-e^2)^{3/2}}.
$$
At $e=1/2$ this is $5/3$ times the printed result. The inverse-radius averages can also be obtained directly with <eccentric anomaly> $E$, $r=a(1-e\cos E)$ and $dt=(1-e\cos E)dE/n$.
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