= Solution
For the bound release orbits, use $e_0=\beta/(1-\beta)$ and
$$
C=r_b\beta^{-4/5}(1-\beta)^{-1/5},\qquad t_\star=\frac{cr_b^2}{GM_\star}.
$$
The <inspiral time under Poynting–Robertson drag> is then
$$
\boxed{\frac{t_{\rm PR}(\beta)}{t_\star}=\frac25\beta^{-13/5}(1-\beta)^{-2/5}\int_0^{\beta/(1-\beta)}\frac{u^{3/5}}{(1-u^2)^{3/2}}\,du,\quad0<\beta<\frac12.}
$$
The limiting branches are
$$
\boxed{\frac{t_{\rm PR}}{t_\star}\simeq\frac1{4\beta}\quad(\beta\to0^+),\qquad
\frac{t_{\rm PR}}{t_\star}\simeq\frac4{5\beta\sqrt{1-2\beta}}\quad(\beta\to\tfrac12^-).}
$$
Thus the formal bound-grain curve, shown in the right panel of the preceding figure, diverges at both ends and has a minimum at an intermediate $\beta$. At $\beta=0$ there is no <Poynting–Robertson drag> and the circular grain survives indefinitely in this collision-free model.
\b[Initially unbound grains do not have a bound-orbit inspiral lifetime.] For $\beta\geq1/2$ the release calculation gives outward trajectories; their residence time in a specified finite observation region is a dynamical escape time, not the averaged $t_{\rm PR}$ above. A numerical escape lifetime cannot be assigned without specifying the outer boundary. Extremely close to $1/2$, energy lost on the first passage can alter even this bound/unbound classification: orbital averaging requires roughly $1-2\beta\gg v_b/c$. Accordingly the divergence is a formal limit of the supplied secular equations, not an exact prediction for marginal grains.
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