= Solution
{c}
For $a=0,\frac12,1,\frac32,2,\ldots$, put $n=2a=0,1,2,3,4,\ldots$. Repeated <integration by parts> gives $\Gamma(n+1)=n!$. The hemispheric thick-limit results are
$$
\begin{array}{c|ccccc}
a&0&1/2&1&3/2&2\\ \hline
F/(\sigma T_s^4)&1&2/(3\tau_0)&1/\tau_0^2&12/(5\tau_0^3)&8/\tau_0^4\\
T_s/T_0&1&(3\tau_0/2)^{1/4}&\tau_0^{1/2}&(5\tau_0^3/12)^{1/4}&(\tau_0^4/8)^{1/4}
\end{array}
$$
The second row assumes fixed surface <temperature>; the third assumes fixed absorbed <radiant flux>. In the vertical-ray approximation the first of these rows becomes $1,1/\tau_0,2/\tau_0^2,6/\tau_0^3,24/\tau_0^4$, and the surface <temperature> factors are the inverse fourth roots.
Thus increasing $a$ makes the <atmosphere> colder at every fixed $0<p<p_s$, decreases outgoing <radiation> at fixed $T_s$, and requires a warmer surface at fixed absorbed <radiant flux>. The case $a=0$ is isothermal and gives $F=\sigma T_s^4$ exactly, including finite <optical depth>.
There is an essential approximation limit. The extension of the <integral> to infinity is valid for fixed $a$ as $\tau_0\to\infty$, and is not uniform for arbitrarily large $a$ at fixed $\tau_0$. The vertical integrand peaks at $t=2a$, so one needs that peak well above the surface in <optical depth>, with a negligible tail beyond $\tau_0$. For example, at $\tau_0=10$ the thick formula predicts $F/(\sigma T_s^4)=20!/(11\,10^{20})\simeq0.00221$ for $a=10$ but approximately $0.04773$ for $a=12$. This artificial increase contradicts the exact monotonicity: in the finite-<optical depth> expression every atmospheric emission factor $(t/\tau_0)^{2a}$ decreases with $a$. At fixed finite $\tau_0$, the exact limiting outgoing <radiant flux> is the transmitted surface term $2\sigma T_s^4E_3(\tau_0)$, rather than factorial divergence.
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