 Solution

For $a=0,\frac{1}{2},1,\frac{3}{2},2,\dots$, put $n=2a=0,1,2,3,4,\dots$. Repeated integration by parts gives $\Gamma (n+1)=n!$. The hemispheric thick-limit results areThe 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 $F/(\sigma T_s^4)=20!/(1110^{20})\simeq 0.00221$ 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 $2\sigma T_s^4{E}_3({\tau }_0)$, rather than factorial divergence.