Solution (source code)

= Solution

For the <Rayleigh-Jeans spectrum of a finite blackbody disk>, the <Rayleigh-Jeans law> replaces $(e^{h\nu/(kT)}-1)^{-1}$ by $kT/(h\nu)$. Here $h$ is the <Planck constant> and $k$ the <Boltzmann constant>. The <multitemperature blackbody disk> therefore has
$$
F_\nu\sim\frac{k}{h}\nu^2\int_{r_*}^{r_{\rm out}}rT(r)\,dr,\qquad\boxed{F_\nu\propto\nu^2.}
$$
For example, setting $R=r_{\rm out}/r_*$ makes its <frequency>-independent coefficient proportional to the finite dimensionless integral
$$
T_{\rm in}r_*^2\int_1^R y^{1/4}(1-y^{-1/2})^{1/4}\,dy.
$$
Although the exact <effective temperature> vanishes at the inner edge, the very narrow cold rim where the <Rayleigh-Jeans law> fails makes a negligible contribution in this limit. More formally, after dividing the integrand by $\nu^2$, the inequality $e^x-1\geq x$ bounds it by $(k/h)rT$, so <dominated convergence> justifies the result even at that edge.

At large <radius>, $T\propto r^{-3/4}$, and the contribution per logarithmic interval is $r^2T\propto r^{5/4}$. \b[The outer disk dominates the low-<frequency> emission] because its much greater area outweighs its lower <effective temperature>.

For intermediate frequencies use the allowed power-law approximation to the <effective temperature> and introduce
$$
x=\frac{h\nu}{kT(r)}=\frac{h\nu}{kT_{\rm in}}\left(\frac r{r_*}\right)^{3/4}.
$$
Then
$$
r\,dr=\frac43r_*^2\left(\frac{kT_{\rm in}}{h\nu}\right)^{8/3}x^{5/3}\,dx,
$$
so
$$
F_\nu\propto\nu^{1/3}\int_{h\nu/(kT_{\rm in})}^{h\nu/(kT_{\rm out})}\frac{x^{5/3}}{e^x-1}\,dx.
$$
The lower limit is much smaller than one and the upper limit much larger than one. Extending them to zero and infinity leaves a constant: near zero the integrand behaves as $x^{2/3}$, and at infinity it decays exponentially. Hence \b[the intermediate spectrum] is
$$
\boxed{F_\nu\propto\nu^{1/3}.}
$$
This is the <one-third spectrum of a multitemperature disk>. Much of the emission comes from radii where $kT(r)$ is of order $h\nu$, moving inward as the <frequency> rises. With the exact inner-edge profile, a broad intermediate interval also requires <frequency> well below $kT_{\max}/h$; the supplied approximation captures its slope away from the hottest annuli.