Maintaining local thermodynamic equilibrium (LTE) in radiating gas with a broad atomic transition line Definitions / Background
In LTE, Kirchoff's law for radiation holds:
$$ \frac{j_{\nu}}{\alpha_{\nu}} = B_{\nu} (T) $$
where $j_{\nu}$ is the specific radiative emissivity, $\alpha_{\nu}$ is the monochromatic radiative absorption, and $B_{\nu} (T)$ is the Planck function evaluated at the temperature $T$. 
Consider a gas of two-level atoms with energies $E_u$ and $E_l$, with $E_u > E_l$, statistical weights $g_u$ and $g_l$, and number densities $n_u$ and $n_l$. The transition between these states has Einstein coefficients $A_{ul}$, $B_{ul}$ and $B_{lu}$ that we can use to write the emissivity and absoprtion of the transition:
$$ j_{\nu} = \frac{h \nu}{4 \pi}  n_u A_{ul} \psi({\nu})$$
$$ \alpha_{\nu} = \frac{h \nu}{4 \pi} [ n_l B_{lu} \phi({\nu}) - n_u B_{ul} \chi({\nu})]$$
where $\psi$, $\phi$, and $\chi$ are line profile functions accounting for line-broadening mechanisms such as thermal motion. 
Then, making use of the standard relations between the Einstein coefficients, we have 
$$ \frac{j_{\nu}}{\alpha_{\nu}} = \frac{2 h \nu^3}{c^2} \frac{\frac{\psi}{\phi}}{\frac{g_u n_l}{g_l n_u} - \frac{\chi}{\phi}}$$
Question
I want to understand how the right-hand side of the last equation simplifies to the Planck function under conditions of LTE, over the entire width of the line, without assuming that the line is narrow.
A standard discussion of this topic, as found in e.g. the astrophysics textbook by Rybicki and Lightman, does not seem to achieve this. In my reading, they proceed by making the following observations and/or assumptions:
1) There is complete redistribution of frequency between absorption and all types of emission, so that $\psi = \chi = \phi$.
2) In LTE at temperature T, the fraction $ g_u n_l / (g_l n_u)$ is equal to $\exp[h \nu_0/(kT)]$, where $\nu_0 = (E_u - E_l)/h$. 
If those are true, then we have
$$ \frac{j_{\nu}}{\alpha_{\nu}} = \frac{2 h \nu^3}{c^2} \frac{1}{\exp[h \nu_0/(kT)] - 1}$$
But this does not have the correct exponential term in the denominator to match the Planck function, because $\nu_0$ is constant (frequency-independent). Or from another perspective, the problem is that $g_u n_l / (g_l n_u)$ is frequency-independent.
So, what is going on? Do we have to break either of the assumptions/observations 1 or 2 above? If so, how? If not, then does Kirchoff's law simply not apply on a frequency-by-frequency basis over the width of a broad line, although it still might apply in a line-averaged sense? Is there some other possibility or detail I have overlooked?
 A: Let's suppose the broadening mechanism is van der Waals or Stark broadening - something where the energy levels of individual atoms are perturbed.
In this case you could use the following argument.
Divide the line profile up into groups of atoms which share the same perturbation and treat each of these as a subpopulation with a different energy gap and hence a perturbed $\nu_0^{\prime}$. You can go through the usual argument involving the relationship between the Einstein coefficients and populate the two levels according to the Boltzmann factor, but with $h\nu_0^{\prime}$ in the exponential argument. At the end of this you find the source function for each subpopulation follows the Planck function at the same temperature $T$. Hence LTE means that the source function equals the Planck function at each frequency.
A: I'm not sure if this explanation is right, so please feel free to point out inconsistencies.
A gas with two energy levels absorbs a range of frequencies centered around $\nu_{0}$ because the atoms of the gas are in motion, making the photons appear appropriately doppler-shifted in their frame of reference. 
In the laboratory frame, where the atoms of the gas are whizzing around, one could interpret this as a manifestation of the spread in energy levels of the gas. Therefore, any gas has fundamentally more than two energy levels. The relationship between Einstein's coefficients of any pair of energy levels (that are separated by a frequency $\nu$) is still:
\begin{equation}
\frac{A_{21}}{B_{21}} = \frac{8\pi h\nu^{3}}{c^{3}} \\
\frac{B_{21}}{B_{12}} = \frac{g_{1}}{g_{2}}
\end{equation}
Note that thermal spreading wouldn't disturb the degeneracies. This spread in energy levels has consequences, as you pointed out in the comments, for the Boltzmann factor. For every frequency difference $\nu$, the Boltzmann factor would now be:
\begin{equation}
\frac{n_{u}}{n_{l}} = e^{-h\nu/KT}
\end{equation}
Therefore, in principle, you could derive the source function for each frequency independently. It would then collectively hold that Kirchoff's law is applicable to any gas in LTE, regardless of what physical mechanisms contributed how much towards line spreading. 
Of course, the following assumption is still utilized for all transition frequencies:
\begin{equation}
\phi(\nu) = \chi(\nu) = \psi(\nu)
\end{equation}
These line profile functions represent three very specific processes: Spontaneous Emission, Absorption, Stimulated Emission. These processes are ideally centered around a specific frequency $\nu_{0}$ so that I would expect thermal broadening to affect them equally. 
