= Solution
Let $m$ be the absorber <mass path> along a ray. If the <spectral line profile> is independent of position, <Beer-Lambert law> gives $\mathfrak T_\nu(m)=e^{-m\kappa_\nu}$. A finite-band mean and the line's <equivalent width> are respectively
$$
\boxed{\overline{\mathfrak T}_{\mathcal B}(m)=\frac1{\Delta\nu_{\mathcal B}}\int_{\mathcal B}e^{-mS\phi(\nu)}\,d\nu,
\qquad E(m)=\int_{-\infty}^{\infty}[1-e^{-mS\phi(\nu)}],d\nu.}
$$
When the band contains all significant <absorption>, $\int_{\mathcal B}\mathfrak T_\nu d\nu\simeq\Delta\nu_{\mathcal B}-E$. An unqualified integral of <Transmittance>[transmission] over the entire infinite line domain diverges because <Transmittance>[transmission] tends to one far from <Spectral line>[line centre]; $E$ is the finite and physically useful line deficit.
This is <nonexponential band transmission>: each monochromatic <Transmittance>[transmission] is exponential, but their average generally is not. Differentiating with respect to <mass path> gives
$$
\overline{\mathfrak T}'=-\langle\kappa e^{-m\kappa}\rangle_{\mathcal B}<0,
\qquad\overline{\mathfrak T}''=\langle\kappa^2e^{-m\kappa}\rangle_{\mathcal B}>0.
$$
More decisively, define the <absorption coefficient> averaged over the transmitted spectrum by
$$
a_{\rm eff}(m)=-\frac{d\log\overline{\mathfrak T}}{dm}
=\frac{\langle\kappa e^{-m\kappa}\rangle}{\langle e^{-m\kappa}\rangle}.
$$
Then $\boxed{a_{\rm eff}'(m)=-\operatorname{Var}_{m}(\kappa)\le0}$. The core saturates first and the surviving radiation increasingly samples weakly absorbing wings. The effective attenuation per additional mass therefore decreases. <Jensen's inequality> also gives $\overline{\mathfrak T}\ge e^{-m\langle\kappa\rangle}$, with equality for a spectrally constant coefficient. Thus the band <Transmittance>[transmission] decays more slowly than the exponential with its initial mean attenuation.
The corresponding <curve of growth> has $E(m)\simeq mS$ for an optically weak line, and sublinear growth after saturation. For the <Thermal Doppler broadening>[Gaussian profile], with $\tau_0=mS/(\sqrt\pi\Delta_D)\gg1$, <absorption> extends to $|\delta\nu|\simeq\Delta_D\sqrt{\log\tau_0}$, so $E\sim2\Delta_D\sqrt{\log\tau_0}$. For a strongly saturated pure <Lorentzian line shape>[Lorentz profile] the wings instead give $E\sim2\sqrt{mS\Delta_L}$. These asymptotes apply while the line remains isolated and its important wings remain within the range where the assumed profile and nearly constant <Planck function> are appropriate. The integrated <absorption> is not an exponential either: $E'>0$ and $E''\le0$.
\Image[/past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2017-line-saturation-and-transmission.png]
{title=Line saturation produces nonexponential band transmission}
{description=An original fixed Gaussian line example shows its opaque core expanding with absorber path. The mean transmission across a fixed band remains above the single exponential determined by its initial mean attenuation.}
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