 Solution

Specify the path convention before integrating the neutral-hydrogen column-density distribution. Write $f_z={\partial }^2\mathcal{N}/(\partial N\partial z)$ and $f_X={\partial }^2\mathcal{N}/(\partial N\partial X)$, with absorption distance satisfyingThe lower panel of the preceding sketch shows the approximate decrease of $f$ with $N=N_{\mathrm{HI}}$ near $z=3$. Ordinary forest columns are roughly described by $f\propto N^{-\beta }$, $\beta \sim 1.5$, but the slope changes through partially shielded Lyman limit systems and high-column damped Lyman-alpha systems. No single slope should be extrapolated through all these regimes. In logarithmic bins the number is $N\ln (10)f$, whereas their neutral mass contribution is proportional to $N^{2}f$; confusing these two weightings would give a wrong baryon inventory.

To derive the neutral gas density from an absorption distribution, a redshift interval $dz$ has proper light-path length $d\ell =cdz/[(1+z)H]$. The sum of neutral hydrogen columns per path length estimates the mean proper number density:The proper mass density is $m_H{\overline{n}}_{\mathrm{HI}}$; division by $(1+z{)}^3$ converts it to a comoving mass density. Thus, with today's critical density ${\rho }_{\mathrm{crit},0}=3H_0^2/(8\pi G)$,A representative finite-sample estimator replaces the last integral by ${\sum }_j{N}_j/\Delta X$, with completeness and selection corrections. This defines a comoving inventory divided by today's critical density; a fraction of the instantaneous critical density has a different normalization. Neutral gas including accompanying helium is obtained by multiplying by a specified mass correction, approximately $1.3$; hydrogen-only ${\Omega }_{\mathrm{HI}}$ must not silently include it.

At $z\sim 3$ the observed neutral inventory is of order $10^{-3}$, far below the cosmological baryon density of order $0.04$–$0.05$ for $h\sim 0.7$. High-column damped Lyman-alpha systems contribute most of this neutral mass although weak forest lines dominate the counts; for $\beta =1.5$ the mass per logarithmic column interval increases as $N^{1/2}$ in the forest regime. The small neutral inventory therefore does not imply a small total gas inventory: the intergalactic medium contains a large ionized baryon reservoir invisible to a neutral-only count. Recovering that reservoir requires an ionization correction and a gas-density model. The survey normalization and high-column contribution are described in the neutral-gas inventory analysis.