= Solution
Use <absorption distance> $X$ and its distribution $f_X$. Summing hydrogen columns per unit $X$ gives $\int N f_X\,dN$. The physical light-path length per $X$ is
$$
\frac{d\ell}{dX}=\frac{c}{H_0(1+z)^3}.
$$
The mean neutral-hydrogen <number density> is therefore $(H_0/c)(1+z)^3\int N f_X\,dN$. Divide by $(1+z)^3$ to obtain its comoving <density>, multiply by hydrogen atomic mass, and divide by today's mass <critical density> $\rho_{\rm crit,0}=3H_0^2/(8\pi G)$. This derives the <neutral gas density from an absorption distribution>:
$$
\boxed{\Omega_{\rm HI}(z)=\frac{H_0m_H}{c\rho_{\rm crit,0}}\int N f_X(N,z)\,dN.}
$$
If “neutral gas” includes <helium>, replace $m_H$ by $\mu_Hm_H$, with a stated abundance correction approximately $\mu_H\simeq1.3$. Equivalently, for the distribution per redshift the prefactor inside the integral is $m_H H(z)/[c\rho_{\rm crit,0}(1+z)^2]$ multiplying $Nf_z$. In the PDF's energy-density convention, both the mass <density> and <critical density> are multiplied by $c^2$, leaving their ratio unchanged. The usual quoted abundance uses comoving mass <density> relative to today's <critical density>; a ratio of physical <density> to the epoch's <critical density> instead multiplies this answer by $(1+z)^3H_0^2/H(z)^2$.
To infer total gas, one needs the neutral fraction $x_{\rm HI}=N_{\rm HI}/N_H$ for each absorber. The <ionized gas density from an absorption distribution> is
$$
\boxed{\Omega_g(z)=\frac{H_0\mu_Hm_H}{c\rho_{\rm crit,0}}
\int\frac{N}{x_{\rm HI}(N,z)}f_X(N,z)\,dN.}
$$
This is an ionization-corrected integral, not an inference available from the observed columns alone. For optically thin, almost fully ionized hydrogen, <photoionization equilibrium> gives $\Gamma n_{\rm HI}=\alpha(T)n_en_{\rm HII}\simeq\alpha(T)n_H^2$. If the absorbing depth $L$ is specified, $N=\alpha(T)n_H^2L/\Gamma$ and
$$
N_H=n_HL=\left(\frac{\Gamma NL}{\alpha(T)}\right)^{1/2}.
$$
Thus <photoionization rate>, <temperature> and depth are all needed.
One useful physical closure sets $L$ to a local <Jeans length>, $L\propto(f_gT/n_H)^{1/2}$, with local gas fraction $f_g$. Solving $N\propto[\alpha(T)/\Gamma](f_gT)^{1/2}n_H^{3/2}$ then gives the <Jeans-length ionization correction for the Lyman-alpha forest>:
$$
N_H\propto N^{1/3}\Gamma^{1/3}\left(\frac{f_gT}{\alpha(T)}\right)^{1/3},\qquad
\Omega_g\propto\Gamma^{1/3}\left(\frac{f_gT}{\alpha(T)}\right)^{1/3}\int N^{1/3}f_X\,dN.
$$
This illustrates why weak, highly ionized forest lines can dominate the total gas inventory even though <damped Lyman-alpha systems> dominate the neutral-column-weighted inventory. It is a model for locally confined optically thin gas; optically thick <Lyman limit systems> and <damped Lyman-alpha systems> require a separate <radiative transfer> treatment. The physical basis and observational application of this closure are developed in https://arxiv.org/abs/astro-ph/0104272[Schaye's absorption-based matter-density calculation].
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