Trying (and failing) to calculate baryon/photon ratio I'm reading through Modern Cosmology by Dodelson and Schmidt 2nd edition on my own, and at the start of Section 4.2 the book says that we can compute the baryon/photon ratio at the time of big bang nucleosynthesis and get
$$ \eta_{\text{b}} \equiv \frac{n_{\text{b}}}{n_{\gamma}} = 6.0\times 10^{-10} \left(\frac{\Omega_{\text{b}}h^{2}}{0.022}\right). $$
I'm trying to figure out how to get to this result.
Attempt
Note that all calculations are done in units having $\hbar = c = k_{B} = 1$.

*

*From (4.5) in the book, $n_{\gamma} = 2T^{3}/\pi^{2}$.

*I took the baryon number density to be $n_{\text{b}} \approx \rho_{\text{b}} / (\text{mass of proton})$.

*From (2.72) in the book, $\rho_{\text{b}} = \Omega_{\text{b}}\rho_{\text{cr}}a^{-3(1+w)}$ where I took $w=0$ for baryons.

*The critical density is
$$ \rho_{\text{cr}} = \frac{3H_{0}^{2}}{8\pi G} = \frac{3h^{2}}{8\pi G}(100\text{ km s$^{-1}$ Mpc$^{-1}$})^{2} = \frac{3h^{2}}{8\pi G}\frac{1}{(0.98\times 10^{10} \text{years})^{2}}. $$

*Lastly, I used the fact that $a\propto 1/T$ to get $a = T_{0}/T$ where $T_{0} = 2.73\text{ K}$ today.

Putting all this together, we get
\begin{align*}
\frac{n_{\text{b}}}{n_{\gamma}} &= \frac{\rho_{\text{b}}/m_{\text{p}}}{2T^{3}/\pi^{2}} = \frac{\pi^{2}\rho_{\text{b}}}{2T^{3}m_{\text{p}}} = \frac{\pi^{2}\Omega_{\text{b}}\rho_{\text{cr}}a^{-3}}{2T^{3}m_{\text{p}}} = \frac{\pi^{2}\Omega_{\text{b}}\rho_{\text{cr}}}{2T_{0}^{3}m_{\text{p}}} \\[1.2ex]
&= \frac{\pi^{2}\Omega_{\text{b}}}{2T_{0}^{3}m_{\text{p}}}\frac{3h^{2}}{8\pi G}\frac{1}{(0.98\times 10^{10} \text{years})^{2}} \\[1.2ex]
&= \Omega_{\text{b}}h^{2} \,\cdot\, \frac{\pi^{2}}{2T_{0}^{3}m_{\text{p}}}\frac{3}{8\pi G}\frac{1}{(0.98\times 10^{10} \text{years})^{2}}.
\end{align*}
Now plugging everything in, using WolframAlpha, and multiplying by whatever factors of $\hbar$, $c$, and $k_{B}$ are needed to get rid of the units, I get a factor of $\approx 3.26\times 10^{-8}$.
The factor should be $6.0\times 10^{-10}/0.022 \approx 2.7\times 10^{-8}$. This in the same order of magnitude, but it is clearly off. What did I do wrong? What is the right way to do the calculation?
 A: Ahh, I see what I did wrong. At the first bullet point, I invoked (4.5), which is an approximation. It involves an approximation of the distribution by a Boltzmann distribution for a classical dilute gas. This is not what we need for photons.
What we need instead is the Bose-Einstein distribution. In that case, we have
\begin{align*}
n_{\gamma} &= 2\int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{e^{p/T} - 1}.
\end{align*}
This is an integral over 3D momentum space. Switch to spherical coordinates whose volume element is $d^{3}p = p^{2}\sin\theta \, dp\,d\theta\, d\phi = p^{2}dp\, d^{2}\Omega$ to get
\begin{align*}
&= 2\int \frac{p^{2} dp\, d^{2}\Omega}{(2\pi)^{3}} \frac{1}{e^{p/T} - 1} \\[1.2em]
&= 2\int d^{2}\Omega \int_{0}^{\infty} \frac{p^{2} dp}{(2\pi)^{3}} \frac{1}{e^{p/T} - 1} \\[1.2em]
&= 2\cdot 4\pi \int_{0}^{\infty} \frac{p^{2} dp}{(2\pi)^{3}} \frac{1}{e^{p/T} - 1} \\[1.2em]
&= \frac{8\pi}{(2\pi)^{3}} \int_{0}^{\infty} dp\, \frac{p^{2}}{e^{p/T} - 1} \\[1.2em]
&= \frac{8\pi}{(2\pi)^{3}} T^{3} \int_{0}^{\infty} dx\, \frac{x^{2}}{e^{x} - 1} \\[1.2em]
&= \frac{T^{3}}{\pi^{2}} \cdot 2\zeta(3) \\[1.2em]
&= \frac{2T^{3}}{\pi^{2}}\cdot \zeta(3).
\end{align*}
We see that the new $n_{\gamma}$ has an extra factor of $\zeta(3)$ in it. In my calculation this extra factor of $\zeta(3)$ goes into the denominator. When I do this, WolframAlpha gives me
$$ 2.712\times 10^{-8} $$
as expected.
I'm not confident in my work (I am self-studying and I have no guidance), so if anyone (who knows their stuff better than me) is reading over this, please tell me whether or not my work here is solid. Thank you.
