Expanding the <relic-neutrino energy density> at $T_\nu/a\ll m_\nu$ gives
$$
\langle v^2\rangle^{1/2}\simeq
\sqrt{\frac{15\zeta(5)}{\zeta(3)}}\frac{T_\nu}{am_\nu}.
$$
The thermal speed decreases as $a^{-1}$ after the neutrinos become nonrelativistic.

Late-time relic-neutrino speed

For one species with two internal states, the <relic-neutrino energy density> obtained from the frozen <Fermi-Dirac distribution> is
$$
\rho_\nu=2\int\frac{d^3p}{(2\pi)^3}
\frac{\sqrt{p^2+m_\nu^2}}{e^{ap/T_\nu}+1}
=\frac1{\pi^2}\int_0^\infty
\frac{p^2\sqrt{p^2+m_\nu^2}}{e^{ap/T_\nu}+1}\,dp.
$$
In the relativistic limit, set $x=ap/T_\nu$ and use the standard <Riemann zeta function> integral
$$
\int_0^\infty\frac{x^3}{e^x+1}\,dx=\frac{7\pi^4}{120}.
$$
This gives
$$
\boxed{\rho_\nu=\frac{7\pi^2}{120}\left(\frac{T_\nu}{a}\right)^4}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 3 / b / Solution

For one relic-neutrino species with two internal states, physical momentum $p$, comoving temperature $T_\nu$, and mass $m_\nu$,
$$
\rho_\nu=\frac1{\pi^2}\int_0^\infty
\frac{p^2\sqrt{p^2+m_\nu^2}}{e^{ap/T_\nu}+1}\,dp.
$$
It approaches $7\pi^2(T_\nu/a)^4/120$ while relativistic and $n_\nu m_\nu$ after becoming nonrelativistic.


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