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

At $T_{\nu }/a\ll m_{\nu }$, the Taylor expansion of the relativistic energy is $\sqrt{p^2+m_{\nu }^2}=m_{\nu }+p^2/(2m_{\nu })+O(p^4/m_{\nu }^3)$. The number density isThe ratio of the next momentum moment to this one isThereforeso $\boxed{F=15\zeta (5)/(2\zeta (3))\simeq 6.47}$.