The number density of fermions of mass $m$ at equilibrium in the early universe with temperature $T$, is given by the integral

$$n=\frac{4 \pi}{h^{3}} \int_{0}^{\infty} \frac{p^{2} d p}{\exp [(\mathcal{E}(p)-\mu) / k T]+1}$$

where $\mathcal{E}(p)=c \sqrt{p^{2}+m^{2} c^{2}}$, and $\mu$ is the chemical potential. Assuming that the fermions remain in equilibrium when they become non-relativistic $\left(k T, \mu \ll m c^{2}\right)$, show that the number density can be expressed as

$$n=\left(\frac{2 \pi m k T}{h^{2}}\right)^{3 / 2} \exp \left[\left(\mu-m c^{2}\right) / k T\right]$$

[Hint: You may assume $\left.\int_{0}^{\infty} d x e^{-\sigma^{2} x^{2}}=\sqrt{\pi} /(2 \sigma), \quad(\sigma>0) .\right]$

Suppose that the fermions decouple at a temperature given by $k T=m c^{2} / \alpha$ where $\alpha \gg 1$. Assume also that $\mu=0$. By comparing with the photon number density at $n_{\gamma}=16 \pi \zeta(3)(k T / h c)^{3}$, where $\zeta(3)=\sum_{n=1}^{\infty} n^{-3}=1.202 \ldots$, show that the ratio of number densities at decoupling is given by

$$\frac{n}{n_{\gamma}}=\frac{\sqrt{2 \pi}}{8 \zeta(3)} \alpha^{3 / 2} e^{-\alpha}$$

Now assume that $\alpha \approx 20$, (which implies $n / n_{\gamma} \approx 5 \times 10^{-8}$ ), and that the fermion mass $m=m_{p} / 20$, where $m_{p}$ is the proton mass. Explain clearly why this new fermion would be a good candidate for solving the dark matter problem of the standard cosmology.