In the ground state of a Fermi system, the chemical potential is identical to the Fermi energy: \((\mu)_{T=0}=\varepsilon\left(p_{F}\right)\). Making use of the energy spectrum \(\varepsilon(p)\) of the previous problem, we obtain

\[(\mu)_{T=0} \simeq \frac{p_{F}^{2}}{2 m}\left[1+\frac{4}{3 \pi}\left(k_{F} a\right)+\frac{4}{15 \pi^{2}}(11-2 \ln 2)\left(k_{F} a\right)^{2}\right]\]

Integrating this result, rederive equation (11.7.31) for the ground-state energy of the system.