Show that the ground-state energy \(E_{0}\) of a relativistic gas of electrons is given by

\[E_{0}=\frac{\pi V m^{4} c^{5}}{3 h^{3}} B(x)\]

where

\[B(x)=8 x^{3}\left\{\left(x^{2}+1\right)^{1 / 2}-1\right\}-A(x),\]

\(A(x)\) and \(x\) being given by equations (8.5.13) and (8.5.14). Check that the foregoing result for \(E_{0}\) and equation (8.5.12) for \(P_{0}\) satisfy the thermodynamic relations

\[E_{0}+P_{0} V=N \mu_{0} \quad \text { and } \quad P_{0}=-\left(\partial E_{0} / \partial V\right)_{N}\]