The ground-state energy of a relativistic gas of electrons is given by

\[
E_{0}=\frac{8 \pi V}{h^{3}} \int_{0}^{p_{F}} m c^{2}\left[\left\{1+(p / m c)^{2} \right\}^{1 / 2}-1 \right] p^{2} d p
\]

Making the substitution (8.5.9), we get

\[
\begin{equation*}
E_{0}=\frac{8 \pi m^{4} c^{5} V}{h^{3}} \int_{0}^{\theta_{F}}(\cosh \theta-1) \sinh ^{2} \theta \cosh \theta d \theta \tag{1}
\end{equation*}
\]

Now the integral

\[
\begin{equation*}
\int_{0}^{\theta_{F}} \sinh ^{2} \theta \cosh ^{2} \theta d \theta=\left.\frac{1}{3} \sinh ^{3} \theta \cosh \theta \right|_{0} ^{\theta_{F}}-\frac{1}{3} \int_{0}^{\theta_{F}} \sinh ^{4} \theta d \theta . \tag{2}
\end{equation*}
\]

Substituting (2) into (1) and making use of eqn. (8.5.12), we get

\[
\begin{equation*}
E_{0}=\frac{8 \pi m^{4} c^{5} V}{3 h^{3}} \sinh ^{3} \theta_{F} \cosh \theta_{F}-P_{0} V-\frac{8 \pi m^{4} c^{5} V}{3 h^{3}} \sinh ^{3} \theta_{F} \tag{3}
\end{equation*}
\]

note that the last term is simply \(N m c^{2}\). Finally, using the definition \(x=\sinh \theta_{F}\), we obtain the desired result.

We observe that eqn. (3) can also be written as

\[
E_{0}+P_{0} V=N m c^{2}\left(\cosh \theta_{F}-1 \right)=N \varepsilon_{F} \equiv N \mu_{0}
\]

To verify that the derivative \(\left(\partial E_{0} / \partial V \right)_{N}\) is equal to \(-P_{0}\), we have to show that

\[
\begin{aligned}
& {[\partial\{V B(x)\} / \partial V]_{\left(V x^{3} \right)}=-A(x), \text { i.e. } \partial\left\{x^{-3} B(x) \right\} / \partial x^{-3}=-A(x) \text {, i.e. }} \\
& x^{4} \frac{\partial}{\partial x}\left[8\left\{\left(x^{2}+1 \right)^{1 / 2}-1 \right\}-x^{-3} A(x) \right]=3 A(x), \text { i.e. } \\
& \partial A(x) / \partial x=8 x^{4}\left(x^{2}+1 \right)^{-1 / 2},
\end{aligned}
\]

which can be readily verified with the help of expression (8.5.13).