By eqns. (8.1.4, 5 and 24 ), the Fermi energy \(\varepsilon_{F}\) is given by

\[
\varepsilon_{F}=\left\{\frac{3}{4 \pi} f_{3 / 2}(z)ight\}^{2 / 3} \frac{h^{2}}{2 m \lambda^{2}}=\left\{\frac{3 \pi^{1 / 2}}{4} f_{3 / 2}(z)ight\}^{2 / 3} k T
\]

With the help of Sommerfeld's lemma (E.17), this becomes

\[
\begin{align*}
\varepsilon_{F} & =k T \ln z\left\{1+\frac{\pi^{2}}{8}(\ln z)^{-2}+\frac{7 \pi^{4}}{640}(\ln z)^{-4}+\ldotsight\}^{2 / 3} \\
& =k T \ln z\left\{1+\frac{\pi^{2}}{12}(\ln z)^{-2}+\frac{\pi^{4}}{180}(\ln z)^{-4}+\ldotsight\} \tag{1}
\end{align*}
\]

To invert this series, we write

\[
\begin{equation*}
k T \ln z \equiv \mu=\varepsilon_{F}\left\{1+a_{2}\left(\frac{k T}{\varepsilon_{F}}ight)^{2}+a_{4}\left(\frac{k T}{\varepsilon_{F}}ight)^{4}+\ldotsight\} \tag{2}
\end{equation*}
\]

and substitute into (1), to get

\(1-a_{2}\left(\frac{k T}{\varepsilon_{F}}ight)^{2}+\left(a_{2}^{2}-a_{4}ight)\left(\frac{k T}{\varepsilon_{F}}ight)^{4}+\ldots=1+\frac{\pi^{2}}{12}\left(\frac{k T}{\varepsilon_{F}}ight)^{2}+\left(\frac{\pi^{4}}{180}-\frac{\pi^{2}}{6} a_{2}ight)\left(\frac{k T}{\varepsilon_{F}}ight)^{4}+\ldots\)

Equating coefficients on the two sides of this equality, we get: \(a_{2}=\) \(-\pi^{2} / 12, a_{4}=-\pi^{4} / 80, \ldots\) Equation (2) then gives the desired result (8.1.35a).

Next, we have from eqns. (8.1.7) and (E.17)

\[
\begin{align*}
\frac{U}{N}= & \frac{3}{5} k T \ln z\left\{1+\frac{5 \pi^{2}}{8}(\ln z)^{-2}-\frac{7 \pi^{4}}{384}(\ln z)^{-4}+\ldotsight\} \\
& \left\{1+\frac{\pi^{2}}{8}(\ln z)^{-2}+\frac{7 \pi^{4}}{640}(\ln z)^{-4}+\ldotsight\} \\
& =\frac{3}{5} k T \ln z\left\{1+\frac{\pi^{2}}{2}(\ln z)^{-2}-\frac{11 \pi^{4}}{120}(\ln z)^{-4}+\ldotsight\} . \tag{3}
\end{align*}
\]

Substituting from eqn. (8.1.35a) into (3), we get

\[
\begin{gather*}
\frac{U}{N}=\frac{3}{5} \varepsilon_{F}\left\{1-\frac{\pi^{2}}{12}\left(\frac{k T}{\varepsilon_{F}}ight)^{2}-\frac{\pi^{4}}{80}\left(\frac{k T}{\varepsilon_{F}}ight)^{4}+\ldots+ight\} \\
\left\{1+\frac{\pi^{2}}{2}\left(\frac{k T}{\varepsilon_{F}}ight)^{2}-\frac{\pi^{4}}{120}\left(\frac{k T}{\varepsilon_{F}}ight)^{4}+\ldotsight\} \\
=\frac{3}{5} \varepsilon_{F}\left\{1+\frac{5 \pi^{2}}{12}\left(\frac{k T}{\varepsilon_{F}}ight)^{2}-\frac{\pi^{4}}{16}\left(\frac{k T}{\varepsilon_{F}}ight)^{4}+\ldotsight\} . \tag{4}
\end{gather*}
\]

The specific heat of the gas is then given by

\[
\begin{equation*}
\frac{C_{\mathrm{V}}}{N k}=\frac{\pi^{2}}{2} \frac{k T}{\varepsilon_{F}}-\frac{3 \pi^{4}}{20}\left(\frac{k T}{\varepsilon_{F}}ight)^{3}+\ldots \tag{5}
\end{equation*}
\]

We note that the ratio of the \(T^{3}\)-term here to the Debye expression (7.3.23) is \((1 / 16)\left(\Theta_{D} / T_{F}ight)^{3}\). For a typical metal, this is \(O\left(10^{-8}-10^{-9}ight)\).