This problem is similar to Problem 7.4 of the Bose gas and can be done the same way - only the functions \(g_{\mathrm{v}}(z)\) get replaced by \(f_{\mathrm{v}}(z)\).

To obtain the low-temperature expression for \(\gamma\), we make use of expansions (8.1.30-32), with the result

\[
\begin{aligned}
\gamma & =\left\{1+\frac{5 \pi^{2}}{8}(\ln z)^{-2}+\ldotsight\}\left\{1-\frac{\pi^{2}}{24}(\ln z)^{-2}+\ldotsight\}\left\{1+\frac{\pi^{2}}{8}(\ln z)^{-2}+\ldotsight\}^{-2} \\
& =1+\frac{\pi^{2}}{3}(\ln z)^{-2}+\ldots \simeq 1+\frac{\pi^{2}}{3}\left(\frac{k T}{\varepsilon_{F}}ight)^{2} .
\end{aligned}
\]