= Solution
{c}
For $\epsilon\to0^+$, the <Taylor expansions> needed in the exact <energy density> are
$$
\sqrt{4+\epsilon^2}=2+\frac{\epsilon^2}{4}+O(\epsilon^4),\qquad
\ln\frac{2+\sqrt{4+\epsilon^2}}{\epsilon}
=\ln\frac4\epsilon+O(\epsilon^2).
$$
Substitution yields
$$
\begin{aligned}
\mathcal U(\epsilon)
&=\frac{E_{\rm ZB}}a\left[\frac23-\frac{\epsilon^2}{4}
-\frac{\epsilon^2}{2}\ln\frac4\epsilon+O(\epsilon^4)\right]\\
&=\frac{E_{\rm ZB}}a\left[\frac23-\frac{\epsilon^2}{4}
\left(1-2\ln\frac\epsilon4\right)+O(\epsilon^4)\right].
\end{aligned}
$$
Therefore
$$
\boxed{\alpha=1,\qquad\beta=2.}
$$
Here <Big O notation> describes the neglected small-coupling terms, and the <logarithm> is natural. The nonanalytic $\epsilon^2\ln\epsilon$ term comes from the formerly degenerate <quantum states> near the <Brillouin zone> boundary; a uniform nondegenerate <perturbation theory> expansion of the integrand would fail there. This is the same mechanism as the <Peierls energy gain>.
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