 Solution

For fixed $0<\gamma <1$, the denominator's first squared term has no zero on the interval and is at least $(1-\gamma {)}^2$. Expand the integrand uniformly in ${\epsilon }^2$ and integrate:The asymptotic expansion is uniform for $\gamma$ in compact subsets below $1$, including the continuous limit $\gamma \to 0$. It is not uniform through $\gamma =1$, where a narrow endpoint peak forms. That transition occurs when $1-\gamma =O(\epsilon )$, as the next part shows.