Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 336 / 2 / a

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.

The small denominator lies near $x=0$ when $\gamma$ approaches $1$. Its two contributions balance when $\gamma -1=O(\epsilon )$ and $x=O(\epsilon )$. Letwith fixed real $\eta$. Then $1-\gamma +\gamma x=\epsilon (s-\eta )+O({\epsilon }^2)$ and $\log (2-x)=L+O(\epsilon s)$ in the contributing region. The endpoint transition of a regularized interior pole givesIndeed, the leading scaled integral is ${\epsilon }^{-1}{\int }_0^{\infty }[(s-\eta {)}^2+L^2{]}^{-1}ds$. The upper endpoint can be sent to infinity at leading order because the omitted tail contributes only $O(1)$ to the original integral. At $\gamma =1$ the result is $\pi /(2\epsilon \log 2)$: only one half of the narrow peak lies in the domain.

This formula provides the required matching. For $\eta \to -\infty$, $\pi /2+\arctan (\eta /L)\sim -L/\eta$, giving $I\sim 1/(1-\gamma )$. For $\eta \to +\infty$ it tends to $\pi$, giving $I\sim \pi /(\epsilon \log 2)$, the $\gamma \downarrow 1$ limit of the interior-peak approximation. These are overlap regions with $\epsilon \ll |\gamma -1|\ll 1$, not assertions that the fixed-$\eta$ remainder is uniform to arbitrarily large $\eta$.

For fixed $\gamma >1$, the linear term $1-\gamma +\gamma x$ vanishes at $x_{\ast }=1-1/\gamma$. Put $L_{\ast }=\log (1+1/\gamma )>0$ and resolve its neighborhood using $x=x_{\ast }+(\epsilon L_{\ast }/\gamma )s$. The integral is dominated by the resulting narrow peak:The local Taylor expansion of the logarithm changes it by a vanishing relative amount on this width, while both endpoints are many peak widths away for fixed $\gamma >1$.

There is no additional breakdown when $\gamma \gg 1$: the peak approaches $x=1$, but its distance from that endpoint is $d=1/\gamma$, whereas its width is $w=\epsilon L_{\ast }/\gamma \sim \epsilon /{\gamma }^2$. Thus $w/d=\epsilon L_{\ast }\sim \epsilon /\gamma \ll 1$. The relative logarithm variation over a width is $O(\epsilon /\gamma )$, also improving as $\gamma$ grows. For $\gamma \ge 1+c$ with fixed $c>0$ these separation conditions hold uniformly, including arbitrarily large $\gamma$. In particularThe only endpoint transition requiring a new distinguished limit is the earlier one near $\gamma =1$, where the peak approaches $x=0$. Moving toward an endpoint is not by itself a reason for nonuniformity; the peak width must be compared with its endpoint distance.