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

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.

First make explicit the usual method of steepest descent hypotheses: the saddle point is simple, $G^{\prime }(z_0)=0$ and $g_2=G^{\prime \prime }(z_0)\ne 0$; a contour deformation leads to the two descent arms with one clockwise half-indentation around the pole; and endpoints or other contour portions contribute smaller terms. The sign of the exponential matters: $|e^{-i\lambda G}|=e^{\lambda \mathrm{Im}G}$, so decay is toward valleys of $\mathrm{Im}G$.

Use a local analytic function $w(z)$ as coordinate. Nondegeneracy allows an analytic square root after factoring the double zero of $G-G(z_0)$, so the coordinate can be chosen withThe sign of $c$ is selected by the orientation of the contour integral, with $w$ running from negative to positive real values. Matching cubic coefficients gives $d/c=-c{G}^{\prime \prime \prime }(z_0)/(6g_2)$. The transformed amplitude therefore has the Laurent seriesThe Cauchy principal value contribution of the pole on the two real arms is zero by oddness. Its clockwise semicircular indentation contributes $-i\pi F(z_0)$, since $\int dw/w=-i\pi$ there. The regular term contributes the Gaussian integral $B\sqrt{2\pi /\lambda }$. Consequently the simple saddle coinciding with a simple pole has the additive expansionThe remainder written here assumes the other contour contributions are negligible to that order. Otherwise the local result has to be supplemented by them. The next local correction is generically $O({\lambda }^{-1/2})$ times the displayed exponential. If $B=0$, it can vanish; odd Taylor terms integrate to zero between the symmetric arms. When $F(z_0)\ne 0$, the leading term gives the printed asymptotic equivalence. When $F(z_0)=0$, asymptotic equivalence to zero is not meaningful: the additive formula is still valid and a regular saddle-point approximation may lead instead.

Nondegeneracy is necessary and is not explicitly printed. For a counterexample to the unrestricted reading, take $F=1$, $z_0=0$ and $G(z)=-i{z}^4$. There is one critical point, but it is degenerate. Choose the contour from $+i\infty$ to $+\infty$, with a clockwise quarter-circle avoiding zero. Both rays are decaying valleys for $e^{-\lambda z^4}$. Their straight-ray contributions cancel, and the indentation gives $J=-i\pi /2$, not $-i\pi$. Thus the requested coefficient requires the simple saddle point and contour deformation assumptions above; the mere existence of one critical point is insufficient.