Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 336 / 1

There are two positive roots for $u>1$: the derivative of $x{e}^{1/x}$ is $e^{1/x}(1-1/x)$, so the function decreases to its minimum $e$ at $x=1$ and then increases. Negative $x$ cannot solve the equation. For the larger root, expansion of the exponential function in $1/x$ givesso series reversion yieldsFor the smaller root, set $y=1/x$ and take a logarithm. Then $y-\log y=u$, so successive substitution gives $y=u+\log u+O(\log u/u)$ andThese are the first two nonzero terms of the two asymptotic expansions. Equivalently, the roots are $-1/W_0(-e^{-u})$ and $-1/W_{-1}(-e^{-u})$, from the real branches of Lambert W.

The three fixed-$a$ estimates below cease to be uniform as the minimum of the phase approaches the endpoint. For the cubic endpoint-to-saddle transition, writeThe Taylor series of the hyperbolic sine gives, for bounded $s$ and fixed $l$,The cubic term controls the tail, so localization of the Laplace integral gives the uniform leading formulaHere $I$ is the cubic Laplace transition integral, equal to $\pi \mathrm{Hi}$ in terms of the Scorer Hi function.

To recover the endpoint regime, let $l\to +\infty$. Scale $s=v/(2^{1/3}l)$ in $I$; the cubic term becomes negligible and $I(-2^{1/3}l)\sim (2^{1/3}l{)}^{-1}$. Thereforewhich agrees with part (i) in the overlap $1\ll l\ll {\nu }^{2/3}$. At $l=0$, the Gamma integral gives $I(0)=3^{-2/3}\Gamma (1/3)$, recovering part (iii).

For $l\to -\infty$, set $M=-2^{1/3}l>0$. The exponent $Ms-s^3/3$ has its maximum at $s=\sqrt{M}$, with second derivative $-2\sqrt{M}$. Thus Laplace's method givesand the transition formula becomesFor $a=1-\eta$ with $\eta \downarrow 0$, part (ii) hasPutting $\eta =(-l){\nu }^{-2/3}$ reproduces both the exponential and its prefactor. For relative agreement of these leading exponentials, one may use the overlap $1\ll -l\ll {\nu }^{4/15}$, which makes $\nu {\eta }^{5/2}\to 0$. Thus the same transition integral connects all three regimes.

Let $S(t)=a\sinh t-t$. For $a>1$, $S^{\prime }(t)=a\cosh t-1>0$ on the integration interval, so its minimum is the endpoint $t=0$. The Taylor series isOn the contributing scale $t=O({\nu }^{-1})$, the higher terms are negligible. The endpoint form of Laplace's method, or the Watson lemma, therefore gives

For $0<a<1$, the phase has an interior minimum atThe inverse hyperbolic cosine selects the positive stationary point. At this point,Expanding the phase to second order and evaluating the resulting Gaussian integral by Laplace's method yieldsThe exponential grows because the minimum of $S$ is negative; the original integral remains convergent because $a\sinh t$ dominates $t$ at infinity.

At $a=1$, the endpoint is degenerate: $S(t)=\sinh t-t=t^3/6+O(t^5)$. The contributing width is therefore $t=O({\nu }^{-1/3})$, rather than the width in part (i). With $v=\nu t^3/6$, the formula for a degenerate endpoint in Laplace's method givesThe coefficient is a Gamma function value obtained by the Gamma integral.