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> gives
$$
\boxed{A_\nu(\nu)\sim\int_0^\infty e^{-\nu t^3/6}\,dt=\frac{6^{1/3}}{3\nu^{1/3}}\Gamma(1/3).}
$$
The coefficient is a <Gamma function> value obtained by the <Gamma integral>.

Solution

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

Suppose a real phase has its unique minimum at the left endpoint and $\phi(t)-\phi(0)\sim ct^m$, with $c>0$ and integer $m\geq1$. For a continuous amplitude $a$ with $a(0)\ne0$, localization followed by $s=(\nu c)^{1/m}t$ gives
$$
\int_0^\infty a(t)e^{-\nu\phi(t)}\,dt\sim a(0)e^{-\nu\phi(0)}\frac{\Gamma(1/m)}{m(\nu c)^{1/m}}.
$$
This uses the <Gamma integral>. It assumes sufficient decay away from the minimum to make the remaining integral negligible. A cubic endpoint has width $\nu^{-1/3}$; an ordinary quadratic endpoint has width $\nu^{-1/2}$.

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 Incoming links: Degenerate endpoint in Laplace's method