Solution
= Solution
Let $S(t)=a\sinh t-t$. For $a>1$, $S'(t)=a\cosh t-1>0$ on the integration interval, so its minimum is the endpoint $t=0$. The <Taylor series> is
$$
S(t)=(a-1)t+\frac a6t^3+O(t^5).
$$
On 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
$$
\boxed{A_\nu(a\nu)\sim\int_0^\infty e^{-\nu(a-1)t}\,dt=\frac1{\nu(a-1)}.}
$$