For $0<a<1$, the phase has an interior minimum at
$$
t_*=\operatorname{arcosh}(1/a),\qquad \sinh t_*=\frac{\sqrt{1-a^2}}a.
$$
The <inverse hyperbolic cosine> selects the positive <stationary point>. At this point,
$$
S(t_*)=\sqrt{1-a^2}-\operatorname{arcosh}(1/a),\qquad S''(t_*)=\sqrt{1-a^2}>0.
$$
Expanding the phase to second order and evaluating the resulting <Gaussian integral> by <Laplace's method> yields
$$
\boxed{A_\nu(a\nu)\sim\sqrt{\frac{2\pi}{\nu\sqrt{1-a^2}}}\exp\left\{\nu\left[\operatorname{arcosh}(1/a)-\sqrt{1-a^2}\right]\right\}.}
$$
The exponential grows because the minimum of $S$ is negative; the original integral remains convergent because $a\sinh t$ dominates $t$ at infinity.

Solution

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

{title2=$\operatorname{arcosh}$}

For $x\geq1$, the inverse <hyperbolic cosine> on its nonnegative real branch is $\operatorname{arcosh}x=\log(x+\sqrt{x^2-1})$.

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 Incoming links: Inverse hyperbolic cosine