Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 31B (Asymptotic Methods)

Put $t=u+iv$ and $\Phi (t)=\sinh t-t$. Sincea steepest-descent curve through the saddle point $t=0$ lies inThe saddle is cubic rather than simple:Thus the curves meet the origin in the three descent directions for which $t^3$ is negative real,One curve is the negative real axis $v=0$, $u<0$. The other two are conjugate curves $v=\pm v_{+}(u)$ with $u>0$ and $0<v_{+}(u)<\pi$. Near the origin, $v_{+}(u)\sim \sqrt{3}u$, while for $u\to +\infty$ the defining equation gives $\sin v_{+}\sim 2v_{+}{e}^{-u}$, so $v_{+}\to \pi$. Their three asymptotes are therefore

Use the real integral representation of the modified Bessel function of the second kindThe unique stationary point is the strict maximumThe Laplace approximation is consequentlyFor fixed $x>0$, as $\nu \to +\infty$,Substitution gives the required large-order asymptotic of the modified Bessel function of the second kind: