 Solution

Write the phase and amplitude asThe saddle point relevant to the contour is $z=0$, since ${\varphi }^{\prime }(z)=\sinh z$ andThe local steepest-descent direction is therefore vertical: with $z=iy$,

The given contour approaches the lines $\mathrm{Im}z=-\pi$ at its left end and $\mathrm{Im}z=\pi$ at its right end. The poles of $f$ are at $z=\pm 4i$, outside the closed strip $|\mathrm{Im}z|\le \pi$. The Cauchy integral theorem therefore permits deformation to the contour consisting of the lower horizontal ray from $-\infty -i\pi$ to $-i\pi$, the vertical segment from $-i\pi$ to $i\pi$, and the upper horizontal ray from $i\pi$ to $\infty +i\pi$. On either horizontal ray,so those two integrals are exponentially small, of order $e^{-x}{x}^{-1/2}$.

The vertical segment, oriented upward, contributesIts real phase has a unique maximum at $y=0$. Sincethe simple-saddle contribution in steepest descent, equivalently the local Gaussian integral, givesThe factor $i$ comes from the upward tangent $dz=idy$ of the deformed contour.