= Solution
Write the phase and amplitude as
$$
\phi(z)=\cosh z,
\qquad
f(z)=\frac1{z^2+16}.
$$
The <saddle point> relevant to the contour is $z=0$, since $\phi'(z)=\sinh z$ and
$$
\phi(0)=1,
\qquad
\phi''(0)=1.
$$
The local <method of steepest descent>[steepest-descent] direction is therefore vertical: with $z=iy$,
$$
\phi(iy)=\cos y=1-\frac{y^2}{2}+O(y^4).
$$
The given contour approaches the lines $\operatorname{Im}z=-\pi$ at its left end and $\operatorname{Im}z=\pi$ at its right end. The poles of $f$ are at $z=\pm4i$, outside the closed strip $|\operatorname{Im}z|\leq\pi$. 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,
$$
\cosh(p\pm i\pi)=-\cosh p,
$$
so those two integrals are exponentially small, of order $e^{-x}x^{-1/2}$.
The vertical segment, oriented upward, contributes
$$
i\int_{-\pi}^{\pi}\frac{e^{x\cos y}}{16-y^2}\,dy.
$$
Its real phase has a unique maximum at $y=0$. Since
$$
\frac1{16-y^2}=\frac1{16}+O(y^2),
$$
the <simple-saddle contribution in steepest descent>, equivalently the local <Gaussian integral>, gives
$$
I(x)
\sim \frac{i e^x}{16}
\int_{-\infty}^{\infty}e^{-xy^2/2}\,dy
=\boxed{\frac{i e^x}{16}\sqrt{\frac{2\pi}{x}}}
\qquad(x\to\infty).
$$
The factor $i$ comes from the upward tangent $dz=i\,dy$ of the deformed contour.
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