 Solution

For $t=u+iv$, the exponent isIts imaginary part must remain equal to its value $x$ at the saddle $t=0$. The descending branch connecting the required ends is thereforeAlong it $\sin v=\tanh u$, so the real part is $-x(\cosh u-\mathrm{sech}u)$, strictly negative away from zero for $x>0$. The branch approaches $v=\pm \pi /2$ at the two infinities and has tangent $t=e^{i\pi /4}s$ at the saddle, oriented towards increasing $s$.

Locally $\cosh t=1+t^2/2+O(t^4)$, hence the steepest descent method givesOutside a small saddle neighborhood the descending contour has strictly smaller real exponent; inside it the Gaussian rescaling $s=x^{-1/2}r$ controls the higher terms. Multiplication by $1/(i\pi )$ and taking the real part yieldThis additive asymptotic form specifies the leading term even near the zeros of the oscillating cosine, where a literal pointwise ratio interpretation of the printed $\sim$ is unsuitable.