 Solution

The exponent $\Phi (t)=\frac{i}{2}(t-t^{-1})$ has saddles $t=\pm i$, with $\Phi (-i)=1$ and $\Phi (i)=-1$. Since the integrand is analytic off the origin, deform the contour to the unit circle and rotate $t=-i{e}^{i\theta }$. This gives the exact representationThe dominant saddle at $\theta =0$ is a steepest-descent direction for the method of steepest descent: $\cos \theta =1-{\theta }^2/2+{\theta }^4/24-\cdots$. Set $\theta =s/\sqrt{x}$ and expand the analytic amplitude against $e^{-s^2/2}$. Integrating its Gaussian moments gives the leading factor $e^x/\sqrt{2\pi x}$; the rest of the contour is exponentially smaller than this dominant contribution outside any fixed neighbourhood of the maximum.

To obtain every coefficient efficiently, the integral satisfies $x^2{I}_0^{\prime \prime }+x{I}_0^{\prime }-x^2{I}_0=0$, by differentiating and integrating the derivative of $\sin \theta e^{x\cos \theta }$. Substitute $I_0=e^x{x}^{-1/2}\sum a_n{x}^{-n}/\sqrt{2\pi }$ to get $a_0=1$ and $a_n=(2n-1{)}^2{a}_{n-1}/(8n)$. Therefore the full asymptotic expansion isHere $(-1)!!=1$. For each fixed truncation at $n=N$, the relative remainder is $O(x^{-N-1})$ as $x\to +\infty$. The infinite series is generally divergent, and this all-orders expansion does not purport to resolve exponentially small saddle contributions or their Stokes phenomenon.