Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2008 / A / 3 / i / Solution

Let $b=\pi /4$ and introduceDeform the contour upward to the steepest descent contour $t=e^{i\pi /4}r$ from zero and the endpoint path $t=(b^2+is{)}^{1/2}$ from $b$. More precisely the original integral is the first path traversed forward minus the endpoint path traversed forward. Connect the truncated paths where $\mathrm{Im}(t^2)=R$ and let $R\to \infty$. That connector has exponentially small factor $e^{-xR}$, and no pole of $\tan t$ is crossed: all its poles are real and outside the original segment.

On the first path ${\tan }^{2}t=t^2+O(t^4)$, so its leading contribution isThe last equality follows by $u=x{r}^2$, $\Gamma (3/2)=\sqrt{\pi }/2$.

On the endpoint path $dt=ids/(2\sqrt{b^2+is})$, and ${\tan }^{2}b=1$. Its negative orientation gives the leading endpoint termTaking real parts and inserting $b=\pi /4$ givesThe oscillatory first coefficient can vanish at particular values of $x$; the stated remainder still holds as an additive asymptotic expansion.