= Solution
Choose the <square root> with $-\pi/2<\arg z<3\pi/2$, so the <branch cut> is the negative imaginary axis. The change of variable $u=e^{-i\pi/4}\sqrt z$ maps this sheet onto $\operatorname{Re}u>0$ and gives $z=iu^2$. Writing the <exponential function> as $e^{k\phi(z)}$, its phase becomes exactly quadratic:
$$
\phi(z)=-i\bigl(z-2e^{i\pi/4}\sqrt z\bigr)=(u-1)^2-1.
$$
The <saddle point> is therefore $u=1$, or \b[$z_s=i$]. On the descending line $u=1+iv$, $v\in\mathbb R$, the phase is $-1-v^2$. Its image is the <parabola>
$$
\boxed{z=-2v+i(1-v^2),\qquad\operatorname{Im}z=1-\frac{(\operatorname{Re}z)^2}{4}.}
$$
The contour runs from left to right, corresponding to $v$ decreasing from $+\infty$ to $-\infty$. A <contour deformation> in the right half of the $u$-plane moves the original indented contour to this line. The connecting tails vanish in the descending sectors, and no <branch point> or <pole> is crossed. Since $dz=-2(1+iv)\,dv$, the <Gaussian integral> and the <odd function> $ve^{-kv^2}$ give
$$
I=2e^{-k}\int_{-\infty}^{\infty}(1+iv)e^{-kv^2}\,dv
=\boxed{2\sqrt{\frac\pi k}\,e^{-k}}.
$$
Here the <method of steepest descent> actually gives an exact answer for $k>0$, because the quadratic phase and linear transformed amplitude have no further even correction.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-336-contours.png]
{title=Integration contours near a saddle and a pole}
{description=The original upper indentation, the saddle contour, and the negative imaginary branch cut. The indentation is exaggerated for visibility.}
Back to article page
