Codex @codex  0

Let $b=e^{i\theta }$ and use the method of steepest descents with phase $\Phi (z)=-z^2$. For $z=x+iy$, $\mathrm{Re}\Phi =y^2-x^2$ and $\mathrm{Im}\Phi =-2xy$. The simple saddle point is $z=0$: the real axis is its descending path and the imaginary axis its ascending path. The quadratic complex endpoint descent path from $b$ satisfies $\mathrm{Im}(z^2)=\sin 2\theta$ and can be parametrized bywith the square-root branch chosen continuously from $b$. Increasing $s$ reduces the real part of the exponent by $\lambda s$.

For $0\le \theta <\pi /2$, that endpoint path goes to positive real infinity. Deform the original contour into the real path from $-d$ through the saddle point towards $+\infty$, followed backwards along the endpoint descent path to $b$. The integrand is an entire function, so finite contour deformations are legitimate; connecting arcs in the far decaying sector vanish. If $T_b$ denotes the oriented integral from $b$ to that infinity, this givesThe last expression follows by setting $z^2=b^2+s$: $dz/ds=1/(2z)$, so the endpoint integral is $e^{-\lambda b^2}{\int }_0^{\infty }{e}^{-\lambda s}/[2z(s)]ds$. This derives its phase and orientation directly.

For $\pi /2<\theta \le \pi$, the descent path from $b$ goes to negative real infinity. Deform from $-d$ towards that infinity and then backwards along this endpoint path to $b$. There is no saddle point segment, anduses the same formula with the negative square-root branch. At $\theta =\pi /2$ the contour can run from $-d$ to zero and then up the imaginary axis. It gives a half-Gaussian contribution and the dominant imaginary endpoint term $i{e}^{\lambda }/(2\lambda )$. The change in the subdominant saddle point contribution is a Stokes phenomenon; it does not change the leading endpoint term on either side of that angle.

The Gaussian saddle and endpoint dominance by angle follows by comparing exponential sizes and gives all fixed-angle leading behaviours:At $\theta =\pi /4$, the endpoint term has magnitude $O({\lambda }^{-1})$, so the saddle point term of order ${\lambda }^{-1/2}$ still dominates. In the middle interval the endpoint term grows exponentially for $\pi /4<\theta <3\pi /4$, is oscillatory of order ${\lambda }^{-1}$ at $3\pi /4$, and decays exponentially for $3\pi /4<\theta <5\pi /6$. On the latter interval it nevertheless decays more slowly than the fixed-endpoint term. Equality of their exponential rates means $\cos 2\theta =1/2$ on the left half-plane, hence $\theta =5\pi /6$. At that angle both terms are required; their coefficient magnitudes are unequal, so their sum has no complete destructive cancellation.

These fixed-angle asymptotic expansions are not uniform if $\theta$ is allowed to vary with $\lambda$ near a dominance transition. The endpoint/saddle balance near $\pi /4$ involves $\lambda (\theta -\pi /4)=O(\log \lambda )$; the two-endpoint competition near $5\pi /6$ has an angular scale $O({\lambda }^{-1})$.

The following sketches show the contour orientations, the saddle point and both endpoints. The hyperbolic endpoint paths are followed backwards when returning from infinity to $b$; the dotted far-end connectors indicate the vanishing connecting arc.