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For a Gaussian integral from $-1/\sqrt{2}$ to $e^{i\theta }$, a saddle point contribution dominates for $0\le \theta \le \pi /4$. The moving endpoint contribution in steepest descent dominates for $\pi /4<\theta <5\pi /6$. At $5\pi /6$ the two endpoints have equal exponential rates and both are needed; for larger angles the fixed endpoint dominates. Saddle connectivity switches at $\pi /2$, but that change is subdominant in the leading endpoint regime.

Set $d=1/\sqrt{2}$. At the positive real endpoint the contour is real and crosses the saddle point at zero. The Gaussian integral givesBoth omitted tails are exponentially small. At the negative real endpoint the contour is oriented leftward and does not cross the saddle point:The endpoint contribution in steepest descent, or one integration by parts, gives ${\int }_d^{\infty }{e}^{-\lambda x^2}dx\sim e^{-\lambda d^2}/(2\lambda d)$. The upper-end correction is $O(e^{-\lambda }/\lambda )$ and is much smaller. ThereforeThe sign comes from contour orientation, not from a different Gaussian convention.

Write $w=z-1=X+iY$. The phase and its derivative areso the saddle points arewith $\varphi (z_{+})=2i$ and $\varphi (z_{-})=-2i$. The contour geometry can be drawn fromThe stationary-phase level $\mathrm{Re}\varphi =0$ consists of $X=0$ and $X^2-3Y^2+3=0$, meeting at the saddles. The steepest curves through $z_{\pm }$ are the levels $\mathrm{Im}\varphi =\pm 2$. Far away, sectors with $\cos (3\arg w)>0$ are exponential hills and those with $\cos (3\arg w)<0$ are valleys.

The stated contour deforms through the upper saddle. If $z=z_{+}+s$, thenThe descent tangent has $s=e^{i\pi /4}t$, because $3i{s}^2=-3t^2$. The simple-saddle contribution in steepest descent is therefore

When the contour begins at $z=1$, deform it first from the endpoint into the decaying negative-real direction and then onto the same upper-saddle descent path. Near the endpoint, $s=z-1$ and $\varphi =3s+O(s^3)$, so the endpoint contribution in steepest descent isAdding the saddle and endpoint pieces gives

For $e^{-\lambda z^2}$, the saddle point descent is the real axis and ascent is the imaginary axis. A descent path from a nonzero endpoint $b$ has $z(s{)}^2=b^2+s$, $s\ge 0$, with the continuous square-root branch from $b$. It ends at positive or negative real infinity according to the sign of $\mathrm{Re}b$. The oriented outward endpoint contribution in steepest descent is $e^{-\lambda b^2}/(2\lambda b)$.