Solution

Write the contour near the simple saddle as

z = z_{0} + e^{i α} s, s \in R,

with its given orientation. Since

ϕ^{'} (z_{0}) = 0

ϕ (z) = ϕ (z_{0}) + \frac{1}{2} ϕ^{''} (z_{0}) e^{2 i α} s^{2} + O (s^{3}) .

Along a steepest-descent tangent the quadratic coefficient is real and negative, so

ϕ^{''} (z_{0}) e^{2 i α} = - ∣ ϕ^{''} (z_{0}) ∣, ar g ϕ^{''} (z_{0}) + 2 α = π (mod 2 π) .

Moreover

d z = e^{i α} d s

. Only a neighbourhood of width

O (x^{- 1/2})

contributes at leading order, and there

f (z) = f (z_{0}) + o (1)

. The local integral is consequently

f (z_{0}) e^{x ϕ (z_{0}) + i α} \int_{- \infty}^{\infty} exp (- \frac{x ∣ ϕ ^{''} ( z _{0} ) ∣}{2} s^{2}) d s .

I (x) \sim f (z_{0}) \frac{2 π}{x ∣ ϕ ^{''} ( z _{0} ) ∣} e^{x ϕ (z_{0}) + i α} .

Solved by gpt-5.6-sol high.

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