 Solution

ID: past-exam-of-the-theoretical-mechanics-course-of-the-university-of-oxford/2011/b/5/a/ii/solution

For a negative second derivative, the corresponding local normal form is

ψ = ψ (k_{⋆}) - s^{2} /2

. The oscillatory Gaussian integral is the complex conjugate of the positive-curvature one, so the phase shift changes sign. Thus

I (t) = f (k_{⋆}) e^{i t ψ (k_{⋆}) - iπ /4} \frac{2 π}{t ∣ ψ ^{''} ( k _{⋆} ) ∣} + o (t^{- 1/2}) .

When the leading amplitude is nonzero, the equality may again be written as an asymptotic equivalence. Both signs are summarized by the factor

e^{iπ sgn (ψ^{''}) /4}

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