 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-336/1/b/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

For

0 < a < 1

, the phase has an interior minimum at

t_{*} = arcosh (1/ a), sinh t_{*} = \frac{1 - a ^{2}}{a} .

The inverse hyperbolic cosine selects the positive stationary point. At this point,

S (t_{*}) = 1 - a^{2} - arcosh (1/ a), S^{''} (t_{*}) = 1 - a^{2} > 0.

Expanding the phase to second order and evaluating the resulting Gaussian integral by Laplace's method yields

A_{ν} (a ν) \sim \frac{2 π}{ν 1 - a ^{2}} exp {ν [arcosh (1/ a) - 1 - a^{2}]} .

The exponential grows because the minimum of

S

is negative; the original integral remains convergent because

a sinh t

dominates

t

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