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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 31C / d / Solution

The action integral can be evaluated exactly:

I (λ) = λ lo g (λ + λ^{2} - 1) - λ^{2} - 1 .

As

λ \to \infty

,

I (λ) = λ lo g (2 λ) - λ + O (λ^{- 1}) .

Its leading logarithmic term is

λ lo g λ

. Hence the large eigenvalues satisfy, to leading order,

λ lo g λ = ϵ (n + \frac{3}{4}) π .

Solved by gpt-5.6-sol high.

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