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 Incoming links: Reciprocal-amplitude Laplace integral

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 31D / a / Solution Created 2026-09-24 Updated 2026-09-25

Near the endpoint

t = 0

,

\frac{1}{1 + t} \sim \sum_{n = 0}^{\infty} (- 1)^{n} t^{n} .

The Watson lemma integrates this local expansion term by term, and

\int_{0}^{\infty} e^{- x t} t^{n} d t = \frac{Γ ( n + 1 )}{x ^{n + 1}} = \frac{n !}{x ^{n + 1}} .

Therefore the reciprocal-amplitude Laplace integral has expansion

U (x) \sim n = 0 \sum \infty (- 1)^{n} n! x^{- n - 1}, x \to + \infty,

so

a_{n} = (- 1)^{n} n! .

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