 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-3/30a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

f (t) \sim \sum_{n = 0}^{\infty} a_{n} t^{λ_{n} - 1} (t ↓ 0), 0 < λ_{0} < λ_{1} < \dots,

and the Laplace integral has suitable growth control away from zero, then

\int_{0}^{b} e^{- x t} f (t) d t \sim \sum_{n = 0}^{\infty} a_{n} Γ (λ_{n}) x^{- λ_{n}} (x \to + \infty) .

sin (t^{2}) = \sum_{k = 0}^{\infty} \frac{( - 1 ) ^{k}}{( 2 k + 1 )!} t^{4 k + 2} .

Termwise application of Watson lemma therefore gives the asymptotic expansion

I (x) \sim k = 0 \sum \infty \frac{( - 1 ) ^{k} Γ ( 4 k + 3 )}{( 2 k + 1 )! x ^{4 k + 3}} = \frac{2}{x ^{3}} - \frac{120}{x ^{7}} + \frac{30240}{x ^{11}} - \dots .

In particular, the leading asymptotic approximation is

I (x) \sim 2 x^{- 3}

 New to topics? Read the docs here!