 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/32c/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Repeated integration by parts starts with

si (x) = \frac{c o s x}{x} - \frac{1}{x} \int_{1}^{\infty} \frac{c o s ( x t )}{t ^{2}} d t = \frac{c o s x}{x} + \frac{s i n x}{x ^{2}} - \frac{2}{x ^{2}} \int_{1}^{\infty} \frac{s i n ( x t )}{t ^{3}} d t .

Continuing alternates sine and cosine and multiplies by successive integers. After finitely many steps the remainder is bounded by a constant times the next inverse power, so the fixed-lower-limit sine-integral expansion is

si (x) \sim cos x \sum_{n = 0}^{\infty} (- 1)^{n} (2 n)! x^{- 2 n - 1} + sin x \sum_{n = 0}^{\infty} (- 1)^{n} (2 n + 1)! x^{- 2 n - 2} .

Therefore

a_{n} = (- 1)^{n} (2 n)!, b_{n} = (- 1)^{n} (2 n + 1)! .

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