 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-4/12a/b/solution

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

Split the transform into periods and translate each interval:

L {g} (s) = \sum_{j = 0}^{\infty} \int_{j T}^{(j + 1) T} e^{- s t} g (t) d t = \sum_{j = 0}^{\infty} e^{- s j T} \int_{0}^{T} e^{- s u} g (u) d u .

Summing the geometric series yields

L {g} (s) = \frac{L { g _{T} } ( s )}{1 - e ^{- s T}} .

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