 Solution

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

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

The Heaviside factor restricts the integral to

t \geq α

. With

u = t - α

L {f (t - α) H (t - α)} (s) = \int_{α}^{\infty} e^{- s t} f (t - α) d t = e^{- α s} \int_{0}^{\infty} e^{- s u} f (u) d u = e^{- α s} F (s) .

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