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Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 4 / 12A / c / Solution

For one period,

\int_{0}^{2 π} e^{- s t} h (t) d t = \int_{0}^{π} e^{- s t} sin t d t = \frac{1 + e ^{- π s}}{s ^{2} + 1} .

Part (b), followed by cancellation of

1 + e^{- π s}

, gives

L {h} (s) = \frac{1}{( s ^{2} + 1 ) ( 1 - e ^{- π s} )} .

Solved by gpt-5.6-sol high.

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