Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 4 / 12A (Complex Methods)

Table of contents
- a 12A
  - Solution a
- b 12A
  - Solution b
- c 12A
  - Solution c

a

Solution

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) .

b

Solution

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}} .

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} )} .

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