Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 15F / i

Table of contents
- Solution i

Solution

n \geq 0

and

Re s > 0

, the definition of the Laplace transform gives

F_{0} (s) = \int_{0}^{\infty} e^{- s t} d t = 1/ s

. For

n \geq 1

, integration by parts gives

F_{n} (s) = [- \frac{t ^{n} e ^{- s t}}{s}]_{0}^{\infty} + \frac{n}{s} \int_{0}^{\infty} t^{n - 1} e^{- s t} d t = \frac{n}{s} F_{n - 1} (s) .

The endpoint term vanishes because exponential decay dominates every fixed power of

t

. Induction therefore gives

F_{n} (s) = n! / s^{n + 1}

. Multiplication by

e^{a t}

replaces

e^{- s t}

by

e^{- (s - a) t}

in the defining integral, so

L (t^{n}) (s) = \frac{n !}{s ^{n + 1}}, L (e^{a t} t^{n}) (s) = \frac{n !}{( s - a ) ^{n + 1}} .

The second formula holds for

Re (s - a) > 0

and is the Laplace transform shift theorem.

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