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{c}

If $X\sim\operatorname{Exp}(\lambda)$, then direct integration gives
$$
\mathbb E[e^{-sX}]=\frac{\lambda}{\lambda+s},
\qquad \operatorname{Re}s>-\lambda.
$$


Laplace transform of an exponential distribution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 211 / 2 / b / Solution

Laplace transform of an exponential distribution

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Laplace transform of an exponential distribution

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