Connection between the Laplace transform and generating functions

This is a special case of the moment generating property of the Laplace transform $^1$:

$$ \mathfrak{L}\Big\{f\Big\} = \int_{-\infty}^{\infty} \Big\{e^{-st} \Big\} f(t) \operatorname{dt} = \int_{-\infty}^{\infty} \Big\{\sum_{n=0}^{\infty}(-1)^n \frac{s^n t^n}{n!} \Big\} f(t) \operatorname{dt}$$ $$= \sum_{n=0}^{\infty}(-1)^n \frac{s^n}{n!} \int_{-\infty}^{\infty}t^n f(t)\operatorname{dt} = \sum_{n=0}^{\infty}(-1)^n \frac{s^n}{n!} m_n $$

where $m_n$ is the $n$-th moment of $f$ about $0$ (one assumes these exist).

In particular, if $f$ is the PDF of some random variable $X$, then $M_X (-s)$, the reflected moment generating function, is the Laplace transform of $f$.

see M.J. Dupré, «Transforms and Moment Generating Functions» for a more in-depth treatment. (PDF accessible as of 29/2/2015)

Note 1: The double-sided Laplace transform is used here because it supposedly translates better into the probabilistic case. It seems there is no significant difference between using the regular $\mathfrak{L}_+$ versus the two-sided $\mathfrak{L}$. However, I am not well-acquainted with its use in probability, so don't take my word for it.

Note: I just noticed that, expanding $f$ as a Taylor series, one arrives at the same equality:

$$ \mathfrak{L}\Big\{\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}t^n\Big\} = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\mathfrak{L}\Big\{t^n\Big\} = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\frac{n!}{s^{n+1}} = \frac{1}{s} \sum_{n=0}^{\infty} f^{(n)}(0)\frac{1}{s^n} $$