Laplace transform
Relation to moments

The quantities μ n = ∫ 0 ∞ t n f (t) d t {\displaystyle \mu _{n}=\int _{0}^{\infty }t^{n}f(t)\,dt}

are the moments of the function f. If the first n moments of f converge absolutely, then by repeated differentiation under the integral, (− 1) n (L f) (n) (0) = μ n. {\displaystyle (-1)^{n}({\mathcal {L}}f)^{(n)}(0)=\mu _{n}.} This is of special significance in probability theory, where the moments of a random variable X are given by the expectation values μ n = E {\displaystyle \mu _{n}=\operatorname {E} }. Then, the relation holds μ n = (− 1) n d n d s n E (0). {\displaystyle \mu _{n}=(-1)^{n}{\frac {d^{n}}{ds^{n}}}\operatorname {E} \left(0).}
combination:repeat_prompt, startend:quotation