Why is the $F_{X}(x)={\mathcal {L}}^{-1}\left\{{\frac {1}{s}}E\left[e^{-sX}\right]\right\}(x)$?

Question

According to Wikipedia, the cumulative distribution function can be obtained from the Laplace transform as follows: $F_{X}(x)={\mathcal {L}}^{-1}\left\{{\frac {1}{s}}E\left[e^{-sX}\right]\right\}(x)$

Where does this result come from?

Answer 1

Use the Time-domain integration property: $${\mathcal {L}}\left\{\int_0^x f(t) dt\right\}(s)=\frac{{\mathcal {L}}\left\{f(x)\right\}(s)}{s}.$$ Hence $${\mathcal {L}}\left\{F_X(x)\right\}(s)={\mathcal {L}}\left\{\int_0^x f(t) dt\right\}(s)=\frac{{\mathcal {L}}\left\{f(x)\right\}(s)}{s}=\frac{1}{s}\int_0^{+\infty} f(x)e^{-sx} dx={\frac {1}{s}}E\left[e^{-sX}\right]$$ where $f$ is the probability density function of the random variable $X$.