Laplace–Stieltjes transform

Remarks

Example: let $a > 0$ and $g(t) = 0$ for $t < a$, $g(t) = 1$ for $t \ge a$. Then $\int_0^T e^{-s t} \, dg(t) = e^{-a s}$ for every $T > a$, so $\mathcal{L}^*\{g\}(s) = e^{-a s}$ for every complex $s$. If $g(t) = \int_0^t f(\tau) \, d\tau$ with $f$ continuous, then $dg = f(t) \, dt$ and $\mathcal{L}^*\{g\}$ is the Laplace transform of $f$. Non-example: $g(t) = e^{t^2}$ is of bounded variation on each $[0, T]$, but $\int_0^T e^{-s t} \, dg(t) = \int_0^T 2 t e^{t^2 - s t} \, dt$ diverges as $T \to \infty$ for every $s$, so no transform exists at any point. In probability one often uses the Lebesgue–Stieltjes form $\int_{[0, \infty)} e^{-s t} \, dG(t)$ for a distribution function $G$, which counts a possible jump at $t = 0$. With $G(t) = 0$ for $t < 0$ it equals $\mathbb{E}[e^{-s X}]$ for $X \ge 0$ with distribution function $G$. If $g(0) = 0$ and $\abs{g(t)} \le M e^{a t}$ for all $t \ge 0$, integration by parts gives $\mathcal{L}^*\{g\}(s) = s \, \mathcal{L}\{g\}(s)$ for $\operatorname{Re} s > a$.