Two-sided Laplace transform

Remarks

Example: for $f(t) = e^{-\abs{t}}$, the part over $[0, \infty)$ is $1/(1 + s)$ for $\operatorname{Re} s > -1$, and the part over $(-\infty, 0]$ is $1/(1 - s)$ for $\operatorname{Re} s < 1$. So $\mathcal{B}\{f\}(s) = \frac{2}{1 - s^2}$ for $-1 < \operatorname{Re} s < 1$. Non-example: for the constant $f = 1$, the part over $[0, \infty)$ needs $\operatorname{Re} s > 0$ and the part over $(-\infty, 0]$ needs $\operatorname{Re} s < 0$, so the integral converges for no $s$. If $f(t) = 0$ for $t < 0$, the two-sided transform is the one-sided Laplace transform of $f$ restricted to $[0, \infty)$. For a random variable $X$ with density $p$, $\mathcal{B}\{p\}(-s) = \mathbb{E}[e^{s X}]$ is its moment generating function wherever it converges.