Bilateral Laplace transform of a probability law

Example

Let $\mu$ be a probability measure on $\R$, for example the law of a real random variable $X$. Its two-sided Laplace transform is $$\mathcal{B}\mu(s) = \int_{\R} e^{-sx} \, d\mu(x) = E\left[e^{-sX}\right],$$ for the complex $s$ at which the integral converges absolutely. Since $\abs{e^{-sx}} = e^{-x \operatorname{Re} s}$, convergence depends only on $\operatorname{Re} s$, and it always holds on the line $\operatorname{Re} s = 0$, where the integrand has modulus $1$. For real $t$, the moment-generating function is $M_X(t) = \mathcal{B}\mu(-t)$.

Take $\mu$ the standard normal distribution, with density $(2\pi)^{-1/2} e^{-x^2/2}$. Completing the square gives $-sx - x^2/2 = -(x + s)^2/2 + s^2/2$. So $$\mathcal{B}\mu(s) = e^{s^2/2} \cdot \frac{1}{\sqrt{2\pi}} \int_{\R} e^{-(x+s)^2/2} \, dx = e^{s^2/2}$$ for every $s \in \C$, where the last integral equals $\sqrt{2\pi}$ by a shift of the line of integration. Here the strip of convergence is the whole plane.

Take $\mu$ the exponential distribution with rate $\lambda > 0$, with density $\lambda e^{-\lambda x}$ on $[0, \infty)$. Then $\int_0^\infty e^{-sx} \lambda e^{-\lambda x} \, dx = \lambda / (\lambda + s)$ when $\operatorname{Re} s > -\lambda$, and the integral diverges when $\operatorname{Re} s \le -\lambda$. Here the strip of convergence is the half-plane $\operatorname{Re} s > -\lambda$.