Analyticity of the bilateral Laplace transform of a law

Remarks

Part 1 holds because $\sigma \mapsto e^{-\sigma x}$ is convex, so $e^{-\sigma x} \le e^{-a x} + e^{-b x}$ for $a \le \sigma \le b$; and $0 \in I$ because $\mu$ has total mass $1$. Part 2 holds because $\abs{e^{-sx}} = e^{-x \operatorname{Re} s}$. For part 3, fix $a < \operatorname{Re} s < b$ in $I$; then $\abs{x}^n e^{-x \operatorname{Re} s} \le C_n (e^{-a x} + e^{-b x})$ for a constant $C_n$, so differentiation under the integral sign is allowed by dominated convergence. On the real axis this says that the moment-generating function $M(t) = \mathcal{B}\mu(-t)$ is real analytic on the interior of its domain, with $M^{(n)}(t) = \int x^n e^{tx} \, d\mu(x)$. Example: for the exponential law with rate $\lambda$, $I = (-\lambda, \infty)$ and $\mathcal{B}\mu(s) = \lambda/(\lambda + s)$, which is holomorphic for $\operatorname{Re} s > -\lambda$ and has a pole at $s = -\lambda$. The open strip is needed in part 3: the derivative formula uses room on both sides of $\operatorname{Re} s$, and at an end point of $I$ the integral $\int \abs{x} e^{-x \operatorname{Re} s} \, d\mu(x)$ may diverge. If $I = \set{0}$, as for the Cauchy law, the strip is empty and the statement says nothing.

Going deeper

For a student

This is the reason why a moment-generating function that is finite near $0$ produces all the moments, as its derivatives at $0$. It is also the starting point of large deviation theory, where the convex function $\log M$ and its derivatives are used. The domain of finiteness is the strip of convergence of the bilateral transform. The same proof gives analyticity of the two-sided Laplace transform of any measure on its open strip. On the imaginary axis the transform is the characteristic function, which is always defined but need not be analytic.