Suppose $X$ is a random variable and $g$ is its moment-generating function. The following appears to hold:
$$E[(s-X)^{-1}]=(\mathcal{L}g)(s)$$
For example, let $X\sim \operatorname{Beta}(\frac{1}{2},1)$, we have:
$$E[(s-X)^{-1}]=(\mathcal{L}g)(s)=\frac{\coth ^{-1}\left(\sqrt{s}\right)}{\sqrt{s}}$$
In other words, Stieltjes transformation of a random variable is the Laplace Transform of its moment generating function. Is this statement generally true? What's the easiest way to prove it?
dist = BetaDistribution[1/2, 1];
mgf = MomentGeneratingFunction[BetaDistribution[1/2, 1], t]
LaplaceTransform[mgf, t, s] // FullSimplify
Expectation[1/(s - y), y \[Distributed] dist]
Changing order of integration still works
pdf = 1/(2 Sqrt[y]);
Integrate[
Integrate[pdf Exp[t y] Exp[-s t], {y, 0, 1}], {t, 0, \[Infinity]}]
Integrate[
Integrate[pdf Exp[t y] Exp[-s t], {t, 0, \[Infinity]}], {y, 0, 1}]
Numeric integration matches symbolic for $s=3.$
Block[{s = 3.},
NIntegrate[pdf Exp[t y] Exp[-s t], {t, 0, \[Infinity]}, {y, 0, 1}]]
(* 0.380173 *)
ArcCoth[Sqrt[s]]/Sqrt[s] /. s -> 3.
(* 0.380173 *)
Best Answer
By the hints in the comments, I meant the following: \begin{align*} \int_0^{ + \infty } {{\rm e}^{ - sw} M_X (w){\rm d}w} &= \int_0^{ + \infty } {{\rm e}^{ - sw} \int_{ a }^{ b } {{\rm e}^{tw} f_X (t){\rm d}t}\, {\rm d}w} \\ & =\int_{ a }^{ b } {f_X (t)\int_0^{ + \infty } {{\rm e}^{ - (s - t)w} {\rm d}w} \,{\rm d}t} \\ &=\int_{ a }^{ b } {\frac{{f_X (t)}}{{s - t}}{\rm d}t} = E\left[ {(s - X)^{ - 1} } \right] , \end{align*} where $[a,b]$ is the support of $f_X(t)$ and $s>b$.