Relationship between Stieltjes transformation and Moment-Generating Function

Question

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 *)

Answer 1

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 $s>b$.

Answer 2

This is a manifestation of the well known property of the Stieltjes transform of a measure of local finite variation and its Laplace transform: The Stieltes transform is an iteration of the Laplace transform. To be more precise,

Theorem: Suppose $\alpha$ is a right-continuous function of local bounded variation on $[0,\infty)$ and set $\alpha(t)=0$ for $t<0$. Let $\mu_\alpha$ denote the Stieltjes-Lebesgue measure induced by $\alpha$ ($\mu_\alpha$ is the unique complex Borel measure on $[0,\infty)$ such that $\mu_\alpha((a,b])=\alpha(b)-\alpha(a)$ for all $0\leq a\leq b$, and $\mu_\alpha(\{0\})=\alpha(0)$). If the integral $$ f(s)=\int^\infty_0\frac{\mu_\alpha(dt)}{s+t}$$ converges for some $s\in\mathbb{C}\setminus\{z\in\mathbb{C}: z\in\mathbb{R}, \, z\leq 0\}$, then $f(s)$ converges for all $s\in\mathbb{C}\setminus\{z\in\mathbb{C}: z\in\mathbb{R}, \, z\leq 0\}$, and $$ f(s)=\int^\infty_0 e^{-st}\phi(t)\,dt,\qquad \mathfrak{R}(z)>0$$ where $$\phi(t)=\int^\infty_0 e^{-tu}\mu_\alpha(du),\qquad \mathfrak{t}>0$$

Here I am using a more standard form of the Laplace transform according to analysis: $\mathcal{L}\mu(t)=\int^\infty_0 e^{-tu}\mu(du)$ where $\mu$ is a complex Borel measure on $[0,\infty)$. Bilateral results (where the integral is taken over $\mathbb{R}$ instead of $[0,\infty)$ can also be obtained).

For complex Borel measures with densities (w.r.t.Lebesgue measure) and compact support the result in the theorem is easy to prove by a direct application of Fubini's theorem as in Gary's answer.The example in the OP follows along these lines since the Beta distribution has density and its support is $[0,1]\subset [0,\infty)$.

A good presentation of the result I am presenting is the classic book Widder, D. V., The Laplace Transform, Dover Publications, 2010 (unabridged republication of the original 1941 version, Princeton University Press), Chapter 8.