Iterated Laplace transform equals Stieltjes transform

Remarks

The proof is one application of Fubini's theorem. For $x, t > 0$ one has $\int_0^\infty e^{-xs} e^{-st}\, ds = \frac{1}{x + t}$. So $\int_0^\infty \int_0^\infty e^{-xs} e^{-st} \abs{f(t)}\, dt\, ds = \int_0^\infty \frac{\abs{f(t)}}{x + t}\, dt < \infty$ by Tonelli, and Fubini allows the exchange of the two integrals. The inner integral $\int_0^\infty e^{-st} \abs{f(t)}\, dt$ is then finite for almost every $s$, and since it decreases in $s$ it is finite for every $s > 0$. The hypothesis cannot be dropped: for $f = 1$ the right side diverges for every $x$, while $\mathcal{L} f(s) = 1/s$ is not integrable against $e^{-xs}$ near $s = 0$.

Example: $f(t) = t^{-1/2}$. Then $\mathcal{L} f(s) = \sqrt{\pi / s}$ and $\mathcal{L}(\mathcal{L} f)(x) = \sqrt{\pi}\, \Gamma(1/2)\, x^{-1/2} = \pi\, x^{-1/2}$. Directly, the substitution $t = x u^2$ gives $\int_0^\infty \frac{t^{-1/2}}{x + t}\, dt = \frac{2}{\sqrt x} \int_0^\infty \frac{du}{1 + u^2} = \frac{\pi}{\sqrt x}$.

Going deeper

For a student

This identity lets one move facts between the two transforms. Properties of the Stieltjes transform, such as analyticity in $x$ and its behaviour as $x \to 0^+$ or $x \to \infty$, follow from the corresponding properties of the Laplace transform applied twice. Conversely, inversion formulas for the Laplace transform, such as Post's inversion formula, can be applied twice to recover $f$ from its Stieltjes transform.