that Stieltjes functions are completely montone is proved in [@SchillingSongVondracek12].

[[@SchillingSongVondracek12 Theorem 2.2]]{} Stieltjes functions are completely monotone. A completely monotone function is a Stieltjes function if and only if the measure $\mu$ in is absolutely continuous on $(0,\infty)$ and its Radon–Nikodym derivative is completely monotone.

It turns out that any Stieltjes function has an analytic extension to the cut complex plane ${\mathbb C}\setminus (-\infty,0]$. This property gives a complete characterisation of the class of Stieltjes functions. Let ${\mathbb C}_+=\{z\in {\mathbb C}\colon \operatorname{Im}z>0\}$ and ${\mathbb C}_-=\{z\in {\mathbb C}\colon \operatorname{Im}z<0\}$.

[[@SchillingSongVondracek12 Corollary 7.4]]{} \[thm:Stieltjes extensions\] Let $g$ be a positive function on $(0,\infty)$. Then $g$ is a Stieltjes function if and only if the limit $\lim_{x\searrow 0} g(x)$ exists in $[0,\infty]$ and $g$ extends analytically to ${\mathbb C}\setminus (-\infty, 0]$ such that $\operatorname{Im}z \cdot \operatorname{Im}g(z)\le 0$.

Note that positive constant functions are examples of Stieltjes functions. It follows easily by basic properties of analytic functions that a nonconstant Stieltjes function maps ${\mathbb C}_+$ to ${\mathbb C}_-$. Note also that if $g$ is not identically $0$, then $1/g(z)$ is a Nevanlinna function (also known as Herglotz or Pick functions). The corresponding function $1/g(x)$ is then a complete Bernstein function, see [@SchillingSongVondracek12].

It 