in a part of book of Prüss (Evolutionary integral equations and applications pg. 99) it say..."Since $b$, $c \in \mathcal{BF}$ (Bernstein functions) by Bernstein's Theorem exist a function $\beta \in BV_{loc}(\mathbb{R}_+)$ (variation bounded functions) with $\int_{1}^{\infty}d\beta(s)/s<\infty$ such that $$\widehat{db}(z)=b_0+\int_0^{\infty}\frac{1}{z+\mu}d\beta(\mu), \ z>0;$$ where $$\widehat{db}(z)=\int_0^{\infty} e^{-zt}db(t),$$ and $b_0=\lim_{t\to0}b(t)$." But the Bernstein's Theorem says that a $C^{\infty}$-function $f:(0,\infty)\to \mathbb{R}$ is completely monotonic iff there is a nondecreasing function $b:\mathbb{R}_+\to \mathbb{R}$, such that $$f(\lambda)=\int_0^{\infty}e^{-\lambda t}db(t), \ \lambda>0.$$ Moreover,
$$ (-1)^nf^{n}(\lambda)=\int_0^{\infty}e^{-\lambda t}t^ndb(t), \ \lambda>0, n \in \mathbb{N}\cup \{0\}.$$ Can someone see the relation between these two affirmations?
Prüss starts with a function $b\in BV_{loc}(\mathbb R_+)$ by which the Laplace transform $$ f(\lambda)=\widehat{db}(\lambda)=\int_0^\infty e^{-\lambda t}db(t) $$ is completely monotone. Then he assumes that $b$ is a Bernstein function which is equivalent to its derivative $b'$ being completely monotone. By the Bernstein theorem this means that $b'$ is a Laplace transform $$ b'(t)=\int_0^\infty e^{-t\mu}d\beta(\mu) $$ of a function $\beta\in BV_{loc}(\mathbb R_+)\,.$ Putting together and writing $db(t)=b'(t)\,dt$ we get \begin{align} \widehat{db}(\lambda)&=\int_0^\infty\int_0^\infty e^{-(\lambda+\mu) t}\, dt\,d\beta(\mu)\,. \end{align} The inner integral is equal to \begin{align} \frac{-e^{-(\lambda+\mu) t}}{\lambda+\mu}\Bigg|_{t=0}^{t=\infty}=\frac{1}{\lambda+\mu}\,. \end{align}