How can one write $z^{-1}$ as a Stieltjes function? I would like to write $f(z) = z^{-1}$, for $z \in \mathbb{C} \setminus (-\infty,0]$, as a Riemann-Stieltjes integral of the following form:
$$
f(z) = \int_{0}^{\infty}{\frac{\text{d}\mu(t)}{t+z}},
$$
where $\mu$ is positive, monotonically increasing, and satisfies
$$
\int_{0}^{\infty}{\frac{\text{d}\mu(t)}{t+1}}<\infty.
$$
That is, I would like find a $\mu$ with the above properties.
I've played around with $\mu$ of the form $(t+a)^{-\alpha}$, but after splitting into partial fractions, I have to deal with the integral $  \int_{0}^{\infty}{\frac{\text{d}t}{t+z}}$, which does not converge.  As I have little experience with Stieltjes functions, I'm not sure what else to try beyond this.
Additional info:
The following holds for $z^{-\alpha}$, where $\alpha \in (0,1)$:
$$
z^{-\alpha} = \dfrac{\sin(\alpha \pi)}{\pi} \int_{0}^{\infty}{\dfrac{t^{-\alpha}}{t+z}\text{d}t},
$$
making $ \mu_{\alpha}(t) = \dfrac{\sin(\alpha \pi)}{\pi(1-\alpha)} t^{1-\alpha}$ in this case.  Perhaps the $\mu$ I seek can be generalized from these $\mu_{\alpha}$ somehow, through a limiting procedure.

A friend of mine solved this for me.  As he does not own an account, I will post his answer here.
Quite simply, $\mu(t) = H(t)$, where $H$ is the Heaviside function.  The proof is immediate, since $\text{d}\mu(t) = \text{d}\delta(t)$ would give that 
$$
\int_0^{\infty}\frac{\text{d}\mu(t)}{t+z} = \int_0^{\infty}\frac{\text{d}\delta(t)}{t+z} = \frac{1}{z}.
$$
