My function is defined as
$$f(z) = \int_0^\infty \frac{\rho(x) dx}{x-z } , $$
where $\rho(x)$ is a real function. It is easy to see that $f$ has a branch cut along the positive real axis. Now let us do analytic continuation, i.e., let $z$ make an anticlockwise circle around the origin starting from $x - i 0^+ $. The problem is, for what $\rho$, when $z$ comes back to $x + i 0^+$, the value of $f$ will agree with the value defined above? In short, for what $\rho$, can $f$ extend into a nice two-leaf function?
To avoid bothering with convergence issues, I'll assume that $\rho(x)$ is continuous and $\displaystyle\int_0^\infty |\rho(x)|\,\text{d}x$ converges.
What you can do is to look at
$\displaystyle\int_0^\infty \dfrac{\rho(x)}{x-z}-\dfrac{\rho(x)}{x-\overline{z}}\,\text{d}x$
$=\displaystyle\int_0^\infty \dfrac{(z-\overline{z})\rho(x)}{|x-z|^2}\,\text{d}x$
which, for $z=a+bi$ where $a>0$ and $b>0$, can be rewritten as
$=\displaystyle\int_0^\infty \dfrac{2bi\rho(x)}{(x-a)^2+b^2}\,\text{d}x$
which tends to $2\pi i\rho(a)$ as $b\to0^+$.