Analytically continuing an integral into the lower half plane This is a generalization of the questions (1) and (2). Let
$$f(z) = \int_0^\infty dx \frac{\rho(x)}{x-z}$$
where we assume that $\rho(x)$ is nice enough so that $f(z)$ is analytic in the upper half-plane. Note that $f(z)$ has a branch cut along the positive real axis. I would like to analytically continue $f(z)$ across the branch cut into the lower half plane. If I denote the expression for the analytical continuation in the lower half plane $f^\downarrow(z)$, I could simply write $f^\downarrow(z) = f(z) - 2\pi i \rho(z)$. This follows from algebra similar to that of the answer of (1).
Note that there is still a branch cut in the continued function along the negative real axis (i.e. $f(z)$ and $f^\downarrow(z)$ match along the positive real axis, but not along the negative real axis.) This is fine with me. In some sense, I am simply asking how to define $f(z)$ so that the branch cut is moved to the negative real axis.
However, I unfortunately do not know $\rho(x)$ off the real axis. One attempt I had, similar to the work in the answer of (2), is $f^\downarrow(z) = f(\bar{z})$. This, by construction, matches $f(z)$ on the real axis (i.e. $f(\omega+i\delta) = f^\downarrow(\omega-i\delta)$.) However, this $f^\downarrow(z)$ is a function of $\bar{z}$ and is therefore obviously not analytic.
The other thought I had was to analytically continue $\rho(x)$ into the lower half-plane numerically, by solving the Cauchy-Riemann PDEs. However, this problem is equivalent to solving Laplace's equation with a Cauchy boundary condition, and is well-known to be an ill-posed problem.
As such, I would love if there was an integral form for $f^\downarrow(z)$ similar to the definition of $f(z)$. Thank you for your help!
 A: There seem to be a series of misconceptions to go over.

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*Any continuous function defined on a contour, creates an analytic function for all $z$ not on that contour, $f(z)=\int_\gamma \frac{p(x)}{x-z}dx$ is analytic for all $z$ not on $\gamma$. These are known as Cauchy-Type integrals (Theorem 2.4.5, Marsden and Hoffman).


*Analytic continuation only makes sense when you are talking about moving from one region to another continuously. It makes no sense to talk about "continuing" over a line where we are not defined. For example, consider a $J$ function, which is analytic on the upper half plane and is discontinuous on the whole real line. Such function cannot be "continued" across the real line in any meaningful fashion.


*If $f(z)$ is analytic, then so is $\overline{f(\overline{z})}$, automatically.  Furthermore, if $f(z)$ and $\overline{f(\overline{z})}$ are continuous on and agree on a line segment of the real line, then they are the same function (that is, one could be called an analytic continuation of the other). This is known as the Schwarz reflection principle. In general, a real valued function (like in your case) can ONLY be analytically continued across the upper/lower half planes using this double conjugation trick. Another way to say this is that a reflection is unique.
In summary, choosing $p(x)$ to be only continuous automatically means $f(z)$ is analytic everywhere but $[0,\infty)$. This means $f(z)$ is automatically analytic on $(-\infty,0)$. It could be the case that $f(z)$ can be analytically continued to part or all of $[0,\infty)$, but that is a case-by-case. Likewise, stating that $p(z)$ is real-valued immediately means that $f(z)=\overline{f(\overline{z})}$ everywhere it is analytic.
