Converting a scalar sum jump to a difference jump ($V=W/\sqrt{z(z-c)}$)

A Cauchy integral solves a prescribed boundary difference, not a prescribed sum. If $W_{+}+W_{-}=g$ on $(0,c)$, choose $\rho (z)=\sqrt{z(z-c)}\sim z$ at infinity. Its boundary values satisfy ${\rho }_{-}=-{\rho }_{+}$, so $V=W/\rho$ has difference $V_{+}-V_{-}=g/{\rho }_{+}$. Thus $W=\rho [C(g/{\rho }_{+})+H]$, where $C$ is the Cauchy integral with coefficient $1/(2\pi i)$ and $H$ has no jump. Endpoint integrability and normalization at infinity determine permissible regularizations and homogeneous terms. Simply using $C(g)$ for a sum jump is incorrect.