= Converting a scalar sum jump to a difference jump
{title2=$V=W/\sqrt{z(z-c)}$}
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.
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