Finite Hilbert transform inversion ($f(x)=-\frac{i}{\pi \rho (x)}\mathrm{PV}{\int }_0^1\frac{\rho (t)G(t)}{t-x}dt,\rho (x)=\sqrt{x(1-x)}$)

For $G\in C^1([0,1])$, the displayed expression solves $Tf=G$, where $T=H_{(0,1)}/i$ and $H_{(0,1)}$ is the finite Hilbert transform with denominator $t-x$. To derive it, put $S(z)=\sqrt{z(z-1)}\sim z$ at infinity, take $h$ to be the Cauchy transform of $i\rho G$, and set $w=h/S$. The Sokhotski–Plemelj formula gives $w_{+}+w_{-}=G$ and the displayed jump $f=w_{+}-w_{-}$. The endpoint behavior is at most inverse square root, and $w=O(z^{-2})$ at infinity. Cauchy reconstruction from a jump across a finite interval then gives $w=C[f]$ and hence $Tf=G$. Adding any constant multiple of $1/\rho$ gives another solution in the integrable endpoint class.

Multiplying an auxiliary sign-changing factor by a growing analytic factor can invalidate this reconstruction. For example, using $S$ instead of $1/S$ with $G=1$ produces the constant function $w=1/2$ and zero jump. The boundary sum alone is not enough: the entire remainder in reconstructing $w$ from its jump must also be controlled.