= Finite Hilbert transform inversion
{title2=$f(x)=-\dfrac{i}{\pi\rho(x)}\operatorname{PV}\int_0^1\dfrac{\rho(t)G(t)}{t-x}\,dt,\quad\rho(x)=\sqrt{x(1-x)}$}
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.
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