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There are two distinct inconsistencies in the printed formulas. First the initial Cauchy integral has coefficient $1/(2\pi )$, whereas the stated boundary limits require $1/(2\pi i)$. Second the final integral is a solution of a difference jump problem, although the intervening equation for $W$ is a sum jump problem. We derive the valid identities, show precisely where that final deduction fails, and give the corrected construction. Take $c>0$ and, initially, a Hölder continuous density on the interior with integrable endpoints, so the boundary limits and principal values exist.

Define the Cauchy principal value byFor the corrected normalization $C_f(z)=(2\pi i{)}^{-1}{\int }_0^{c}f(t)/(t-z)dt$, subtract $f(x)$ from the numerator near $x$. The subtracted integral has the common principal-value limit, while integration of the remaining constant density produces boundary arguments differing by $2\pi$. This proves the Sokhotski–Plemelj theoremIn particular $f=(C_f{)}_{+}-(C_f{)}_{-}$ and $(C_f{)}_{+}+(C_f{)}_{-}=(\pi i{)}^{-1}\mathrm{pv}\int f(t)/(t-x)dt$. With the literal printed integral $w=i{C}_f$ instead, the limits are $\pm if/2+(2\pi {)}^{-1}\mathrm{pv}\int f/(t-x)$, and the next stated boundary equation would have right side $i\gamma$.

Use the corrected normalization from now on and write $w=C_f$, $a=\alpha +\beta$, $b=\alpha -\beta$. The original singular integral equation then becomesIf $\tilde{w}$ satisfies the printed homogeneous relation $a{\tilde{w}}_{+}+b{\tilde{w}}_{-}=0$, division of $w=\tilde{w}W$ gives, correctly,A nonvanishing factor is needed as well as the printed nonvanishing of $a,b$; its zeros would require separate treatment.

The next printed step is invalid because a Cauchy integral of $g$ has boundary difference $g$, not boundary sum $g$. A particularly direct counterexample is $\alpha =1$, $\beta =0$, $\gamma =1$. The original equation is simply $f=1$. Choose $\tilde{w}=\rho (z)=\sqrt{z(z-c)}$, with $\rho (z)\sim z$ at infinity and cut $[0,c]$. Its boundary values are ${\rho }_{\pm }(x)=\pm i\sqrt{x(c-x)}$, so it satisfies precisely the printed homogeneous sum relation. But the printed proposed integral yieldsThe integral follows by $t=c(1+\cos \theta )/2$, or the usual elementary integral ${\int }_0^{\pi }d\theta /(A+B\cos \theta )$. Its proposed $w$ is constant, has jump zero, and is not the Cauchy transform of $f=1$; it also fails the required decay at infinity. If the printed $I(x)$ is interpreted as principal value in its last formula, that value is zero, giving $f=0$ and the same contradiction.

To solve the sum jump correctly, introduce this sign-changing factor $\rho$. Set $V=W/\rho$, soThe corrected local Cauchy representation is consequentlyHere $H$ has no jump across the open interval; its possible endpoint singularities and polynomial part are determined by endpoint admissibility, absence of unwanted poles, and $w(z)=O(1/z)$ at infinity. Some choices of factor require a weighted or regularized integral at the endpoints. Thus nonvanishing of $a,b$ alone does not determine a unique global solution without a function class and growth conditions.

Equivalently put $X=\tilde{w}\rho$. It satisfies the homogeneous difference relation $a{X}_{+}-b{X}_{-}=0$. If $g=\gamma /(a{X}_{+})$, $J(z)={\int }_0^{c}g(t)/(t-z)dt$, and the integral is admissible, thenTaking its jump gives the usable density formulaThis formula, together with the endpoint and infinity conditions, replaces the last printed equation. Alternatively, if the homogeneous relation in the question is changed from plus to minus, its unweighted Cauchy representation becomes a valid particular solution, but the density must still be obtained from this jump calculation and homogeneous terms still require normalization. The statements cannot all hold simultaneously as printed.

Orient $\Gamma$ and label its left side plus and its right side minus. Assume the density is Hölder continuity near the point in question, or has sufficient regularity for the boundary limits below. The source does not explicitly state this regularity assumption. Parametrize locally by oriented arclength $\gamma (s)$, with $\gamma (0)=t$. Define the Cauchy principal value by removing equal arclengths on the two sides:Subtract $f(t)$ from the numerator near $t$. The Hölder continuity remainder is locally bounded in absolute value by a constant times $|s{|}^{\nu -1}$ for some $\nu >0$, so its integral has the same limit from the two sides. The constant part contributes half the residue, with sign determined by the side of the small semicircle. This proves the Sokhotski–Plemelj theoremIn particular $f=w_{+}-w_{-}$ and the Cauchy principal value term with coefficient $1/(\pi i)$ equals $w_{+}+w_{-}$. Substituting these two identities into the singular integral equation gives

Choose a canonical homogeneous solution $\tilde{w}$ whose boundary values do not vanish, and account for any permitted zeros or poles when forming a quotient. The assumptions $a\pm b\ne 0$ alone do not ensure that an arbitrary homogeneous solution has this property. Its jump relation isDividing the inhomogeneous relation by $(a+b){\tilde{w}}_{+}$ therefore givesThe Cauchy integral solves this additive jump. More generally every solution in the chosen analytic class haswhere $H$ has no jump across $\Gamma$ and obeys the applicable growth, endpoint and divisor conditions. Setting $H=0$ gives exactly the intermediate formula printed in the source.

Let $C\rho =(2\pi i{)}^{-1}\mathrm{PV}{\int }_{\Gamma }\rho (\zeta )/(\zeta -t)d\zeta$. Applying Sokhotski–Plemelj theorem to the displayed solution and subtracting boundary values yieldsUseThe normalized particular solution is consequentlyand the full solution isThus the printed expression is a particular solution; homogeneous terms need separate normalization. To see why “general” cannot hold without qualification, take the positively oriented unit circle, $a=(1+z)/2$, $b=(1-z)/2$, and $c=0$. Then $a+b=1$ and $a-b=z$ are nonzero on the contour. The analytic pair ${\tilde{w}}_{+}=1$, ${\tilde{w}}_{-}=1/z$ satisfies the homogeneous jump and decays on the exterior side. Its nonzero density $f=1-1/z$ has these very Cauchy boundary values and solves the original homogeneous equation, whereas the source formula with $c=0$ gives zero. This is homogeneous freedom in a scalar singular integral equation; endpoint and index conditions determine which jump-free additions are permitted.