= Solution
{c}
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:
$$
\operatorname{PV}\int_\Gamma\frac{f(\zeta)}{\zeta-t}\,d\zeta
:=\lim_{\eta\downarrow0}
\int_{\Gamma\setminus\gamma((-\eta,\eta))}
\frac{f(\zeta)}{\zeta-t}\,d\zeta.
$$
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 theorem>
$$
\boxed{w_\pm(t)=\pm\frac{f(t)}2+
\frac1{2\pi i}\operatorname{PV}\int_\Gamma
\frac{f(\zeta)}{\zeta-t}\,d\zeta.}
$$
In 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
$$
\boxed{(a+b)w_++(b-a)w_-=c.}
$$
Choose a canonical homogeneous solution $\widetilde w$ whose boundary values do not vanish, and account for any permitted zeros or <poles> when forming a quotient. The assumptions $a\pm b\neq0$ alone do not ensure that an arbitrary homogeneous solution has this property. Its jump relation is
$$
(a+b)\widetilde w_+=(a-b)\widetilde w_-.
$$
Dividing the inhomogeneous relation by $(a+b)\widetilde w_+$ therefore gives
$$
\left(\frac w{\widetilde w}\right)_+
-\left(\frac w{\widetilde w}\right)_-
=\rho,\qquad
\rho=\frac{c}{(a+b)\widetilde w_+}.
$$
The <Cauchy integral> solves this additive jump. More generally every solution in the chosen analytic class has
$$
w(z)=\widetilde w(z)
\left[\frac1{2\pi i}\int_\Gamma
\frac{\rho(\zeta)}{\zeta-z}\,d\zeta+H(z)\right],
$$
where $H$ 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 $C\rho=(2\pi i)^{-1}\operatorname{PV}\int_\Gamma\rho(\zeta)/(\zeta-t)d\zeta$. Applying <Sokhotski–Plemelj theorem> to the displayed solution and subtracting boundary values yields
$$
f=(\widetilde w_+-\widetilde w_-)C\rho
+\tfrac12(\widetilde w_++\widetilde w_-)\rho
+(\widetilde w_+-\widetilde w_-)H.
$$
Use
$$
\widetilde w_+-\widetilde w_-
=-\frac{2b\widetilde w_+}{a-b},
\qquad
\widetilde w_++\widetilde w_-
=\frac{2a\widetilde w_+}{a-b}.
$$
The normalized particular solution is consequently
$$
f_{\rm p}(t)=
-\frac{b(t)\widetilde w_+(t)}{a(t)-b(t)}
\frac1{\pi i}\operatorname{PV}\int_\Gamma
\frac{c(\zeta)}{(a(\zeta)+b(\zeta))\widetilde w_+(\zeta)}
\frac{d\zeta}{\zeta-t}
+\frac{a(t)c(t)}{a(t)^2-b(t)^2},
$$
and the full solution is
$$
\boxed{f=f_{\rm p}-\frac{2b\widetilde w_+}{a-b}H.}
$$
Thus the printed expression is a \b[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 $\widetilde w_+=1$, $\widetilde w_-=1/z$ 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.
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