Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / B / 5 / i

Orient the contour so that the plus side is on the left; for a positively oriented simple closed contour this is its interior. If $\zeta (s)$ is a smooth arclength parametrization with $\zeta (0)=t$, define the Cauchy principal value by deleting a symmetric interval $|s|<\eta$ and taking $\eta \downarrow 0$. This pairs the two singular sides at equal local arclength.

Near $t$, write $f(\zeta )=f(t)+[f(\zeta )-f(t)]$. The second term divided by $\zeta -t$ is regular because $f$ is analytic there. For the constant term, indent the contour on either side of $t$ by a small semicircle. Its limiting contribution to the Cauchy integral is $\pm f(t)/2$, while the remaining contour tends to its principal value. This proves the Sokhotski–Plemelj theorem:In particular $w_{+}-w_{-}=f$ and $w_{+}+w_{-}=g$ for the given integral equation.

Define $W=w$ inside and $W=-w$ outside. Its jump is $W_{+}-W_{-}=w_{+}+w_{-}=g$, and it vanishes at infinity. ThereforeIndeed, the difference between the two sides has no jump, extends analytically across the smooth contour, and vanishes at infinity; Liouville's theorem makes it zero. Applying the Sokhotski–Plemelj theorem again,Thus the normalized closed-contour singular integral operator is its own inverse.