Cauchy-type integral

Remarks

The function $\Phi$ is holomorphic on $\C \setminus \Gamma$, and $\Phi(z) \to 0$ as $z \to \infty$. Example: let $\Gamma$ be the unit circle, oriented counterclockwise, and $\varphi(\tau) = \tau^{-1}$. For $\abs{z} < 1$, the integrand $\frac{1}{\tau(\tau - z)} = \frac{1}{z} \left( \frac{1}{\tau - z} - \frac{1}{\tau} \right)$ gives $\Phi(z) = \frac{1}{z}(1 - 1) = 0$; for $\abs{z} > 1$ it gives $\Phi(z) = \frac{1}{z}(0 - 1) = -\frac{1}{z}$. So the boundary values from inside and outside are $0$ and $-\tau^{-1}$, and their difference is $\varphi(\tau)$. Non-example: for $z \in \Gamma$ the formula is not a convergent integral in general. The reason is that $\frac{1}{\tau - z}$ is not integrable along $\Gamma$ near $\tau = z$. On $\Gamma$ one uses instead the Cauchy principal value, as in singular integral equations. Let $\Gamma$ be a closed curve bounding a domain $D$, and let $\varphi$ be the boundary value of a function $F$ holomorphic on $D$ and continuous on its closure. Then $\Phi = F$ on $D$ and $\Phi = 0$ outside $\overline{D}$, by Cauchy's integral formula and Cauchy's theorem. The boundary values of $\Phi$ for Hölder continuous $\varphi$ are given by the Sokhotski–Plemelj formulae.