Let $D \subseteq \mathbb{C} $ be bounded and simply connected, $\Gamma:= \partial D \in C^2 $, $\phi, \psi \in C^{0,\alpha}(\Gamma)$,
$$
f(z):= \frac{1}{2\pi i} \int\_{\Gamma} \frac{\phi(\zeta)}{\zeta - z} d\zeta,\quad g(z):= \frac{1}{2\pi i} \int\_{\Gamma} \frac{\psi(\zeta)}{\zeta - z} d\zeta，\quad z \in \mathbb{C} \setminus \Gamma
$$
and
$$
f\_{+}(z) := \lim\_{h \to +0} f(z+ hv(z)), \quad f\_{-}(z) := \lim\_{h \to +0} f(z - hv(z)), \quad z \in \Gamma \\
$$
where $v(z)$ denotes the unit exterior normal at point $z$.  

Why
$$
\int\_{\Gamma} (f\_{-}g\_{-} - f\_{+}g\_{+}) dz = - \int\_{\vert z \vert = R} f(z)g(z) dz 
$$
holds for any sufficiently large $R$?


Note: This is one step in [Kress, Linear integral equations, P114].



Limiting behaviour of Cauchy integral near boundary
You did not say what $v$ is so I make a guess: $v(z)$ is the outer normal to $\Gamma$. Your $f,g$ are piecewise analytic:
inside $D$ they are equal to $f\_-, g\_-$ while outside $D$ they
are equal to $f\_+,g\_+$.
Then
$$\int\_{\Gamma}f\_-g\_-=0$$
by Cauchy theorem (apply it first to some small deformation of $\Gamma$
to the inside of $D$, and then pass to the limit, while a similar argument gives
$$\int\_{\Gamma}f\_+g\_+=-\mathrm{res}\_\infty f\_+g\_+=\int\_{|z|=R}f\_+g\_+=\int\_{|z|=R}fg,$$
when $R$ is large enough.
Your formula follows.


Remark. The ``Cauchy-type integral''
$$f(z)=\frac{1}{2\pi i}\int\_\Gamma\frac{\phi(\zeta)d\zeta}{\zeta-z}$$
which is defined for $z\not\in\Gamma$ is a standard way to
represent a (non-analytic) function $\phi$ on $\Gamma$ as a difference of (boundary values of) two functions $f\_+$ analytic outside $\Gamma$ and $f\_-$ analytic inside.
This simple and important theorem due to Julian Sochocki was taught in all standard undergraduate complex variables courses in Soviet Union, but for some reasons it is not included in textbooks in other countries. So I refer
on [Wikipedia](https://en.wikipedia.org/wiki/Sokhotski%E2%80%93Plemelj_theorem).


