Cauchy path integral as a linear operator: kernel and image? Let $\mathcal O(\Omega)$ be the algebra of functions holomorphic on the open set $\Omega\subset\mathbb C$. For $\gamma$ a simple compact curve in $\mathbb C$ consider the linear operator given by path-integrating against the Cauchy kernel $$ \Gamma : \mathcal O(\mathbb C) \longrightarrow \mathcal O(\mathbb C\setminus \gamma) \\ f\longmapsto \left(z\mapsto \int_\gamma \frac{f(w)}{w-z}\mathrm{d}w\right)$$ 
In the trivial case where $\gamma$ is closed then $\Gamma$ is injective (apply Cauchy formula to all $z$ in the bounded connected component of $\mathbb C\setminus \gamma$) and its range consists in those functions that are zero outside $\gamma$ and the restriction of an entire function inside. 
My question: what happens if $\gamma$ is not closed (i.e. $\mathbb C\setminus \gamma$ connected)? Is it possible to characterize the range of $\Gamma$, besides the obvious $O(|z|^{-1})$ towards infinity?
(In this case $f$ may only be a germ near $\gamma$ of a holomorphic function, but keep it simple if necessary.)
I'd find it hard to believe that injectivity would fail (although I don't quite have an argument). 
If this question has already a well-documented answer (which is fairly probable but I didn't know which keywords to look for), please accept my apologies. A pointer to a reference would be great!
 A: Such an integral over a simple (non-closed) curve is called the Cauchy type
integral. It is convenient to define
$$F(z)=\frac{1}{2\pi i}\int_\gamma \frac{f(\zeta)d\zeta}{\zeta-z},\quad z\not\in\gamma.\tag{1}$$
The curve $\gamma$ is oriented, so for our function $F$ defined in $C\backslash\gamma$ and $z\in\gamma$ different from an endpoint,
we can talk of the right limit $F^+(z)$ and left limit $F^-(z)$.
Then we have Sokhotski formula
$$F^-(z)-F^+(z)=f(z),\quad z\in\gamma$$
As in addition $F(z)\to0,\; z\to\infty$, this shows that your map is injective. (If we have two functions with the same jump, their difference is entire, and since it tends to zero at $\infty$ it must be zero).
The image consists of those analytic functions in $C\backslash\gamma$ which
tend to $0$ as $z\to\infty$, have limits on both sides of $\gamma$ and the
jump between these limits is an entire function (analytic in $C$).
(At the end-points, the limits do not exist, and we have logarithmic singularities.)
With some regularity of $\gamma$ and $f$ (non necessary analytic) we have
$$F^+(z)=F^*(z)-\frac{1}{2}f(z),\quad F^-(z)=F^*(z)+\frac{1}{2}f(z),$$
where $F^*(z)$ is defined for $z\in\gamma$ using the principal value of the integral in (1).  The above formula is obtained by subtracting these two relations. But when $f$ is analytic on $\gamma$,
the curve can be slightly deformed (preserving its ends), and you get rid of
the regularity condition by approximating it with a smooth (or piecewise-linear)
curve with the same ends.
If the curve happens to be closed then $F^+=0$ and we recover the result for
the closed curve which you cite.
It is interesting that the curve $\gamma$ plays little role in the description
of the image: functions from the image have analytic continuation along
any curve starting and ending at $\infty$ and not passing through the endpoints of $\gamma$.
References. This material is standard and is included to every Russian undergraduate Complex Variables text, but for some strange reasons none of them is translated into English. A good reference in English which comes to my mind is a translation from the Japanese:
MR1026013
Kaneko, A.
Introduction to hyperfunctions.
Kluwer, Dordrecht, Tokyo, 1988.
