Let $f : \Bbb C \setminus \{0 \} \longrightarrow \Bbb C$ be a holomorphic function such that $\int_{\gamma} f(z)\ dz = 0$ for any closed curve $\gamma$ in $\Bbb C \setminus \{0 \}.$ Does there exist an entire function $F : \Bbb C \longrightarrow \Bbb C$ such that $F'(z)=f(z)$ for all $z \in \Bbb C \setminus \{0 \}$?
I think it may not be true. If it is true then $f$ has a removable singularity at $z=0$ since $F'(z)$ is also entire as $F$ is so. But there exist functions which are holomorphic in $\Bbb C \setminus \{0 \}$ and contour integration of those functions over any closed contour in $\Bbb C \setminus \{0 \}$ is zero but they don't have removable singularity at $z=0.$ For instance we may take the function $f(z) = \left (e^{\frac {1} {z}} \right )' = - \frac {1} {z^2} e^{\frac {1} {z}},\ z \in \Bbb C \setminus \{0 \}.$
Is my reasoning correct at all? Would anybody please verify it?
Thank you very much.
That is correct. Under the given condition there exists an antiderivative $F$ in $\Bbb C \setminus \{0 \}$, but that is not necessarily an entire function. You already gave one example, another example is $f(z) = \frac{1}{z^2}$ which is the derivative of $F(z) = -\frac{1}{z} + C$ for any constant $C$.
Generally, If $f$ is holomorphic in $\Omega$ with $\int_{\gamma} f(z)\ dz = 0$ for all closed curves $\gamma$ in $\Omega$, then (and only then) $f$ has an antiderivate $F$ in $\Omega$. If $f$ has an isolated singularity at some point $z_0$ then $F$ has an isolated singularity at $z_0$ of the same nature (essential singularity, pole, or removable singularity). That can easily be seen from the Laurent expansion of $F$ at $z_0$.