Even holomorphic function on the punctured disk has a primitive Let $f$ such that $f$ is holomorphic on $\{z|0<|z|<1\}$, and $f$ is even.
I need to show that f has a primitive. Any ideas?

Let $0 < r < 1$. Then we have
$$\int_{\partial B_r(0)} f(z)\, dz = 
\int_0^{2\pi} f(re^(it)ire^{it}\, dt
=  \int_0^{\pi} f(re^{it})ire^{it}\, dt + \int_\pi^{2\pi} f(re^{it})ire^{it}\, dt.$$
For the second integal we have
$$ \int_\pi^{2\pi} f(re^{it})ire^{it}\, dt
=  \int_0^{\pi} f(re^{i(t+ \pi)})ire^{i(t+ \pi) }\, dt
=  -\int_0^{\pi} f(-re^{it})ire^{it }\, dt=
-\int_0^{\pi} f(re^{it})ire^{it }\, dt,$$
by the evenness of $f$. 
By homotopy, every simple loop avoiding 0 integrates to 0.   Now pick a any point $z_0$ in the punctured disk.  Define $F(z) = \int_\gamma f(z)\, dz,$
where $\gamma$ is a path connecting $z_0$ to $z$.  This is well-defined because of the fact for that every loop in the punctured disk, the integral along the loop integrates to zero.
