In the annulus $0<|z-c|<R$, $$f'(z)=\sum^\infty_{k=-\infty}a_k(z-c)^k$$ where $a_{-1}\ne0$.
Because the series is uniformly convergent within the annulus of convergence, after performing term-by-term indefinite integration, one obtains (for $0<|z-c|<R$):
$$f(z)=K+a_{-1}\log(z-c)+\sum^\infty_{k=-\infty,\,k\ne0}\frac{a_{k-1}}k(z-c)^k$$ where $K$ is the constant of integration.
But, how can I determine the branch cut of the logarithm? Is it arbitrary?
My attempt:
In complex analysis, the antiderivative of $f'(z)$ should be written formally as $$f(z)=\int f'(z)dz=\int_0^1f(\gamma(t))\gamma'(t)dt$$ where $\gamma(t):[0,1]\to\mathbb C\setminus\{c\}$, $\gamma(0)=\text{arbitrary constant } L$, $0<|L-c|<R$, $\gamma(1)=z$.
If $\gamma$ goes around $z=c$ positively, doing the integration would give us one of the possibilities of $f(z)$, call it $f_1(z)$. If $\gamma$ goes around $z=c$ negatively, we obtain another function $f_2(z)$, with $f_2(z)=f_1(z)-2\pi ia_{-1}$.
It appears to me that if we walk from $L$, go around $c$ positively once, and go back to $L$, we would get an extra $2\pi ia_{-1}$, thus the branch cut must connect $c$ and $L$.
This argument is quite vague. Moreover, I've seen a theorem that guarantees the region of convergence of an uniformly convergent series would remain unchanged after term-by-term integration. However, in this case, any branch cut would remove continuity of a line of points, and $f(z)$ no longer has the original size of annulus of convergence.
Can you help me to determine the relation between $L$ and the choice of branch cut, and help me to solve the above contradiction?
Thanks in advance.
You can write the derivative of the intended $f$ in the form $$f'(z)={a_{-1}\over z-c}+\sum_{k\ne-1}a_kz^k={a_{-1}\over z}+g'(z)\ .$$ Here $g'$ is the derivative of a bona fide analytic function $g:\>\Omega\to{\mathbb C}$, where $\Omega$ is the given annulus. It follows that $$f(z)=a_{-1}\ {\rm ''}\log(z-c)\>{\rm ''}+g(z)+K\ ,$$ whereby we now have to make some sense of the $\ {\rm ''}\log(z-c)\>{\rm ''}$ appearing here. After fixing an initial (admissible) value at some point $z_0\in\Omega$ the $\log(z-c)$ picks up an additive constant $2\pi i$ with every complete loop around the point $c$. In order to obtain a true analytic function $f$ we have to shrink $\Omega$ to a subdomain $\Omega'$ that does not contain such loops. How you do this is arbitrary (you could even draw a logarithmic spiral with center $c$). Make the cut that is most practical in the case of your concrete situation. You should be forced to "look over the cut" as little as possible.