Integrating Laurent series termwise I have some trouble understanding termwise integration of a Laurent series. Say that $f$ is a complex valued function, which is analytic on some annulus $D = \{z\in \mathbb{C}\mid R<|z-z_0|<S\}$. Then it follows that $f$ may be expanded in a Laurent series as follows:
$$f(z) = \sum_{n=-\infty}^{\infty}a_n(z-z_0)^n,\quad z\in D.$$
I understand that since the convergence is uniform on any compact annulus contained in $D$ it follows that for any curve $\gamma$ in $D$ we have that
$$\int_\gamma f(z)dz = \sum_{n=-\infty}^{\infty}a_n\int_\gamma (z-z_0)^n.$$
But say we want to integrate something where the path is not given. If we are given $z_1,z_2\in D$ then what is
$$\int_{z_1}^{z_2}f(z)dz?$$
In general this is not defined since the integral is not path independent. This brings me to my question which is related to another question on the forum:
Primitive of $f(x)$ on the punctured plane where $f(x)$ is analytic
Proposition Suppose that $R = 0$ and $S=\infty$ above. Then there is a constant $A$ such that $f(z)-A/z$ has a primitive on $D$. 
As is directly clear the problem occurs for the index $a_{-1}$ since $z^{-1}$ does not have a primitive defined on the whole of $D$. So if we let $A = a_{-1}$ then each term in the series has a primitive on $D$. But how do we integrate termwise formally? The integral
$$\int_{z_1}^{z}(f(\zeta)-a_{-1}/\zeta) d\zeta$$
need not be well defined? Do we integrate first and then conclude that this integral is indeed path independent?
Path dependence
 A: Let's write
$$g(z) = f(z) - \frac{a_1}{z-z_0} = \sum_{n \neq -1} a_n(z-z_0)^n.$$
We can use that for every path $\gamma$ from $z_1$ to $z$ in $D$ the Laurent series converges uniformly on the trace of $\gamma$ to pull the sum out of the integral, and obtain
$$\int_{\gamma} g(\zeta)\,d\zeta = \sum_{n \neq -1} a_n \int_{\gamma} (\zeta - z_0)^n\,d\zeta. \tag{1}$$
Since every integrand on the right hand side of $(1)$ has a primitive on $D$, we know that each of the integrals there is path-independent, and hence the total integral is path-independent. Thus the integral
$$\int_{z_1}^z g(\zeta)\,d\zeta$$
is well-defined, and hence gives a primitive of $g$.
Alternatively, one can look at the Laurent series
$$G(z) = \sum_{m \neq 0} \frac{a_{m-1}}{m}(z - z_0)^m.\tag{2}$$
Since the Laurent series of $f$ converges absolutely and locally uniformly on $D$, it follows that the series $(2)$ converges absolutely and locally uniformly on $D$. Thus $G$ is a holomorphic function on $D$, and by the properties of Laurent series we can differentiate termwise, obtaining
$$G'(z) = \sum_{m \neq 0} a_{m-1}(z-z_0)^{m-1} = \sum_{n \neq -1} a_n (z-z_0)^n = g(z).$$
Thus we have a primitive of $g$ on $D$, and hence it follows that
$$\int_{z_1}^z g(\zeta)\,d\zeta = G(z) - G(z_1)$$
is independent from the path connecting $z_1$ and $z$.
