I was trying to show that the function $f(z) = \frac1{z(1-z^2)}$ does not have an indefinite integral on the annulus $\mathbb{A} = \{z \in \mathbb{C} : 0 < |z| < 1 \}$
Indefinite integrals must be path independent, so by nature, any closed curve in the domain must be $0$ if $f$ is to have an antiderivative. Thus, we need to find a curve $C$ such that:
$$\oint_C \frac1{z(1-z^2)}dz \ne 0$$
and we are done.
My professor chose to split this integral up into two parts:
$$\oint_C \frac1{z(1-z^2)}dz = \oint_C \frac1{z} dz + \oint_C \frac{z}{1-z^2}dz$$
and chose $C$ to the disk of radius $1/2$ centered at the origin.
His claim was that the $\frac{z}{1-z^2}$ is analytic in $\mathbb{A}$, and so the closed contour integral must be $0$ by Cauchy's theorem.
As I have learned the theorem, we need $f$ to be analytic on a simply-connected domain, but here, $\mathbb{A}$ is not simply connected. How can we justify this? Can we somehow argue that the integral doesn't change if we were to add the hole in at $z = 0$ to the domain?
Then, he said that we can apply Cauchy's integral formula:
$$2 \pi i f(z_0) = \oint_C \frac{f(z)}{z-z_0}dz$$
with $f(z) = 1$ and $z_0 = 0$ to get $$\oint_C \frac1{z} dz = 2 \pi i$$
Again, the Cauchy integral theorem as I have learned it depends on having a simply-connected domain. Moreover, it requires that $z_0$ be an interior point on the domain, but $z_0 = 0$ is not contained in the domain. Why does this work?
If we assume both are true, we get
$$\oint_C \frac1{z(1-z^2)}dz = 2 \pi i \ne 0$$
Which is what we want, but I'm very troubled by these inconsistencies. Alternative proofs are always nice, but I'm much more interested in why his method works (assuming it does) so that I can get a better understanding of the scope of these theorems.
If it is at all relevant to your explanation, please note that I do not know Residue theorem.
Best Answer
Let $f: \mathbb{A} \to \mathbb{C}$ be defined as above, and $C$ be the circle of radius $1/2$ centred at the origin. We want to show that the integral
$$ \oint_C f(z) dz = \oint_C \frac1{z} dz + \oint_C \frac{z}{1-z^2}dz$$
is nonzero. Firstly, the value of the integral only depends on the values $f$ takes along $C$, and so for the purposes of evaluating the integral we can forget that $f$ used to be defined on $\mathbb{A}$, and treat it as being defined on (a neighbourhood of) $C$.
By applying Cauchy's integral formula to the function $g(z) = 1$ with $z_0 = 0$, on the simply-connected domain $\mathbb{C}$, we can find that
$$2 \pi i = \oint_C \frac1{z} dz$$
Since the value of the contour integral only depends on the values that $1/z$ take along the circle $C$, this result is still valid in our case.
For the remaining integral, notice that the function $h: \mathbb{D} \to \mathbb{C}$, $h(z) = \frac{z}{1-z^2}$ (where $\mathbb{D}$ is the open unit disk centred at the origin) takes the same values as the integrand along the curve $C$. Therefore,
$$ \oint_C h(z) dz = \oint_C \frac{z}{1 - z^2} dz$$
Since $h(z)$ is analytic on the simply-connected domain $\mathbb{D}$, by Cauchy's integral theorem the integral is $0$.