0
$\begingroup$

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.

$\endgroup$
19
  • $\begingroup$ Actually, if I got your problem right, the only thing bothering you is $\int_C \frac{z \mathrm{d}z}{1-z^2}$. But the limit of the integrand when $z \to 0$ is $0$, so by a classical complex analysis theorem $\frac{z}{1-z^2}$ can be extended to an holomorphic function on all of the open disk, which is simply connected. There you can use the result that the integral is zero. $\endgroup$
    – Junkyards
    Sep 28, 2017 at 19:36
  • $\begingroup$ @Junkyards: I am not aware of this theorem (though I can see the intuition for it), but it's good to know such a thing exists and that the result was justified! I am also bothered by the other integral $\int_c \frac1{z}dz$. Why is it that we can apply the Cauchy integral formula here to get that it is $2 \pi i$ when the domain is not simply-connected and $z_0 = 0$ is not an interior point of the domain? $\endgroup$ Sep 28, 2017 at 19:45
  • $\begingroup$ $\frac{z}{1-z^2} = \sum_{k=0}^\infty z^{2k+1}$ is analytic on the simply connected domain $|z| < 1$. Moreover $\int_C z^{2k+1}dz = \frac{z^{2k+2}}{2k+2}|_{C(0)}^{C(1)} = 0$ so that for $C \subset |z| \le 1-\epsilon$ (to ensure everything converges absolutely) $$\int_{C} \frac{z}{1-z^2}dz = \int_{C} \sum_{k=0}^\infty z^{2k+1} dz = \sum_{k=0}^\infty\int_{C} z^{2k+1} dz=\sum_{k=0}^\infty 0 = 0$$ Also you really need to prove by hand that $\int_C \frac{1}{z}dz = 2i\pi$ (using that $\log z$ is a local primitive) $\endgroup$
    – reuns
    Sep 28, 2017 at 19:56
  • $\begingroup$ @reuns: I understand that it is analytic on the simply connected domain $|z| <1$, and like your way of thinking about it, but would that alone imply that $\int_{0 < |z| < 1} f(z)dz = \int_{|z|<1} f(z)dz$? My problem is you still seem to be using Cauchy's theorem (now on $z^{2k+1}$), but we are still not on a simply-connected domain as the theorem requires $\endgroup$ Sep 28, 2017 at 19:57
  • $\begingroup$ You can see @reuns proof on another simpler way to see why the function is holomorphic, though I find quite a bother to recalculate the integral. For $\int_C \frac{1}{z} \mathrm{d}z$, you're making an honest mistake : it is not about $z \mapsto \frac{1}{z}$ being holomorphic on the open disk, it is about $z \mapsto 1$ being holomorphic on the open disk $\endgroup$
    – Junkyards
    Sep 28, 2017 at 19:58

1 Answer 1

-1
$\begingroup$

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 $0$.

$\endgroup$
2
  • $\begingroup$ Thank you for the clarification. So it seems that we can apply Cauchy's theorem if $f$ is analytic on $C$ and its interior, regardless of whether or not the domain we are working on is simply-connected, as long as $C$ does not cross any holes in the domain (and similar statement for Cauchy's integral formula). Is this correct? $\endgroup$ Sep 29, 2017 at 2:25
  • $\begingroup$ @infinitylord It's more the fact that if two functions $f$ and $g$ are defined on a domain that includes $C$, and $f(c) = g(c)$ for all $c \in C$, then necessarily $\oint_C f = \oint_C g$. If one of $f$ or $g$ happens to be defined on a domain with nice properties, then you can bring all the theorems you know to the study of that function; the contour integrals will always be equal. $\endgroup$
    – Joppy
    Sep 29, 2017 at 2:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .