## Contour integration

#### §1. Definition of the contour integral

One of the topics studied in complex analysis is (not surprisingly) integrals in the complex plane. To be more specific, integration of complex-valued functions of a complex variable — along “closed paths” in the complex plane, that is, curves whose initial and terminal points coincide. This is known as contour integration. Trust me, the basic concept is quite easy.

Curves in the plane are described by nothing more than functions $\gamma : [a,b] \to \mathbb{C}$ which are in some sense “well-behaved” (I won’t go into the details since I want to be quick with this). So we define the contour integral of a function $f : \Omega \to \mathbb{C}$ around the curve $\gamma$ as follows:

$\displaystyle \oint_\gamma f(z) \, dz := \int_a^b f(\gamma(t)) \gamma'(t) \, dt$.

Of course, this only makes sense if firstly $\gamma([a,b]) \subseteq \Omega$: the function needs to be defined on the whole image of the curve $\gamma$ (for convenience, people often refer to the image of $\gamma$ simply as $\gamma$ itself, even though technically $\gamma$ is a function). $f$ itself also has to satisfy an integrability condition, of course. Also, you may be scratching your head wondering if this is well-defined — well, it turns out that the value of the quantity above does not change under reparametrization of the curve in question, so we’re good :D.

Anyways, luckily we don’t need to nitpick about small things like “integrability of $f$“, since we’re usually interested in integration of functions that are holomorphic (complex-differentiable) almost everywhere (that is, except for a couple of bad points — singularities).

Plot of the magnitude of the gamma function, showing poles

#### §2. Singularities — the bad guys

So what, you ask, is a singularity? Well, the easiest way to explain it is as follows. It turns out that similar to the way we can perform Taylor expansions of a “nice” function on some disc around one of its points, we can do something similar on an annulus around a point $a \in \Omega$. We write $\mathcal{A}(a,r_1,r_2)$ for an annulus centered at $a$ with radii $0 \leq r_1 < r_2$. It is defined like this:

$\displaystyle \mathcal{A}(a,r_1,r_2) := \{ z \in \mathbb{C} : r_1 < |z-a| < r_2 \}$.

Note that an annulus around $a$ never actually contains $a$. If $r_1 = 0$ then you obtain what is called a punctured disc of radius $r_2$ centered at $a$.

The expansions we can perform on annuli are not in general Taylor expansions, however, but things called Laurent expansions. These differ from Taylor expansions in the sense that we can have negative-power terms. So, in general, a Laurent series looks like

$\displaystyle \sum_{n=-\infty}^\infty a_nz^n$.

So, it’s not quite a Taylor series, but I guess it’s the next best thing we could ask for.

We then define the different kinds of singularities (in ascending order of atrocity): removable singularities, poles, and essential singularities. A function is never defined at a singularity, hence why they are called singularities. Anyways, a removable singularity is a point where the function is undefined, such that we can choose a value for the function to take at that point which will cause the function to remain holomorphic there. These are kind of like the “holes” people talk about in one-variable calculus. A function is said to have a pole of order $m$ at a point $a$ if the coefficients in the Laurent expansion about $a$ all become zero for powers of $(z-a)$ less than $-m$. These would be similar to the “vertical asymptotes” of one-variable calculus. The point $a$ is finally called an essential singularity if the Laurent expansion about $a$ has an infinite number of nonzero negative-power coefficients (this is really, really bad). In fact, I don’t think there’s anything quite so awful in single-variable calculus. The function completely spazzes out near an essential singularity. In any punctured disc around an essential singularity, it not only takes on every single value in the entire complex plane except one (!!!), but it hits all these values infinitely many times. It’s hard to even picture this kind of erratic behaviour.

#### §3. Residues

A very surprising result known as Cauchy’s integral theorem tells us that the contour integral of a holomorphic function defined on an open, simply connected set $\Omega \subseteq \mathbb{C}$ around any closed path is 0. But often, the functions we want to integrate aren’t going to be holomorphic in the whole interior of the curve. There’s going to be some singularities enclosed by the curve. It turns out that when this is the case, the integral can be easily calculated, by inspecting the behaviour of the function near these singularities.

By “behaviour”, I mean a very important quantity associated to a singularity of a given function, known as the residue of $f$ at that point $a$, denoted $\mathrm{Res}(f,a)$. The definition is simple:

The residue of $f$ at $a$ is merely the coefficient of $(z-a)^{-1}$ in the Laurent expansion of $f$ at $a$.

We have the following theorem, known as the residue theorem:

$\displaystyle \oint_\gamma f(z) \, dz = 2\pi i \sum_{a} \mathrm{Res}(f,a)$,

where the sum is taken over all singularities $a$ in the interior of the curve. This theorem is basically a life-saver for evaluating integrals, since it tells us that a contour integral around a curve which encloses some singularities does not actually depend on the shape of the curve, but only on local information attached to the singularities.

just another guy trying to make the diagrams commute.
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### 3 Responses to Contour integration

1. (Reading archived articles on your blog at work is my new hobby now, so don’t hate =P)

Anyways, just a few things:

1. Your definition of a Laurent expansion seems to just be centered around 0 rather than a.

2. What’s the difference between a holomorphic function and analytic function (both defined on some closed interval [a,b]) ?

3. @Simply connected: I don’t have any background in topology, but is the simply connected property only possible on n-manifolds that are topological equivalent (homeomorphic) to n-spheres by the Poincare [conjecture] theorem? If so, taking the contour integral of a function defined on an open unit sphere (I’m not sure if you can do this) over a closed path always equals 0?

4. Are there any cool functions that contain a (or several) essential singularities? 😀

@Residue theorem: … Wow. Although I’m not too sure what would be more tedious, doing an [contour] integral or finding the singularities, finding the Laurent series and the coefficients (through another contour integral] and then summing them.

2. dx7hymcxnpq says:

1. You’re right. I should probably change that.
2. We are talking about complex-valued functions defined on open subsets of the complex plane here. It turns out that a function is holomorphic on such a region if and only if it is analytic there (the two things have different meanings; the former refers to complex-differentiability and the latter refers to the ability to locally represent it as a power series).
3. I wrote a post on simple connectedness too. I don’t have experience with manifolds, but you basically answered your own question. The statement of the Poincare conjecture says that (in 3 dimensions at least), if a manifold satisfies some hypotheses and is simply connected, then it’s homeomorphic to a sphere, etc. As I said in complex analysis you usually look at functions defined on subsets of the complex plane… so I’m confused about your “integral over a region of a sphere”.
4. Yeah, for sure. I think sin(1/z) has an essential singularity at 0, same with exp(1/z). An essential singularity occurs if and only if the Laurent expansion about that point has infinitely many nonzero negative-degree coefficients. You can probably figure out more functions just by using that fact, and what you know about power series already (try plugging in 1/z to certain functions with non-terminating Taylor series etc…)

The Residue theorem and Cauchy’s integral formula are actually ridiculously invaluable for computing contour integrals, since doing it from the definition often tends to be pretty much impossible…

3. @3: That was probably my geometric intuition screwing with me there. I was thinking it was possible to embed some kind of topology on a complex “2-sphere” (like spherical geometry) sitting in $\mathbb C^3$ space and take contour integrals so that they would always be 0 for any closed path on the sphere… okay, so that was just a bit of my insanity slipping through there, but maybe I’ll make sense of it once I learn topology and complex analysis. =P