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I'm learning some basic complex analysis and came across this integral $$\int_{-i}^{i} \frac{dz}{z}.$$

First of all, Wolfram can't calculate it, but it might be because he treats $i$ like a real parameter (BTW, is there a way to tell Wolfram how to calculate contour integrals along a specified contour?).

Since the function is analytic on any domain that doesn't contain the origin, the integral doesn't depend on path choice there. But there's a problem with the antiderivative, $\log z$, which is multi-valued, and I don't know exactly how to deal with that. So I'm not sure if I can use the fundamental thm. of calculus there. Evaluating by direct parametrization, for instance choosing an anti-clockwise circular path from $-i$ to $i$ gives the answer $i \pi$. But, unless I made an error, the same integral along the clockwise circular path gives the answer $-i \pi$...?

The excercise says to use Cauchy's integral formula, which gives me that ($C$ being a clockwise circular contour around the origin) $$2 \pi i = \oint _C \frac{dz}{z} $$

but this only gives me the same thing I got via parametrization... Is the excercise posed like this on purpose to get you thinking, or am I doing a mistake somewhere?

Spine Feast
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    It entirely depends on the contour. If it's meant to be a straight line, then the integrand is certainly not holomorphic - in fact, the integral won't even exist in any "nice" sense, since it's similar to integrating $\frac 1 x$ between $-1$ and $1$. –  Sep 09 '13 at 06:53
  • The value of the integral is uniquely determined only up to integer multiple of $2\pi i$. If the domain of $z^{-1}$ is restricted to a simply connected domain not containing the origin, then the integral has a unique value. – Sangchul Lee Sep 09 '13 at 06:57
  • Just compare with the integral $\int_{1}^{z}\frac{dz}{z}$. – Mhenni Benghorbal Sep 09 '13 at 07:11

1 Answers1

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The integral DOES depend on the path choice, precisely because there is a singularity at $0$ ; that is why Wolfram will not calculate it, because you didn't specify what this integral means. As it is, your integral is not well defined.

The reason for it is this : given a path which goes from $-i$ to $i$, assuming it does not loop around $0$, if it goes to the left of $0$ it will give you some answer $x$, and if it goes to the right of $0$ it will give you $x - 2\pi i$. (I think $x$ is $i \pi$ but I didn't bother ; the point is they will not be the same.)

So you need a path to compute this integral.

Hope that helps,

Patrick Da Silva
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  • I made this picture: http://i.imgur.com/DEqLLfI.jpg . What you're saying is that on $D_1$ the value of the integral doesn't depend on path choice, and likewise on $D_2$, but the values are different ? Why exactly does this happen then? Is it only because of the singularity or is it because of the multi-valued antiderivative? – Spine Feast Sep 09 '13 at 07:19
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    @DepeHb : Both the contours you drew on this picture give you an integral of zero, because the function $1/z$ is analytic inside the contours you have drawn. What I said in my answer is given a path from $-i$ to $i$, I did not say given a contour which surrounds both $i$ and $-i$ (and you drew such countours). The integral you wrote above usually denotes a certain curve $C$ who is parametrized by $\gamma : [0,1] \to \mathbb C$ such that $\gamma(0) = -i$ and $\gamma(1) = i$. In other words, it is not a loop, it has no "inside". Feel free to ask for more details. – Patrick Da Silva Sep 09 '13 at 07:47
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    What I drew were meant to be domains, not controus. And in these domains, one can then consider appropriate contours. Then given any two points in, say, $D_1$, the integral of the function evaluated between them doesn't depend on the contour connecting them? – Spine Feast Sep 09 '13 at 07:49
  • @DepeHb : I think you are misunderstanding the result ; if two points are in a domain $D$ which is as you drawn, then two points in this domain $D$ relied by a contour can be integrated and the integral does not depend on the choice of the contour, assuming that this contour stays within $D$. If you change the domain $D$ and take a contour in say $D'$ where your two points also belong, you might have changed the value of the integral, as in this example. – Patrick Da Silva Sep 09 '13 at 23:25
  • That's exactly what I wanted to know, thanks. So for arbitrary $z_1, z_2$, not just $\pm i$, the integral $\int_{z_1}^{z_2}dz/z$ is path-dependent, right? One more question: what property of $1/z$ makes it so? Is it just the singularity or is it the multi-valued antiderivative, or some combination of both? I still don't exactly understand the relationship between $1/z$ and $\log z$. – Spine Feast Sep 10 '13 at 07:30
  • So $1/z$ is analytic (i.e. admits a local Taylor series expansions around any point $a \neq 0$ of the form $\sum_{k \ge 0} a_k (z-a)^k$ which is valid for some $|z-a|<r_a$ ball). But at $z=0$ it loses this property ; this is because a function which is analytic at $a$ satisfies the property that the integral around some small circle surrounding $a$ is zero, and $1/z$ does not satisfy that. (You should try to see why path-independence of the integral is equivalent to having zero integral around some contour.) This is why $0$ is a "singularity" of $1/z$. – Patrick Da Silva Sep 10 '13 at 23:28