# How shall we call these types of integrals in Complex Analysis?

1. Sep 5, 2011

### jackmell

$$\mathop\int\limits_{\infty} \log[(z-1)(z+1)]dz=A(z)\biggr|_0^0=4\pi i$$

The infinity symbol below the integral is a positive-oriented, closed, and differentiable path over the function looping around both branch-points and A(z) is the antiderivative of the integrand. I mean would that hold for all of them? You know, like 2 pi i times the number of branch-points? Don't know but probably depends on what direction you're going around them. Integrals like this I think are interesting but not really discussed in classes right? I mean, when have you ever seen something like this anyway? Not really in books and yet it is perfectly consistent with the Fundamental Theorem of Calculus. Isn't it? I think we should explicitly identify them as a particular type of integral more than just a 2-dimensional contour integral. Maybe a 3-D contour integal but that's too wordy. What do you guys think? Have they already been named and I just don't know?

Last edited: Sep 5, 2011
2. Sep 5, 2011

### HallsofIvy

Staff Emeritus
What do you mean by "looping around both of them"? Do you mean, for example, a circle that contains both of them or are you thinking about an actual "figure 8" path so that one pole is contained in one lobe and the othe pole in the other lobe?

In the first case, the integral is $2\pi i$ times the sum of the residues. In the second case, because we are going around one pole counter-clockwise and the other clockwise, the integral is $2\pi i$ times the difference of the residues. (The one you go around clockwise subtracted from the one you go about counter clockwise.)

You might not realize that you have seen such things, and no special name is given to them, because the simplest thing to do is, as I did, to "break" the figure 8 at the cross over point and do it as two separate integrals.

3. Sep 5, 2011

### jackmell

Hall, those are not poles so thus have no residues so I believe the Residue Theorem cannot be applied here. I'm using strictly the value of the antiderivative evaluated at the end point minus the start point to arive at the value given:

$$A(z)\biggr|_0^0=\biggr(-2z-\log(1-z)+\log(1+z)+z\log(z^2-1)\biggr)_0^0=4\pi i$$

Anoying isn't it? That's part of the point I wish (hope) to make.

Last edited: Sep 5, 2011