# Thread: Residue Example in Palka: Is it correct?

1. ## Residue Example in Palka: Is it correct?

I've read through Example 3.10 of Bruce Palka's text "An Introduction to Complex Function Theory" (p. 337). The example uses the residue theorem to compute $\displaystyle \int_{0}^{\infty}{\frac{dt}{t^{\lambda} (t+b)}}$ where $\displaystyle 0 < \lambda < 1$ and $\displaystyle b > 0$.

Because of the contour he chose, he uses a holomorphic branch of the log function on $\displaystyle \matbb{C} \backslash [0,\infty)$ but then proceeds to take limits and claims to compute the integral on the positive real axis, precisely where the branch that he chose is discontinuous. I'm confused; how can this be justified?

2. I suppose he uses the keyhole contour?

It is perfectly legitimate to integrate along a branch cut as long as you know what you're doing! You have to remember that you are, in fact, integrating on the Riemann surface of the logarithm, which is a helicoid. Thus the two parts of the contour which lie above and under the real axis, although they appear to be very close to each other as drawn on the plane, actually lie on two different sheets of the helicoid; they just happen to both lie above the same line segment in the plane.

3. Palka's contour is more like an annulus minus a sector that straddles the postive real axis and the limit is taken so the sector vanishes. I also saw a similar proof where the contour resembled a keyhole (as in your link) but I thought it had the same gap. Thank you for the idea and the link! I don't quite see it yet but I'll pour through it in the next couple days and also see if I can clean up Palka's example.

4. Well, I spent the day on this post. I now learned a little about the helicoid construct of log(z) which is termed a Riemann surface. Also, the keyhole clearly works but Palka's modified keyhole (annulus minus sector) gives the right answer but the lines of the sectorized-annulus do not approach the branch cut uniformly; an important step involves uniform convergence of the limits that arise when the lines approach the branch cut so that the limit can "pass thru" the integral. So this may still be a gap for the Palka keyhole.

I'm talking myself out of the problem now given the keyhole contour. Cauchy's Theorem applies on the entire contour; the paths as the limits are taken are all homologous in $\displaystyle \mathbb{C} \backslash \{0,-b\}$. All the separate path integrals are well defined; the lines approach the branch cut and the lower one approaches the discontinuity but the limit is well defined. The integrals over the circular arcs that are "nearly circles" in the limits go to zero. It seems right. I thought there was a problem due to the branch cut but now I see no gaps in the logic of the analysis. If you see a problem, please let me know. However, ...

The comment by Bruno J. opened up an whole other line of reasoning; how is this integral actually related to this Riemann surface; in some isomorphic sense? I looked through my cpx analysis texts (including Ahlfors) and didn't find a discussion on it. The comment by Bruno J. is that we are actually integrating on this Riemann surface but I would say that we are actually integrating on the contour in $\displaystyle \mathbb{C}$ and using analysis to find the value of the integral (via limits and Cauchy's Theorem). The contour in $\displaystyle \mathbb{C}$ has a corresponding contour on the Riemann Sphere. But why would we say that we are integrating the function $\displaystyle f(z) = \frac{1}{z^{\lambda} (x+b)}$ on the Riemann Surface of $\displaystyle \log(z)$? Any comments or pointers to a good reference that deals with the details of this would be greatly appreciated.

5. ## Post getting stale; one last look

This post is getting stale with no new replies. I'm hopeful that someone can comment on the finale of the June 25th post; at least suggest where I can find detailed background. I'll give this a couple more days then mark this as resolved. Thank you.

6. Originally Posted by huram2215
If you see a problem, please let me know.
Seems fine!

Originally Posted by huram2215
The comment by Bruno J. opened up an whole other line of reasoning; how is this integral actually related to this Riemann surface; in some isomorphic sense?
I don't know what you mean by this.

Originally Posted by huram2215
I looked through my cpx analysis texts (including Ahlfors) and didn't find a discussion on it.
Most complex analysis textbooks do not even go as far as to introduce the concept of a Riemann surface; there is so much to say that they are usually treated separately, in books which assume a certain familiarity with complex analysis on the Riemann sphere (which is just the simplest Riemann surface).

Originally Posted by huram2215
The comment by Bruno J. is that we are actually integrating on this Riemann surface but I would say that we are actually integrating on the contour in $\displaystyle \mathbb{C}$ and using analysis to find the value of the integral (via limits and Cauchy's Theorem).
Well, both! On the one hand, you can consider just a fixed branch of the function and remember not to pass through the branch cut. Or you can just consider the function as defined on an appropriate surface on which it is single-valued, and work with this. It's pretty much equivalent for your purposes. The second viewpoint has the advantage of making a multivalued function's behaviour geometrically intuitive.