Built on Facts

A quick contour integral.

Note to the reader: this post is relatively stiff mathematically. For those not mathematically inclined, I think you might enjoy reading it anyway and enjoying it as you would a tour of a widget factory; even if you’re not worried about the details of the nuts and bolts, it’s fun to see it done.

While watching football yesterday, I got a phone call from a great friend whom I first met in our halcyon high school days. He’s a fellow physics student a few years younger than me, calling to discuss a problem in contour integration. Most people like watching football more than they like doing contour integrals, but not me. And I like watching football*, so that’s saying something.

This one is Arfken and Weber 6.3.4. It effectively asks us to compute this integral:

i-d18063683dcc0d42b9be45451a84d1e3-3.png

Where C is a circle in the complex plane, centered on the origin, with a radius strictly greater than 1. (Note: the problem is given before the section on the residue theorem, so we’re going to evaluate the integral directly rather than by the more direct method of residues.)

A person just encountering this integral might think “Ah! It’s a closed curve, so Cauchy’s integral theorem guarantees the answer is 0.” Unfortunately the theorem only holds for a function which is analytic within the closed curve, and this one isn’t. The denominator is zero for z = 0 and z = -1. Both of those are inside the curve, so the function isn’t analytic in our region and the theorem doesn’t apply. We can fix this problem by finagling our region in the following way: instead of just that one circle, we can dent the circle such that we dodge around those bad points and have a region inside which our function is analytic. The integral for our modified contour is thus 0 by Cauchy’s theorem. I propose some dents that look like this:

i-d3fdbda776403268fadccbd8598f89a3-graph3.png

This modified contour is composed of three sub-contour circles labeled C1, C2, and C3. They’re connected by straight lines. The complete contour fails to enclose any of the non-analytic points, so the integral over this modified contour is certainly 0 no matter how weird it looks.

Now the paths connecting the three circle regions have an interesting and important property. In the limit that the lines come very close together leaving the C1, C2, and C3 circles gap-free, the paths will overlap each other going in opposite directions. This means whatever the value of the integral over the ingoing line happens to be, it will be exactly the negative of its corresponding outgoing line integral. Thus the total contribution to the integral from the lines is zero. As such we can scrap the lines completely, as our single weird path is this exactly the same as the sum of these three sub-paths:

i-d8107f2fa66bfd8e5e6f522cd5b441bc-graph4.png

As a quick summary of what’s important so far: the total integral of the weird path is zero, and the straight lines contribute nothing at all. Thus the sum of the the integrals over the circles is zero. C1 + C2 + C3 = 0,if we take each term to be the value of the integral over that path. Important: notice that to be consistent with the weird path, the interior cicles are integrated in the opposite direction as the exterior circle. This is easy, it’s just an overall minus sign.

So let’s evaluate the interior circle integrals. First, C2. Let’s make our lives easier by breaking up the fraction in the integral by means of partial fraction expansion. (You probably already know how to do this; if not, don’t worry. It’s in any calculus book and it’s super easy.)

i-bb5c2c6b0452df43a61e3974bd9b473f-5.png

The second integral on the right is zero over the contour C2. Why? It’s analytic in that region: C2 doesn’t contain z = -1, which is where that integral fails to be analytic. Thus we only have to evaluate the first integral on the right.

It’s a fundamental fact in complex analysis that a complex number z can be written in a polar form, where:

i-1a518f2655b02f9d30ddb0c8226ad8eb-1a.png

Where r is the distance from the origin and theta is the angle with respect to the positive x axis, as usual. We’re going to want to make a substitution, so we also differentiate this to find:

i-ec6170d6ae0330319bd539b5264503e8-2a.png

We’re integrating over a circle, so theta runs from 0 to 2 pi. The radius r of the tiny circle we’re free to specify, but it won’t matter. Substitute into the integral:

i-5e62ab89294132669fadf517e98dfbb8-3a.png

The negative sign is because we’re integrating clockwise, remember? Now do the same thing for C3:

i-31a7f984f6bf3bd3a24b94b10ad187cf-4a.png

As with the similar argument for C2, the first integral on the right is zero – because C3 doesn’t enclose the bad point z = 0. Therefore we have only to evaluate that second term on the right. To do it, make a substitution. Let u = z + 1, which will also give us du = dz. This turns this integral into the exact same one as the integral we just did, modulo the extra minus sign from the partial fraction. Minus a minus is plus, so:

i-0e4dc42b12d67a787c04bfcd843a4212-5a.png

All right. We’re almost done. We know C1 + C2 + C3 = 0. We know C2 = -2πi. We know C3 = 2πi. Therefore:

C1 -2πi + 2πi = 0

And thus C1 = 0. C1 is the same contour as our original un-dented C, so the answer to our original problem is 0. This looks difficult, but that’s just because I’ve explained it in laborious detail. As far as contour integral problems go this one is pretty elementary. People write dissertations on the hard ones. Seriously – a person in my group is about to get a Ph.D. largely from work on a particularly hair-raising contour integral in the theory of classical electromagnetism.

*All three of my favorite teams – LSU, Texas A&M, and the New Orleans Saints – are undefeated at the moment. I don’t expect this to last, but I’ll take it while I’ve got it.

Comments

  1. #1 rob
    September 21, 2009

    complex analysis is only fun until someone loses an i.

  2. #2 Neill Raper
    September 21, 2009

    Aha, a fellow paper bagger. Well actually I couldn’t give a shit about sports but I have a mild loyalty built up from my dads rabid support of the saints. Although I guess if they are undefeated no paper bag is necessary.

  3. #3 kiwi
    September 21, 2009

    Or, writing R for the radius of the circle R,

    |1/(z^2 + z)| is less than 1/(R^2 – R)
    length of the contour is 2piR

    so the value of the integral is less than 2piR/(R^2 – R) which tends to zero as R tends to infinity.

  4. #4 Anonymous
    September 21, 2009

    Nice write up.

    Over the years I have repeatedly forgetten the signs for the residue theorem with respect to the integral orientation (especially as a math dabbler and not a professional). Doing it directly, as you call it, isn’t really that much different than using the residue theorem itself, since the polar substutition is so simple it’s not really much more work.

    Your way with a picture and direct substutition is what I’d call the residue theorem with enforced sign correctness.

  5. #5 andy
    September 21, 2009

    This kind of messing around is exactly why I found the residue theorem so awesome.

  6. #6 Chris H.
    September 21, 2009

    Awesome, cool. There are no good explanations in my book so I’m glad you went over it. If I fall into a huge pile of money I’ll be sure to send you a fast car or an island or something.

  7. #7 kérberos
    May 30, 2010

    So is it correct to say that the residues of the integrand at z=0 and z=-1 are respectively -2pi*i and 2pi*i? I’m a second year undergraduate pure mathematics major who’s playing around with complex analysis on his spare time. Haven’t got this far in my book just yet, but it certainly looks quite interesting.

  8. #8 Ragib
    November 6, 2010

    Nice, this really helped me out revising for my exam. Very plain and simple, note dense and terse like a lot of books do it.