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Complex integration, branch cuts

  1. Mar 23, 2010 #1
    Hi guys,
    I need to show that:
    [tex] \int_{0}^{\infty } \frac{x^{a}}{(x+1)^2} \dx = \frac{\pi a}{\sin(\pi a)} [/tex]
    , where -1<a<1.
    The problem is, that although a hint is given,the path of integrating it, I have difficulty what they really mean with "cut line, branch points, multivalued functions" etc.

    In the hint , the path of integration is the following, which I don't understand why they have chosen it like this, could some explain their reasoning? For example, why do they exclude z=-1?
    http://img143.imageshack.us/img143/151/acf1.jpg [Broken]

    I have now calculated this integral using my own , simpler, contour, which is basically the same as this one, except it includes the point z=-1 ( Which I still don't understand why my book excludes..).
    By the way; the whole idea of branch cuts etc basically means that if you wounded one time around the origin, you have to use z=r*exp(i(theta+2pi)); is this right? I calculated using this and I got the correct answer..
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 23, 2010 #2
    z = -1 does not need to be excluded.

    There are two different approaches to attack these sorts of problems. The first approach invloves carefully chosing the branch cut, in the second approach you can more or less ignore the precise details about the branch cuts.

    First, you need to understand what branch points and branch cuts are. You should study that from your complex analysis books. A simple explanation is that branch points are points where the function is non-analytic, such that moving around that point leads the value of the function to jump.

    This then means that you have to account for this when developing complex function theory rigorously. The function z^a clearly has a branch point at z = 0. What this means is that you cannot apply the residue theorem as long as z = 0 is inside the contour.

    Another thing is that whenever a point in the comlex plane is specified, you need to know what the value of the function is using some unambiguous prescription. But if going around the origin changes the value of the function, specifying that e.g. f(z) = z^a is not enough.

    You need to be able to distinguish between a point z and the "same point" when you have gone around the origin. That you can do by introducing polar coordinates. At this point you have two main choices to proceed.

    The first choice is to work with single valued functions. Thgis means that you have to prevent the function from jumping, so when going around the origin at some polar angle, you set the polar angle back by letting it jump by minus 2 pi. This then defines the so-called branch cut, it can be chosen arbitrarily.

    Now, when you do contour integration, you can only apply the residue theorem if the function is analytic. This then means that not only do you have to make sure that there are no branch points inside thecontour, the contour also cannot intersect any branch cuts (as the function jumps there, so it discontinuous and thus not analytic). In your case, this means that you have to choose the branch cut along the positive real axis.

    Another approach is to choose the branch cut in some arbitrary way and then to analytically continue the function across the branch cut. You chop up the contour in two pieces, so that the integral along the first contour integrates the original function avoiding its branch cut, while the second contour integral is along the second contour of the analytic continuation which has its branch cut chosen such that it doesn't interect the second contour (it will then intersect the first contour).

    Now this analytic continuation and chopping up of the contour doesn't have to be actually performed. The argument is that you can do it, therefore you don't have to bother with the minutia of choosing the branch cut along the positive real axis and then having to infinitessimally avoiding it.
  4. Mar 23, 2010 #3
    A more instructive problem involving branch points and cuts is to evaluate:

    [tex]\int_{0}^{1}\sqrt[3]{x^2 - x^3}dx[/tex]

  5. Mar 23, 2010 #4
    Sorry to prolong this thread longer than possibly necessary, but how much complex analysis would one need to know to understand the tools used here? I'm fairly comfortable with power series and metric topology, but the extent to which I know complex analysis is basically an equivalence statement regarding the Cauchy-Riemann equations (i.e., I don't know anything).
  6. Mar 23, 2010 #5
    Then you are infinitesimally close to understanding the tools used here. :smile: If f(z) is a complex differentiable function, then:

    [tex]\oint f(z)dz=0[/tex]

    where the contour integral is over any arbitrary contour (Cauchy's theorem). The proof is rather tedious. Once you have arrived at this point, the rest is pretty much trivial. It is easy to show that:

    f(a) = [tex]\frac{1}{2 \pi i}\oint \frac{f(z)}{z-a}dz[/tex] (1)

    where the point a is inside the contour, by applying Cauchy's theorem to the contour that after almost moving around the point z= a moves toward a, encircles it and moves back (see the figure posted by the OP where z = 0 is removed in a similar way). The z = a is no longer in the contour, so the contour integral is zero as the integrand is complex differentiable everywhere inside the contour. You can compute the contour integral with a small radius around the singularity at z = a in the limit that the radius shrinks to zero, which leads to the result.

    Eq. (1) (a.k.a. Cauchy's integral theorem) implies that complex differentiable functions are analytic. The proof is quite simple. It involves studying the derivative w.r.t. a and then noting that what you get is also complex differentiable. Since Eq. (1) is valid for any complex differentiable function, this means that complex differentiable functions are infinitely often differentiable. That f(z) is analytic also follows as a by product of this proof.

    The residue theorem pretty much follows in a similar was as Eq. (1). If you consider functions that are analytic except for poles where they behave like 1/(z-a)^n for some integer n, then you can deal with that by removing the pole as in the proof of Eq. (1). The integral over a small circle encircling z= a of 1/(z-a)^n is zero unless n = 1, in which case it is 2 pi i. he resdue theorem that says that the contour integral is 2 pi i times the sum of all residues at the poles, where residue is at pole z = a is the coefficient of 1/(z-a), then follows.
  7. Oct 18, 2011 #6
    That is an awesome explanation.
  8. Mar 19, 2012 #7
    hey guys, i need to do the following contour integration...
    \int_{-\infty}^{\infty} dz f(z)/\sqrt(z*z+a*a)

    where f(z) is a well behaved function in the complex plane.

    The problem is , how to choose the contour for the evaluation of this integration....
  9. Mar 20, 2012 #8
    You need to start a new thread. I think resurrecting an old thread like this is called "necroposting" and is not allowed.
  10. Mar 20, 2012 #9
    i was not aware of the fact...so my apologies!!!
  11. Mar 21, 2012 #10


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    A branch of a complex function is the restriction of a complex multi-function that turns this multi-function into a well-defined function.

    The standard example is that of the log , in that logz has infinitely-many values. We restrict our domain to be able to define a branch in which log has just one output. Branch cuts are curves that separate branches of a function from each other.
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