Register to reply

Complex (contour indep.) integral of a branched integrand

Share this thread:
ManDay
#1
May14-13, 09:31 AM
P: 155
I'm struggling to reconcile that a complex integrable function may be multi-branched with the statement that its integral is contour-independent. Consider f(z) = z^(1/n), n natural, n-branched and its integral from z_0 to z_1.

On the way from z_0 to z_1 I can take a few detours near the origin where in each evaluation of z^(1/n) I can choose a branch. So I may pass across, say, the real axes at z=1 and choose, for n=2, -1 while I'm still below it and 1 as soon as I have crossed it. On the way back, however, I stick with one branch.

That wouldn't work. So my question is, what is the mathematical correct statement then? It obviously can't be as broad as "the integral of complex integrable function is contour independent".

Should we say "The integral of a complex integrable function is contour independent if its evaluated so that it's differentiable along the path" (thus forbidding any "switch of branch)? What if we consider more esoteric, multi-branched functions where two different branches concide at a point (which would then still allow us to jump from one onto the other).

Or should we say "Branches can always be choosen such that the integral from z_0 to z_1 equals any other integral from z_0 to z_1"?

On a second thought: Perhaps the "function" is defined by an appropriate choice of a branch at each point a-priori. That would clearly make sense. We require the function to be holomorphic, which it only is under a certain choice of branching.
Phys.Org News Partner Mathematics news on Phys.org
Professor quantifies how 'one thing leads to another'
Team announces construction of a formal computer-verified proof of the Kepler conjecture
Iranian is first woman to win 'Nobel Prize of maths' (Update)
Mandelbroth
#2
May14-13, 04:11 PM
Mandelbroth's Avatar
P: 615
Quote Quote by ManDay View Post
On the way from z_0 to z_1 I can take a few detours near the origin where in each evaluation of z^(1/n) I can choose a branch. So I may pass across, say, the real axes at z=1 and choose, for n=2, -1 while I'm still below it and 1 as soon as I have crossed it. On the way back, however, I stick with one branch.

That wouldn't work.
You seem to have figured out that this doesn't work. I don't understand what you are asking.

You say you are having problems with the concept. Perhaps looking at the concepts in more detail may help? I don't know, because your wording seems unclear. Your understanding of multivalued functions appears to be "new", so I don't know how I might approach this with you.
ManDay
#3
May15-13, 02:04 AM
P: 155
No, I'm not really "new" to the subject. I don't understand why you neglect to quote the questions I asked and then claim you wouldn't understand what I was asking, to be honest.

The_Duck
#4
May15-13, 03:32 PM
P: 860
Complex (contour indep.) integral of a branched integrand

The correct statement is something like, "The integral of an analytic function is unchanged if you continuously deform the contour of integration, as long as you don't push the contour across any singularities." The statement is *not* that any two contours give the same answer. Rather, two contours give the same answer if one can be continuously deformed into the other without passing over any singularities.

It may be easiest to think about this first in a context where there are no branches to worry about. For instance, take the classic example of a closed loop integral of 1/z. The value of this integral is determined entirely by the number of times the contour winds around the pole at z = 0. This is because you can continuously deform the contour without changing the value of the integral, as long as you don't push the contour across the pole. If you do push the contour across the pole, the value of the integral changes.

A similar situation holds when the function has multiple branches. Let's first agree that we should never jump discontinously from one branch to another: that doesn't really make sense. Then integrals of z^(1/n) are much like integrals of 1/z: you can continuously deform the contour without changing the value of the integral, as long as you don't push the contour across z = 0. (z = 0 is again a sort of singularity: the integrand is not analytic there).

Note that, contrary to what you suggest, you don't have a whole lot of freedom to decide what branch you want to be on. You can choose what branch you start on, but then as you move along the contour you have to stick to whatever branch you are currently on. You can only get to a different branch by winding around the origin (or, in the general case, wherever the branch cut starts) an appropriate number of times, which gets you to another branch in a continuous way.
ManDay
#5
May16-13, 03:09 AM
P: 155
Thanks for the reply, it answers my question.

I'm aware of the implications of integrating a non-holomorphic function and, admitted, the way I phrased the contour-independence was obviously wrong, now that you mention it.

I do see your point and how it relates to the two examples you gave. But is there an exact statement relating non-analytic points to multi-branched?

Nvm, that wouldn'T make sense


Register to reply

Related Discussions
Complex Contour Integral 2 Calculus & Beyond Homework 6
Complex Contour Integral Calculus & Beyond Homework 6
Complex contour integral, which one do I use? Calculus 28
Contour integral (from complex analysis) Calculus & Beyond Homework 3
Complex Contour Integral Calculus & Beyond Homework 2