2011-06-04, 05:28 PM
Argh, I had a nice post all typed out and my internet went down.
I have, but I hadn't had time to think about the problem. The key concept you are missing out on is branch cuts. Mathematica won't do branch cuts for you automatically (and I don't know how to do them manually either because it's easy enough to ponder what your branch cut will be without actually doing any computations), which explains why you get multiple answers in your Mathematica codes. So anyways, complex exponentials are inherently multivalued. Integrating around a singularity inherently begs the question of where exactly you are in the complex plane, since e^0 = e^i2π = e^i4π etc. A branch cut is an imaginary line you specify, and don't cross. The problem you run into when you integrate method 3 is not only with your parametrization, which you need to be more careful about, but also the fact that you never specify a branch cut. Method 1 sort of implies that you've done a branch cut, by integrating over [0,2π). But you don't even know it.
Ugh, I have Method 3 redone as an example for you at home. When I get my internet back up there I'll post it. Blame comcast. Wikipedia isn't that much help for this concept, if you ask me.
2147483647 Wrote:Hmm... last last time when I saw this:
I thought you had actually studied some of these types of problems. (Not trying to offend or anything.)
I have, but I hadn't had time to think about the problem. The key concept you are missing out on is branch cuts. Mathematica won't do branch cuts for you automatically (and I don't know how to do them manually either because it's easy enough to ponder what your branch cut will be without actually doing any computations), which explains why you get multiple answers in your Mathematica codes. So anyways, complex exponentials are inherently multivalued. Integrating around a singularity inherently begs the question of where exactly you are in the complex plane, since e^0 = e^i2π = e^i4π etc. A branch cut is an imaginary line you specify, and don't cross. The problem you run into when you integrate method 3 is not only with your parametrization, which you need to be more careful about, but also the fact that you never specify a branch cut. Method 1 sort of implies that you've done a branch cut, by integrating over [0,2π). But you don't even know it.
Ugh, I have Method 3 redone as an example for you at home. When I get my internet back up there I'll post it. Blame comcast. Wikipedia isn't that much help for this concept, if you ask me.