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Math Help - Integration along Paths

  1. #1
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    Integration along Paths

    What is the formal definition of "orientation" of a curve? Or is this a topology issue?

    Let \gamma: [a,b] \to \mathbb{C} be a smooth function. And f(z) is continous on \gamma^* where \gamma^* = \{\gamma(t)|t\in [a,b]\}, i.e. the image. And \eta:[c,d]\to \mathbb{C} is another such function. With the property that \eta^* = \gamma^* and have the same orientation. Then show that \int_{\gamma}f(z) dz = \int_{\eta}f(z) dz.

    Basically, what I am trying to show that if we want to integrate a complex function along some path it makes no difference how we parametrize it (as long the orientation is the same).
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  2. #2
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    Quote Originally Posted by ThePerfectHacker View Post
    What is the formal definition of "orientation" of a curve? Or is this a topology issue?

    Let \gamma: [a,b] \to \mathbb{C} be a smooth function. And f(z) is continous on \gamma^* where \gamma^* = \{\gamma(t)|t\in [a,b]\}, i.e. the image. And \eta:[c,d]\to \mathbb{C} is another such function. With the property that \eta^* = \gamma^* and have the same orientation. Then show that \int_{\gamma}f(z) dz = \int_{\eta}f(z) dz.

    Basically, what I am trying to show that if we want to integrate a complex function along some path it makes no difference how we parametrize it (as long the orientation is the same).
    There is really nothing to show (given that the functions \gamma(t) and \eta(t) are
    sufficiently well behaved) as both the integrals are the integral of the same
    function around the same curve in the same sense. All you have done is
    change how you parameterise it.

    RonL
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    Quote Originally Posted by CaptainBlank View Post
    There is really nothing to show (given that the functions \gamma(t) and \eta(t) are
    sufficiently well behaved) as both the integrals are the integral of the same
    function around the same curve in the same sense. All you have done is
    change how you parameterise it.

    RonL
    I would agree that it should not be so easy to show, however I do not know how to show it. Furthermore, my book assumes that it is okay without even stating that it is valid.

    As I mentioned this is not completely true. Take, for example, the semi-unit upper circle. We can express it as \gamma(t) = e^{it} \mbox{ for }0\leq t\leq \pi and \eta(t) = \sin t + i \cos t \mbox{ for }\pi \leq t\leq 2\pi.
    And consider the function f(z)=z.
    Then, (note \gamma^* = \eta^*)
    \int_{\gamma} f(z) dz \not = \int_{\eta} f(z) dz
    What is wrong? The answer is the orientation. One is clockwise and one is counterclockwise.
    The problem is I do not know how to formally define this notion.
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  4. #4
    Super Member Rebesques's Avatar
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    \eta here does not parametrize the same portion of the circle as \gamma, that's why the integrals are different

    Because actually, the only change of orientation on a curve that can occur, is following it from the endpoint to the startpoint. And that just gives you a change of sign in the integral.
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