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Path of integration

  1. Mar 6, 2007 #1
    1. The problem statement, all variables and given/known data
    show that the integral of z^i dz from -1 to 1 is equal to (1+e^(-pi))/x * (1-i)
    where the path of integration is any contour from -1 to 1 above the x axis

    2. Relevant equations



    3. The attempt at a solution
    I wrote z^i as e^iLogz and I think all I really need to do is find the antiderivitive of e^i*logz where -pi/2<argz<3pi/2
    this antiderivative is giving me trouble I tried to break it up into e^i*lnr-theta
    and then get -e^(i*lnr)*[e^(3pi/2)-e^(-pi/2)] but I can't get it to work
    can anyone help?
     
  2. jcsd
  3. Mar 6, 2007 #2

    Dick

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    An antiderivative of x^n is still x^(n+1)/(n+1) even when n is i.
     
  4. Mar 6, 2007 #3

    HallsofIvy

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    Why find an anti-derivative? Why not integrate, as the problem says, on a contour that stays above the real axis- such as the upper half of the unit circle? On that circle, [itex]z= e^{i\theta}[/itex] so [itex]z^i= e^{-\theta}[/itex]. It should be easy to integrate that from [itex]\theta= \pi[/itex] to [itex]0[/itex] (not from 0 to [itex]\pi[/itex]- that would be the lower semi-circle.

    Of course, if [itex]z= e^{i\theta}[/itex] then [itex]dz= ie^{i\theta}d\theta[/itex].
     
  5. Mar 6, 2007 #4
    what's the difference between integrating something and finding the anti-derivative?
     
  6. Mar 6, 2007 #5

    Dick

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    I think Halls is right pedagogically. The problem with using the antiderivative is making sure the answer corresponds to the correct path choice. In this case, choosing the correct logs for the endpoints. But you can do it either way. If you're careful.
     
  7. Mar 6, 2007 #6

    HallsofIvy

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    Good point! What I mean was that you don't have to find an anti-derivative of zi. After you specialize to the contour you get an integral of a real valued function which, hopefully, will have a simpler anti-derivative.

    (There is also the slight matter of showing that the integral of the complex function is independent of the particular path you use. Strictly speaking, to use the anti-derivative directly you must show that the integral is the same for ANY path. Here, it isn't. Because of the "multivaluedness" of zi, the value depends on the choice of branch and the negative i-axis is a cut line for zi. The integral is independent of the path as long as you do not cross the negative i-axis which is the reason for using a path in the upper half plane.)

    I am a bit puzzled by that 'x' in the denominator in your original post. Was that a typo?
     
  8. Mar 6, 2007 #7
    Thanks for that answer.

    I'm not the original poster
     
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