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Homework Help: Find the path assembly

  1. Jan 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Hey guys.
    I have this two paths as you can see in the picture and I need to find their assembly (I hope I said it correctly).
    Which one is correct, the right or the left?

    Thanks.

    2. Relevant equations



    3. The attempt at a solution
     

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  3. Jan 3, 2009 #2

    HallsofIvy

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    Re: Paths

    Well, "assembly" isn't the correct English. "Union" of the two sets or "combination" of the two paths woud be better.

    In any case, the problem, as I interpret this is to integrate some function from -R to R along the real line, then integrate from R to -R along the upper half circle with radius R. On the left, [itex]\gamma1[/itex] seems to be the line y= x or, in terms of complex numbers, t+ it, for t from -R to R. No, that is not at all what is given. But the picture on the right is not clear. You seem to be indicating that [itex]\gamma1[/itex] is raised up to some non-zero y, or in terms of complex numbers, t+ ai for some positive a. That is also not correct. [itex]\gamma1[/itex] is given as t+0i, not t+ some non-zero number times i. You should be showing [itex]\gamma1[/itex] running on the real axis, not above it.
     
  4. Jan 4, 2009 #3
    Re: Paths

    Got ya :smile:
    So it actually the half circle over there together with the diameter, this is my path.

    Thanks a lot and also thank you for the English correction :smile:
     
  5. Jan 4, 2009 #4
    Re: Paths

    Well, the first part of the question ask me to find the integral in red (in the pic).
    Is it right what I did?
     

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  6. Jan 4, 2009 #5

    HallsofIvy

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    Re: Paths

    What you have is correct but this looks more like a problem where you are expected to evaluate the integral around the closed path by using Residues. The integrand has poles of order 1 at i, -i, 2i, and -2i, of which i and 2i are inside the closed path.
     
  7. Jan 4, 2009 #6
    Re: Paths

    Oh, yeah, you right, much easier.
    And the points that are outside of the closed path equals to 0, right?

    Thanks a lot.
     
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