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Math Help - complex path integral

  1. #1
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    complex path integral

    so i am trying to evaluate the complex path integral of 1/z from -1-i to 1+i via the line connecting the two points. This line passes through the singularity and for the life of me i cant seem to create paths who's addition give me something helpful using cauchy's theorem. any help would be greatly appreciated.
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  2. #2
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    I can use the triangular contour shown below to conclude:

    \mathop\oint\limits_{T}\frac{1}{z}=0

    via Cauchy's Theorem and let the diameter of the center indentation go to zero. We know that for a simple pole, when the diameter goes to zero, the integral over the indentation is \alpha i where \alpha is the angular length. For the contour below and in the positive sense, \alpha=-\pi then:

    \lim_{\rho\to 0}\mathop\int\limits_C \frac{1}{z}dz=-\pi i

    I'll do the top line:

    \int\frac{1}{z}dz=\int_1^{-1}\frac{dx}{x+i}=\text{Log}(x+i)\Bigg|_1^{-1}=(\ln \sqrt{2}+3\pi i/4)-(\ln\sqrt{2}+\pi i/4)=\pi i/2

    You should get \pi i/2 for the vertical line. Add them up, then the integral over these lines is \pi i. But over the indented contour was -\pi i which must mean the Principal Value over the diagonal line is zero, that is:

    P.V. \int_{-1-i}^{1+i} \frac{1}{z}dz=0

    which you can show by letting z=t+it, taking antiderivatives, and taking the limit:

    \lim_{\epsilon\to 0}\left\{\int_{-1}^{-\epsilon}\frac{1}{t}dt+\int_{\epsilon}^1 \frac{1}{t}dt\right\}=0

    In all cases, I chose a branch of log analytic over the entire contour:

    \text{Log}(z)=\ln(r)+i\Theta,\quad 0\leq\Theta<2\pi

    You may wish to try this using a lower-triangular contour and using a suitable determination of \log analytic over that contour.
    Attached Thumbnails Attached Thumbnails complex path integral-triangular-contour.jpg  
    Last edited by shawsend; October 13th 2009 at 12:25 PM. Reason: corrected values
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