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Thread: 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:

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

    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 $\displaystyle \alpha i$ where $\displaystyle \alpha$ is the angular length. For the contour below and in the positive sense, $\displaystyle \alpha=-\pi$ then:

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

    I'll do the top line:

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

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

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

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

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

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

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

    You may wish to try this using a lower-triangular contour and using a suitable determination of $\displaystyle \log$ analytic over that contour.
    Attached Thumbnails Attached Thumbnails complex path integral-triangular-contour.jpg  
    Last edited by shawsend; Oct 13th 2009 at 11:25 AM. Reason: corrected values
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