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Math Help - Cauchy's Integral Formula Question

  1. #1
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    Cauchy's Integral Formula Question

    I'm stuck on a question where I'm meant to use Cauchy's integral formula. Here's the question:

    Evaluate
    \int_\gamma \frac{1}{(z-p)z} dz
    along
    \gamma (t) = p + e^{it},\ 0 \leq t \leq 2 \pi ,\ where\ p \ \epsilon \ R ,\ p > 1
    by Cauchy's integral formula. Computing this integral along the path \gamma (t) show that
    \int_0^{2 \pi} \frac{p+\cos{t}}{p^2 + 1 + 2p\cos{t}}dt \ = \ \frac{2\pi}{p}

    This is how I approached it:

    Let f(z) = \frac{1}{(z-p)z}. f(z) has poles at z_{0} = p and z_{1}=0.

    The path \gamma (t) is a circle centered around p with a radius of 1. Consequently, z_{1} is outside of \gamma (t) as p > 1. Therefore, according to Cauchy's integral theorem,
    \int_\gamma f(z) = \oint_{C_{0}} f(z)
    where C_{0} is an arbitrarily small circle around z_{0}. (I'm not sure if this is right, but I'll show the rest of my work assuming it is.)

    Around C_{0}, f(z) is analytic, with
    f(z) = \frac{1}{z}
    Then, using Cauchy's integral formula,
    \int_\gamma \frac{f(z)}{z-a}dz = 2\pi i f(a)
    \oint_{C_{0}} \frac{(\frac{1}{z})}{z-z_{0}}dz = 2\pi i\frac{1}{z_{0}}
    Therefore,
    \int_\gamma f(z) = 2\pi i\frac{1}{z_{0}}= 2\pi i\frac{1}{p}
    as z_{0} = p.

    I've now got no idea how to relate that to the second integral in the question, or if it's even right. Any help is very much appreciated
    Last edited by InfernoZeus; August 16th 2010 at 05:10 PM.
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  2. #2
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    I have not checked your work but here is something that may help with the second part of the question. Once on the curve \gamma, you have:

    \displaystyle\frac{1}{(z-p)z} = \frac{1}{e^{i t}(p+e^{i t})}= \frac{e^{i t}}{p+e^{i t}} = \frac{e^{it}}{(p+\cos(t))-i\sin(t)}

    \displaystyle  = \frac{e^{it}}{(p+\cos(t)-i\sin(t))} \frac{(p+\cos(t)+i\sin(t))}{(p+\cos(t)+i\sin(t))}

    \displaystyle = \frac{e^{it}(p+\cos(t)+i\sin(t))}{p^2+1+2p\cos(t)}
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  3. #3
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    Quote Originally Posted by Vlasev View Post
    I have not checked your work but here is something that may help with the second part of the question. Once on the curve \gamma, you have:

    *snip*
    Awesome, thanks - I'll have a go with that. Hopefully my working for the first bit is correct :P
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  4. #4
    MHF Contributor chisigma's Avatar
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    In such a case the best is to set s=z-p so that the integral is...

    \displaystyle \int_{\gamma} \frac{d s}{s\ (s+p)} (1)

    ... and the path \gamma is the 'unit circle' for which is s=e^{i t} with 0 \le t < 2 \pi. If p>1 the only pole inside the path is at s=0 so that is...

    \displaystyle \int_{\gamma} \frac{d s}{s\ (s+p)} = 2 \pi i\  \lim_{s \rightarrow 0} \frac{1}{s+p} = \frac{2 \pi i}{p} (2)

    ... so that the result found by 'Zeus' is exact! ...

    Now let suppose that we don't remember the Cauchy's formulas, so that we have to find the integral (1) in 'standard way'. In that case we can set s= e^{i t} so that the integral becomes...

    \displaystyle \int_{\gamma} \frac{d s}{s\ (s+p)} = i\ \int_{0}^{2 \pi} \frac{dt}{p+e ^{i t}} = i\ \int_{0}^{2 \pi} \frac{dt}{p + \cos t + i\ \sin t} =

    \displaystyle = i\ \int_{0}^{2 \pi} \frac { p + \cos t - i\ \sin t}{p^{2} + 2\ p\ \cos t + 1}\ dt (3)

    Now if we confront the (2) and (3) it is almost immediate find that is...

    \displaystyle = \int_{0}^{2 \pi} \frac { p + \cos t }{p^{2} + 2\ p\ \cos t + 1}\ dt = \frac{2 \pi}{p} (4)

    It seems to be all right... but 'Zeus' wrote in his initial post...

    \displaystyle = \int_{0}^{2 \pi} \frac { p + \cos t }{p^{2} + 2\ p\ \cos t + 1}\ dt = 2 \pi p

    A little amletic doubt! ...

    Kind regards

    \chi \sigma
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  5. #5
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    Sorry, there was a typo in my original post. As you worked out, it should be \frac{2\pi}{p}. Just keeping you on your toes . Thanks for the answer, seems to solve the problem.
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