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Math Help - Contour Integrals (to Evaluate Real Integrals)

  1. #1
    Senior Member roninpro's Avatar
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    Contour Integrals (to Evaluate Real Integrals)

    Hello. I am attempting to evaluate

    \displaystyle \int_0^\infty \frac{x^{1/4}}{1+x^3}

    using a contour integral on the function \displaystyle \frac{e^{(1/4)\log z}}{1+z^3}. My contour will be a "keyhole contour" suggested by Wikipedia: Methods of contour integration - Wikipedia, the free encyclopedia.

    I tried to follow their example, but could not understand this:

    \displaystyle \int_R^\varepsilon {\sqrt{z} \over z^2+6z+8}\,dz=\int_R^\varepsilon {e^{{1\over 2} \mathrm{Log}(z)} \over z^2+6z+8}\,dz=\int_R^\varepsilon {e^{{1\over 2}(\log{|z|}+i \arg{z})} \over z^2+6z+8}\,dz=\int_R^\varepsilon { e^{{1\over 2}\log{|z|}}e^{1/2(2\pi i)} \over z^2+6z+8}\,dz

    Why is it permissible to substitute \text{arg }z=2\pi? Is it perhaps because they have taken R\to \infty, so this line gets close to the 2\pi position?

    I would appreciate any clarification on this example.
    Last edited by roninpro; January 17th 2011 at 07:02 PM.
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  2. #2
    Super Member Random Variable's Avatar
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    It's true when you let  \epsilon go to 0.
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  3. #3
    Senior Member roninpro's Avatar
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    Ah - I was personally using slightly different notation than Wikipedia, letting the little circle be described by (1/R)e^{it}. In any case, I see what you mean.

    Thanks for the reply.
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