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Thread: [SOLVED] Evaluating residues

  1. #1
    Member Haven's Avatar
    Jul 2009

    [SOLVED] Evaluating residues

    I'm having troubles evaluating this improper integral
    Show: $\displaystyle \int_0^{\infty} \frac{1}{x^3+1} dx = \frac{2\pi}{3\sqrt{3}}$ using residues and a closed contour.

    So, i used a closed contour C that only contains the singularity $\displaystyle z= e^{\frac{i\pi}{3}}$. I let R be the radius of the contour and I let R tend to infinity

    So, $\displaystyle C = C_{+} + C_{R} + C_{\frac{2\pi}{3}} $
    where $\displaystyle C_{+} = \{z=r: 0<r<R, 1<R<\infty \}$
    $\displaystyle C_{R} = \{ z = Re^{i\theta}: 0\leq\theta\leq\frac{2\pi}{3}, 1<R<\infty \}$
    $\displaystyle C_{\frac{2\pi}{3}} = \{z = re^{\frac{i\pi}{3}}: 0<r<R, 1<R<\infty \} $

    I let $\displaystyle I = \lim_{R\to\infty}{\int_0^{R} \frac{1}{z^3 +1}dz}, I_{R} = \int_{C_R}\frac{1}{z^3 +1}dz, I_{\frac{2\pi}{3}} = \int_{C_{\frac{2\pi}{3}}}\frac{1}{z^3 +1}dz$

    By Integral theorems, I get $\displaystyle I_{R} = 0$ as $\displaystyle R\to\infty$ and $\displaystyle I_{\frac{2\pi}{3}} = - e^{\frac{i\pi}{3}}* I $

    By the Cauchy Residue Theorem, I get
    $\displaystyle (1 - e^{\frac{i\pi}{3}})* I = 2\pi i{Res}_{z=e^{\frac{i\pi}{3}}} \frac{1}{z^3+1}$$\displaystyle = 2\pi i\lim_{z\to e^{\frac{i\pi}{3}}} ((1 - e^{\frac{i\pi}{3}})\frac{1}{z^3+1}) $

    but I can't get the answer to evaluate. I just wanna know if there is something I'm missing in complex arithmetic, or is it my set-up that is wrong
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  2. #2
    Super Member
    Jan 2009
    I_{\frac{2\pi}{3}} = - e^{\frac{i\pi}{3}}* I

    I am confused here . isn't it $\displaystyle - e^{\frac{2\pi i}{3} } I $ ?

    and also consider the residue

    $\displaystyle Res_{z = e^{\frac{\pi i}{3}} } \left (\frac{1}{z^3 + 1} \right ) $

    $\displaystyle = \lim_{z \to e^{\frac{\pi i}{3}} } ( z - e^{\frac{\pi i}{3}} )\cdot \left (\frac{1}{z^3 + 1} \right )$

    Use L'hospital to finish it

    Here is my attempt

    $\displaystyle (1 - e^{\frac{2\pi i}{3} })I = 2\pi i \frac{1}{3z^2} |_{z = e^{\frac{\pi i}{3}} } = \frac{2\pi i e^{\frac{\pi i}{3}}}{-3} $

    $\displaystyle e^{\frac{\pi i}{3}} ( -2i \sin( \frac{\pi}{3} )) I = - \frac{2\pi i }{3} \cdot e^{\frac{\pi i}{3}} $

    $\displaystyle \frac{\sqrt{3}}{2} I = \frac{\pi}{3}$

    $\displaystyle I = \frac{2\pi}{3\sqrt{3}}$
    Last edited by simplependulum; Dec 6th 2009 at 11:35 PM.
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  3. #3
    Super Member
    Aug 2008
    That all looks right to me. So then, when you bring everything to one side and calculate the residue, I get:

    $\displaystyle \int_0^{\infty} \frac{1}{1+x^3}dx=\frac{2\pi i}{(1+e^{\pi i/3})(e^{\pi i/3}-e^{-\pi i/3})(1-e^{2\pi i/3})}$

    Not sure about the easiest way to evaluate that by hand, but I just reduced it to a+bi format and got as one step:

    $\displaystyle \frac{2\pi i}{\sqrt{3}i(3/2+\sqrt{3}/2 i)(3/2-\sqrt{3}/2 i)}$

    which eventually reduces down to $\displaystyle \frac{2\pi}{3\sqrt{3}}$.
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