Results 1 to 3 of 3

Math Help - [SOLVED] Evaluating residues

  1. #1
    Member Haven's Avatar
    Joined
    Jul 2009
    Posts
    197
    Thanks
    8

    [SOLVED] Evaluating residues

    I'm having troubles evaluating this improper integral
    Show: \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 z= e^{\frac{i\pi}{3}}. I let R be the radius of the contour and I let R tend to infinity

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


    I let 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 I_{R} = 0 as R\to\infty and I_{\frac{2\pi}{3}} =  - e^{\frac{i\pi}{3}}* I

    By the Cauchy Residue Theorem, I get
    (1 - e^{\frac{i\pi}{3}})* I = 2\pi i{Res}_{z=e^{\frac{i\pi}{3}}} \frac{1}{z^3+1}  = 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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    <br />
I_{\frac{2\pi}{3}} = - e^{\frac{i\pi}{3}}* I<br />


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


    and also consider the residue


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

     = \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

     (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}

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

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

     I = \frac{2\pi}{3\sqrt{3}}
    Last edited by simplependulum; December 7th 2009 at 12:35 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2008
    Posts
    903
    That all looks right to me. So then, when you bring everything to one side and calculate the residue, I get:

    \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:

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

    which eventually reduces down to \frac{2\pi}{3\sqrt{3}}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] evaluating the expression...
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 15th 2010, 08:06 PM
  2. Replies: 4
    Last Post: November 22nd 2009, 02:08 PM
  3. [SOLVED] quadratic residues congruence problem
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: November 19th 2009, 07:10 PM
  4. [SOLVED] evaluating limits
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 28th 2009, 08:08 PM
  5. Replies: 1
    Last Post: May 14th 2009, 11:49 AM

Search Tags


/mathhelpforum @mathhelpforum