Results 1 to 9 of 9

Math Help - Residue Theorem #2

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    33

    Residue Theorem #2

    Using residue theorem, calculate

     \int_{-\infty}^\infty \frac {1}{1+x^4} dx

    How do I find the poles of this function?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2009
    Posts
    220
    Thanks
    1
    Quote Originally Posted by Richmond View Post

     \int_{-\infty}^\infty \frac {1}{1+x^4} dx

    How do I find the poles of this function?
    This function will have poles when the denominator is zero.

    Just solve  x^4 = -1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    33
    so  x^4 = -1

    x^2 = \sqrt i

    x = \pm i^\frac{1}{4}

    Is that right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2009
    Posts
    220
    Thanks
    1
    That is almost right, x^4 =-1 ~ \Rightarrow x^2 = \pm \sqrt{-1} = \pm i
    What contour would you use to evaluate this integral? Which of the singularities lie within this contour and can you calculate the residues?
    Last edited by pomp; July 14th 2009 at 08:39 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2008
    Posts
    33
    The four values are
     i^\frac{1}{4}, -i^\frac{1}{4}, i^\frac{-1}{4},-i^\frac{-1}{4}

    Since the integral limits are -\infty and \infty
    we take the positive part of the integral and use the positive values

    which are
     i^\frac{1}{4} and i^\frac{-1}{4}

    So
     Res(f,i^\frac{1}{4}) = \frac{1}{(x + i^\frac{1}{4})(x + i^\frac{-1}{4}), (x- i^\frac{-1}{4})}
     = \frac{1}{(2i^\frac{1}{4})(i^\frac{1}{4} + i^\frac{-1}{4})(i^\frac{1}{4} - i^\frac{-1}{4})}

    so the integral will be 2\pi i * Res(f,i^\frac{1}{4})
    =

    I think somewhere is wrong.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328
    Quite frankly, the fact that you keep writing " i^{1/4}" indicates you need to brush up on "complex numbers" before trying contour integrals and residues.

    i can be written in "polar form" as e^{i\pi/2}. It's fourth roots are given by e^{\frac{i(\pi/2+ 2\pi k)}{4}} which gives different result for k= 0, 1, 2, and 3.

    With k= 0, a fourth root of i is e^{i\pi/8}= cos(\pi/8)+ i sin(\pi/8). With k= 1, another is e^{5i\pi/8}= cos(5\pi/8)+ i sin(5\pi/8). With k= 2, e^{9i\pi/8}= cos(9\pi/8)+ i sin(9\pi/8). Finally, with k= 3, e^{13\pi/8}= cos(13\pi/8)+ i sin(13\pi/8).

    Two of those four roots have positive real part, two negative real part. Two have positive imaginary part, two negative imaginary part. That means you really only need to look at the residues at two poles, depending on how you set up the contour.
    Last edited by mr fantastic; July 15th 2009 at 08:04 AM. Reason: Fixed a small typo with e^{5i\pi/8}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by Richmond View Post
    so  x^4 = -1

    x^2 = \sqrt i

    x = \pm i^\frac{1}{4}

    Is that right?
    No. You want the fourth roots of 1, or in other words the square roots of \pm i.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
     x^{4} +1 = 0

     x^{4} = -1

     x^{4} = e^{i \pi}

     x = e^{ \frac{i( \pi + 2 \pi k)}{4}} for k = 0, 1, 2, and 3

    so the roots are  e^{\frac{i \pi}{4}}, e^{\frac{3 i \pi}{4}}, e^{\frac{5 i \pi}{4}}, e^{\frac{7 i \pi}{4}}

    For the countour integration you only need the two poles in the upper half of complex plane (  e^{\frac{i \pi}{4}} and  e^{\frac{3 i \pi}{4}} )

     e^{\frac{i \pi}{4}} = \cos(\pi /4) + i \sin(\pi /4) = \sqrt{2}/2+ i \sqrt{2}/2

     ^{\frac{i 3\ pi}{4}} = \cos(3 \pi /4) + i \sin(3 \pi /4) = -\sqrt{2}/2+ i \sqrt{2}/2
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
     \int^{\infty}_{\text{-}\infty} \frac {1}{1+x^{4}}

    let  f(z) = \frac {1}{1+z^4}

    As I mentioned, there are two simple poles in the upper half of the complex plane.

     Res \{f, z_{o}\} = \lim_{z \to z_{0}} (z-z_{0}) \frac {1}{1+z^{4}} = \lim_{z \to z_{0}} \frac {1}{4z^{3}} = \frac {1}{4z_{0}^{3}}

     Res \{f, \sqrt{2}/2+ i \sqrt{2}/2\} = \text{-} \sqrt{2} /8 - i \sqrt{2} /8 (after simplification)

     Res \{f, - \sqrt{2}/2+ i \sqrt{2}/2\} = \sqrt{2} /8 - i \sqrt{2} /8

     \int^{\infty}_{\text{-}\infty} \frac {1}{1+x^{4}} = 2 \pi i \sum (\text{residues in the upper half plane})  = 2 \pi i (\text{-}i \sqrt{2} /4 ) = \pi \frac{\sqrt{2}}{2}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. residue theorem
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: May 17th 2010, 12:21 PM
  2. Residue Theorem
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: July 9th 2009, 11:36 PM
  3. Residue Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 18th 2009, 12:24 PM
  4. Residue theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 09:15 AM
  5. Residue Theorem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: December 7th 2008, 10:56 AM

Search Tags


/mathhelpforum @mathhelpforum