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Integrating with Infinite boundaries

  1. Dec 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that [itex] \int_{-\infty}^{+\infty} \frac{x-1}{x^5-1}dx = \frac{4\pi}{5}sin(\frac{2\pi}{5}) [/itex]

    3. The attempt at a solution

    This is actually a piece of work from a complex analysis module (not sure if it belongs in this part of the forum or in the analysis section)

    I know that for any infinite bounded integrals you separate them to be [tex] \lim_{c\to -\infty} \int_c^0f(x)dx + \lim_{c\to +\infty}\int_0^cf(x)dx [/tex]
    and I have factorised [itex] x^5-1 = (x-1)(x^4 + x^3 + x^2 + x + 1) [/itex]

    Will I be integrating this as normal or is there a different way to do this if so please help me start :(.
     
  2. jcsd
  3. Dec 5, 2013 #2

    vela

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    Since you found this problem in the context of complex analysis, you're probably meant to evaluate the integral by associating it with a complex integral, which you can evaluate using the residue theorem.
     
  4. Dec 5, 2013 #3

    Mark44

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    SALAAH_BEDDIAF, this is the correct forum section. I deleted your post in the other section.
     
  5. Dec 5, 2013 #4
    So I understand I will be evaluating this integral using Cauchy's Residue Theorem so I must find the singularities in the function, this is when [itex] (x-1)(x^4 + x^3 + x^2 + x + 1) = 0 [/itex], from [itex] (x-1)[/itex] we get a singularity to be 1, but I'm having difficulty finding [itex](x^4 + x^3 + x^2 + x + 1) [/itex] to be 0
     
  6. Dec 5, 2013 #5

    Mark44

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    The solutions of x5 - 1 = 0 are the 5th roots of unity. One of the roots is 1, which you already know. The other four are complex.
     
  7. Dec 5, 2013 #6
    so the roots of [itex](x^4 + x^3 + x^2 + x + 1)[/itex] are also the remaining roots of [itex](x^5-1)[/itex]?
     
  8. Dec 5, 2013 #7

    vela

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    Yes.
     
  9. Dec 5, 2013 #8
    Okay, so I have all the roots now, as these roots are very difficult to be represented by fractions, can I use the approximate roots to find the residues or will I have to use the exact value?
     
  10. Dec 5, 2013 #9

    Mark44

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    If you write them in polar form, they're very simple.
     
  11. Dec 6, 2013 #10

    HallsofIvy

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    All n roots of [itex]z^n- 1= 0[/itex] lie on the circle |z|= 1 and are equally spaced around it. Since the denominator is fifth degree (odd) it has roots 1 and -1 but no roots on the imaginary axis. Of course, to use residues you need to integrate around a closed path. I would recommend integrating from "-iR" to "iR", for R> 1, along the imaginary axis, then on a semicircle back to "-iR". Finally, let R go to infinity.
     
  12. Dec 6, 2013 #11

    Mark44

    Staff: Mentor

    -1 isn't one of the roots. If n were even, it would be, but n is odd in this equation.
     
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