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Homework Help: Integral using Residue Theorem

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    the integral of 1/(1+x^4) from -infinity to +infinity

    2. Relevant equations

    Residue theorem.

    3. The attempt at a solution

    1/(1+z^4) so z^4 = -1

    I know I should be using the residues at z = -sqrt(i) and z= i*sqrt(i)

    I am getting a complex number as an answer which makes no sense

    residue at z = -sqrt(i) = 1/(4*i*sqrt(i))
    and at z = i*sqrt(i) = 1/(4*sqrt(i))

    and therefore integral of (1/1+z^4) = 2pi*i* sum of those residues

    Am I on the right track?
  2. jcsd
  3. Oct 28, 2009 #2
    This is a tricky. When else does [tex] z^4+1 =0 [/tex]? Maybe at [tex] z=e^{i\frac{\pi}{4}} [/tex].
    Can you get it from here?
    Last edited: Oct 28, 2009
  4. Oct 28, 2009 #3


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    Homework Helper

    been a while since I've done these, but as a start, rather than working with sqrt i, which i don't think is unique, I would find the singularities by first letting
    [tex] z = r e^{j \theta} [/tex]

    whilst for some integer n
    [tex] -1 = e^{j\pi(2n+1)} [/tex]

    then the singularities can be found by
    [tex] z^4 = r^4 e^{j 4 \theta} = e^{j\pi(2n+1)}[/tex]

    [tex] \theta = \frac{\pi(2n+1)}{4} = \pi(n/2+1/4)[/tex]

    which I think will give 4 unique singularities for n = 0, 1, 2, 3
  5. Oct 28, 2009 #4
    OOOh, I see, I used the singularities and I get pi/sqrt2 which I think is right. Thank you guys so much. I actually have another related question now though. First of all, why does sqrt(i) not work when I use it as z?


    I'm now trying to do:

    integral (0 to infinity) of sin(x^2) using residue theorem as well. I can't figure out how to make the substitution.

    I dunno when/if you guys will answer that question but thanks so much for your help just now!
  6. Oct 28, 2009 #5


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    Homework Helper

    well first you didn't find all the singularities (2 out of 4) & sqrt(i) is not unique. There are 2 unique answers, note we try and solve
    [tex] z^2 = i [/tex]

    so as before
    [tex] (e^{i \theta})^2 = e^{i 2 \theta} = e^{(2n+1)\pi} [/tex]

    so the 2 solutions in the \theta range [0, 2pi) are
    [tex]\theta = (n+1/2)\pi} = \pi/2, 3\pi/2 [/tex]

    compare it with solving
    [tex] z^2 = 1 [/tex]
    there is a plus & minus solution

    that said if i remember residues correctly, you probably only need 2 of them anyway

    not too sure about your sin question, but it may be worth trying writing the sin in terms of the sum/differnce of 2 complex exponentials
  7. Oct 28, 2009 #6
    I agree. Except when you make the semicircle, the poles n=0, 1 are the only ones on the upper half plane.
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