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Use suitable contours in the complex plane and the residue theorem to show that

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Use suitable contours in the complex plane and the residue theorem to show that

    integral from -infinity to +infinity of [1/(1+(x^4))] dx=pi/(sqrt(2))

    Fix R > 1, and consider the counterclockwise-oriented contour C consisting of the upper half circle of radius R centered at the origin (call it γ) and the real axis for x in [-R, R].

    So, ∫c dz/(1 + z^4) = ∫γ dz/(1 + z^4) + ∫(x = -R to R) dx/(1 + x^4).
    ---------------------------------
    (i) Compute ∫c dz/(1 + z^4) by the Residue Theorem.

    The integrand has two singularities in the upper half plane (where C is located):
    1 + z^4 = 0 ==> z^4 = -1 = e^(πi) ==> z = e^(πi/4), e^(3πi/4); both are simple poles.

    Computing their residues:

    At z = e^(πi/4), we have
    lim(z→e^(πi/4)) (z - e^(πi/4))/(1 + z^4)
    = lim(z→e^(πi/4)) 1/(4z^3)
    = lim(z→e^(πi/4)) z/(4z^4)
    = e^(πi/4)/(4 * -1)
    = (-1/4) e^(πi/4).

    Similarly, the residue at z = e^(3πi/4) equals (-1/4) e^(3πi/4).

    Therefore, ∫c dz/(1 + z^4) = 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2.

    but why 2πi [(-1/4) e^(πi/4) + (-1/4) e^(3πi/4)] = π/√2?
     
  2. jcsd
  3. Sep 8, 2011 #2

    micromass

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    Re: Use suitable contours in the complex plane and the residue theorem to show that..

    Well, try to write [itex]e^{\pi i/4}[/itex] in it's usual form:

    [tex]e^{\pi i 4}=\cos(\pi /4)+i\sin(\pi /4)[/tex]

    work that out abd substitute that in. Do the same with [itex]e^{2\pi i/4}[/itex]. You'll see that it equals [itex]\pi/\sqrt{2}[/itex].
     
  4. Sep 8, 2011 #3
    Re: Use suitable contours in the complex plane and the residue theorem to show that..

    Expand your result using Euler's Formula and be dazzled. ;) Complex contour integration is a truly beautiful application of complex analysis, as is seen in the result of this integral.
     
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