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Trouble with use of the Residue Theorem

  1. Sep 20, 2009 #1
    I have just learned the residue theorem and am attempting to apply it to this intergral.

    [tex]
    \int_{0}^{\infty}\frac{dx}{x^3+a^3}=\frac{2\pi}{3\sqrt{3}a^2}
    [/tex]

    where [itex]a[/itex] is real and greater than 0. I want to take a ray going out at [itex]\theta=0[/itex] and another at [itex]\theta=\frac{2\pi}{3}[/itex] and connect them with an arc at infinity to evaluate the integral using the residue theorem; but so far I am having problems getting the correct result. I have shown

    [tex]\left|\frac{1}{x^3+a^3}\right|=\frac{1}{|x|^3}\frac{1}{\left|1+\frac{a^3}{x^3}\right|}\leq \frac{1}{|x|^3}\frac{1}{\left|1-\frac{|a|^3}{|x|^3}\right|}[/tex]

    by the triangle inequality. If we can say that [itex]\frac{|a|}{|x|}<\frac{1}{2^{1/3}}[/itex] then it follows that

    [tex]\left| \int_{\text{arc}}\frac{dx}{x^3+a^3}\right|\leq \left(\frac{2\pi R}{3}\right)\frac{2}{R^3}=\frac{4\pi}{3R^2}\longrightarrow 0[/tex]

    Thus the arc integral goes to 0 as the radius become infinite as needed. For the ray along [itex]\theta=\frac{2\pi}{3}[/itex] using a parametrization of [itex]x=(R-t)e^{2\pi i/3}[/itex] which gives [itex]dx=-e^{2\pi i/3}dt[/itex] the integral becomes

    [tex] \int_{\text{ray}_2}\frac{dx}{x^3+a^3}=\int_{\text{ray}_2}\frac{-e^{2\pi i/3}dt}{(R-t)^3e^{-2\pi i}+a^3}\longrightarrow 0 [/tex]

    again as needed. This would mean to me that the integral should just be the sum of residues lying inside my contour (times [itex]2\pi i[/itex] of course), which is just the one pole [itex]x=ae^{i\pi/3}[/itex]. This would give

    [tex]\text{Res}f(ae^{i\pi/3})=\lim_{x\rightarrow ae^{i\pi/3}}(x-ae^{i\pi/3})\left(\frac{1}{x^3+a^3}\right)=-\frac{e^{i\pi/3}}{3a^2}[/tex]

    and since it is a first order pole this is also the residue. Therefore the answer I am getting is

    [tex]\int_{0}^{\infty}\frac{dx}{x^3+a^3}=-\frac{2\pi i e^{i\pi/3}}{3a^2}[/tex]

    I'm not sure where I went wrong--any help would be greatly appreciated.
     
  2. jcsd
  3. Sep 20, 2009 #2

    lurflurf

    User Avatar
    Homework Helper

    Your residue is right, but you should know that the integral along z/|z|=exp(2*pi*i/3) is not zero.
    let I be the given integral
    You will see
    ray integral=C*I (for some costant C find it)
    then
    closed integral=given+arc+ray=(1-0-C)*I=2pi*i*res(f,z0)=-2pi*i*exp(2pi*i/3)/(3a^2)
    recall
    sin(z)=[exp(iz)-exp(-iz)]/(2i)
     
  4. Sep 20, 2009 #3
    Ah!! Thank you very much for that insight, it hit me as soon as I read what you had to say. I haven't used that trick in awhile so I think I temporarily forgot about it.
     
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