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Help evaluating a rational integral using residue

  1. Oct 27, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to compute [tex]\int_0^\infty \frac{dx}{x^3+a^3}.[/tex]


    2. Relevant equations

    If f = g/h, then [tex]Res(f, a) = \frac{g(a)}{h'(a)}.[/tex]

    3. The attempt at a solution
    In the first I've used a semicircular contour in the upper plane that is semi-circular around the pole at -a. So I calculate the integral to be

    [tex]1/2 (2\pi iRes(f, -ae^i{2\pi/3}) - \pi i Res(f, -a)) = 1/2 (\frac{2\pi i}{3a^2 e^i{4\pi/3}}-\frac{\pi i}{3a^2}) = \cdot \frac{-\pi i}{3\sqrt{3}a^2+6a^2}.[/tex] The 1/2 comes from those residues being the calculation of the integral over the whole real line, the first term comes from the integral over the entire contour, which contains the pole at [itex]-a\zeta_3[/itex], while the second term from the part of the contour around x=-a. The answer is listed as [tex]\frac{2\pi}{3\sqrt{3}a^2},[/tex] so I'm not sure where I went wrong.
     
    Last edited: Oct 27, 2012
  2. jcsd
  3. Oct 27, 2012 #2

    haruspex

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    Your integration path is not clear to me. Can you be more precise?
    There seems to be an i missing in the locations of the complex poles.
     
  4. Oct 27, 2012 #3
    Sure. My contour is the upper half of a circle of radius r, the segment from -r to -a-e, the upper half of a circle of radius e around -a, and the segment from -a+e to r. I took the limits r -> infinity and e -> 0. The integral over the arc goes to zero, so I end up with the integral over the real line + pi*i*residue at -a = 2*pi*i*residue inside the contour.

    I notice that I forgot to add the i into the exponent of zeta_3, I'll edit the original post. I did use it in my calculations.
     
  5. Oct 27, 2012 #4

    haruspex

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    But the desired integral is only over the positive real line.
     
  6. Oct 27, 2012 #5
    That's why I multiplied by 1/2. But now that I think about it, that wouldn't work since the function isn't symmetric about the y-axis, right?
     
  7. Oct 27, 2012 #6

    haruspex

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    Right.
     
  8. Oct 27, 2012 #7
    I don't see which contour would be best to use in this situation.
     
  9. Oct 28, 2012 #8
    Tell you what, first get it working for say [itex]a=2[/itex] by introducing a branch cut along the real axis and considering:

    [tex]\oint \frac{\log(z)}{z^3+8}dz[/tex]

    around a key-hole contour with the key-hole along the positive real axis. One minor issue is the residue in the fourth quadrant. Note the analytic-extension of [itex]\log(z)[/itex] over that contour into that quadrant is actually [itex]\text{Log}(z)+2\pi i[/itex] right? Ok, get that one then generalize it to any [itex]a[/itex].

    Edit:

    Afraid I made this more complicated than it needed to be: Easier probably to integrate directly [itex]1/(z^3+8)[/itex] over a [itex]2\pi/3[/itex] wedge contour bordering the positive real axis and in that way, only encircle one pole in the first quadrant and not even need to worry about dealing with the log branch-cut. Sorry.
     
    Last edited: Oct 28, 2012
  10. Oct 28, 2012 #9

    haruspex

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    ... and the trick then is to express the integral over the line r e2πi/3 as a constant times that over (0, ∞) portion, yes?
     
  11. Oct 28, 2012 #10
    Yes:

    [tex]\mathop\int\limits_{C} \frac{1}{z^3+8}dz=\int_{\infty}^0 \frac{1}{r^3 e^{6\pi i/3}+8} e^{2\pi i/3} dr[/tex]

    You can see how to get that right? Hope that's not an insulting question to ask.

    Then surely, I think, we can do smaller wedges to figure out:

    [tex]
    \int_0^{\infty} \frac{1}{z^n+a^n}dz[/tex]

    You may wish to look at that one to get the general principle down.
     
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