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Difficult integral

  1. Aug 30, 2010 #1
    I'm trying to sort out the integral:

    [tex]\int x^{-2}e^{ikx}dx[/tex].

    At first, I thought I would have to solve it numerically. But I am hoping to exploit the orthogonality of the plane waves:

    [tex]\int e^{ikx}dx = 2\pi\delta(k)[/tex]

    It has been a while since I have dealt with integral manipulation, so any help would be appreciated.
  2. jcsd
  3. Aug 30, 2010 #2
    what are the limits?
  4. Aug 30, 2010 #3
    The limits are at minus infinity and infinity.

    [edit]-actually good point. If the integral is infinite then I will have to introduce limits.
  5. Aug 30, 2010 #4
    The integral is not convergent in the Riemann sense. It has a pole of order 2 at x = 0. Nevertheless, you can formally try this trick. Consider the integral as a function of the parameter k:

    f(k) \equiv \int_{-\infty}^{\infty}{x^{-2} \, e^{i \, k \, x} \, dx}

    Take two derivatives with respect to k. What do you get?
  6. Aug 30, 2010 #5
    Hmmm... My guess would be

    [tex]f''(k) = \int^{\infty}_{-\infty} e^{ikx}dx[/tex]

    which is infinite, right?

    It's funny that you mention f(k) because that is precisely what I would need, as the integral will be nested in a larger, numerical integral over k, and I was hoping to save on a lot of computational time by getting f(k) analytically.
  7. Aug 30, 2010 #6
    You made a sign error and the integral you got is of the same form as one of the integrals you had already posted.
  8. Aug 30, 2010 #7
    Ah yes, I forgot to bring down the 'i' twice (I'm horribly rusty).

    So I would get:

    [tex]f''(k) = -\int^{\infty}_{-\infty}e^{ikx}dx = -2\pi\delta(k)[/tex]

    Though is there a way I can implement this? Would I be allowed to, say, do the following:

    [tex]f(k) = \frac{2\pi k^2}{2}\delta(k)[/tex]

    I.e. re-integrate f''(k)
  9. Aug 30, 2010 #8
    Yes, you are allowed to re-integrate, but your integration is incorrect.
  10. Aug 30, 2010 #9
    Forgot completely that delta was a function of k.

    So would these steps be correct?

    [tex]f'(k) = 2\pi[/tex]

    [tex]f(k) = 2\pi k[/tex]

    [edit]-Oops, sign error. But is the form correct?
  11. Aug 30, 2010 #10

    Consider the Heaviside step function:

    \theta (k) = \left\{\begin{array}{rcl}
    1 &,& k \ge 0 \\

    0 &,& k < 0

    What is the derivative of this function?
  12. Aug 30, 2010 #11
    Hmmm.... The derivative is a delta function. But would that mean I would have to manipulate something like:

    [tex]\theta(\infty) = \int^{\infty}_{-\infty} \delta(k)dk [/tex]


    [tex]f''(k) = -2\pi\delta(k) [/tex]

    [tex]f'(k) = -2\pi \int^{\infty}_{-\infty} \delta(k)dk = -2\pi\theta(\infty)[/tex]

    [edit]- Have to run but will be back tomorrow. Thanks a million. This should save me weeks of cpu time.
  13. Aug 30, 2010 #12


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    Homework Helper

    When you integrate f''(k) you want a function of k again, which you don't get if you integrate over the entire k range. If you integrate over the entire range of k there's no k left.

    [tex]f'(k) = f'(-\infty) - 2\pi \int_{-\infty}^k dk'~\delta(k') = -2\pi \theta(k)[/tex]
    assuming f and f' vanish at the boundaries. You then do something similar for the next integration.
  14. Aug 31, 2010 #13
    I have read up a little on the method you have used (dummy variable?) and I think I have the hang of it. But would it be valid to assume that f(k) and f'(k) are bounded? Would this need to be tested?

    The integral has shown up due to the fourier transform:

    [tex]|r-r'|^{-1} = (1/2\pi^2)\int x^{-2} e^{ix(r-r')}dx[/tex]

    My intuition is telling me that, as (r-r') gets larger, the integral should get smaller, as |r-r'|^-1 will get smaller. Or, to use the variable I have been using up to now, |k|^-1 goes to 0 as k goes to infinity. Would this also imply that f(k) goes to infinity as k goes to 0? If it does then I am an idiot, as I am integrating an expression that looks like:

    [tex]\frac{1}{2\pi^2}\int^{a}_{b}sin(Cr)sin(Dr')[\int x^{-2} e^{ix(r-r')}dx]sin(Er)sinF(r')drdr'[/tex]

    and the purpose of the nested integral was ultimately to avoid an infinity when r = r'. I am just copping that the function f(k) might be infinite at k = 0? In that case, I am assuming adding finite limits to the nested integral would give me a finite approximation of f(k) at k = 0 which would suit my purposes fine.

    Thanks to both you and Dickfore for your help so far, as I think I better understand how to manipulate the integral. My (hopefully) last question is, is the same line of logic you have shown me valid if the limits are finite?

    [edit]- Actually, would I be right in thinking it would be easier, as all I would need is an expression as a function of k and two arbitrary bounds?
    Last edited: Aug 31, 2010
  15. Aug 31, 2010 #14
    I have tried integrating the intral in mathematica, and it has produced the expression:

    [tex]\int x^{-2} e^{ikx}dx = ikEI(ikx) - \frac{e^{ikx}}{x}[/tex]

    Ei(ikx) has a simple enough expression that I can approximate with a series expansion. Here is hoping that it will then simply be a case of inserting the two arbitrary bounds in place of x.
    Last edited: Aug 31, 2010
  16. Aug 31, 2010 #15
    What limits did you use though?
  17. Aug 31, 2010 #16
    For the last integration:

    f(k) = f(-\infty) + \int_{-\infty}^{k}{f'(t) \, dt}

    f(k) = -2 \, \pi \, \int_{-\infty}^{k}{\theta(t) \, dt}

    I would use integration by parts.
  18. Aug 31, 2010 #17
    I assumed the limits could be inserted the usual way, to get the expression:

    [tex]f(k) = ik(Ei(ika)-Ei(ikb)) + (\frac{e^{ikb}}{b}-\frac{e^{ika}}{a})[/tex]

    Where a and b are arbitrary limits.
  19. Aug 31, 2010 #18
    Hmmm... I was hoping that, as the interval |a-b| approaches infinity, the integral would approach |k|^-1

    But when I plot f(k) from my previous post, the real component oscillates, and the imaginary component appears to be quadratic in form. Increasing |a-b| only increases the frequency of oscillation.

    I will give the integration by parts a try.
  20. Sep 1, 2010 #19
    They way I've approached it:

    [tex]\int^{\infty}_{-\infty}\theta(t)dt = \int^{0}_{-\infty}\theta(t)dt + \int^{k}_{0}\theta(t)dt[/tex]

    And since [tex]\theta(k) = 1[/tex] for all k > 0:

    [tex]\int^{0}_{-\infty}\theta(t)dt + \int^{k}_{0}\theta(t)dt = 0 + \int^{k}_{0}1dt = k[/tex]

    There is a fudge: I have divided the integral (-inf,inf) into (-inf,0] (0,inf] so 0 technically isn't a member of the second interval.
  21. Sep 1, 2010 #20
    Why do you take the upper limit of the integral to be [itex]\infty[/itex]?
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