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Fourier transform of Logarithm ?

  1. Aug 1, 2009 #1
    does anyone know how to calculate (in the sense of distribution) the Fourier transform of

    [tex] f(x)= ln|x| [/tex]

    that is to obtain the integral [tex] \int_{-\infty}^{\infty} dx ln|x|exp(iux) [/tex]
  2. jcsd
  3. Aug 1, 2009 #2


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    I guess the absolute value is the problem?

    \ln|x| = \begin{cases} \ln ( x), & \mbox{if } x \ge 0 \\ \ln (-x), & \mbox{if } x < 0. \end{cases}

    Now you can split up the integral in a part that goes from -infinity to 0 and from 0 to infinity.
  4. Aug 1, 2009 #3
    Do I remember the definition of a distribution sense Fourier transform correctly, when I think that you want to learn something about the mapping

    s\mapsto \int\limits_{-\infty}^{\infty}\Big(\int\limits_{-\infty}^{\infty} s(x)e^{iux} dx\Big) \log|u| du,

    where [itex]s:\mathbb{R}\to\mathbb{R}[/itex] is a Schwartz test function?

    Looks pretty difficult task to me. Do you have some reason to believe that there exists something that could be done with these integrals?


    The expression I wrote is precisely the same thing as this:

    \lim_{R\to\infty} \int\limits_{-\infty}^{\infty} s(x) \Big(\int\limits_{-R}^R \log|u| e^{iux} du\Big) dx

    So it could be that the definition using Schwartz test function looks like unnecessarily complicated. It's really only about the old fashioned "integrate first, take limit last"-stuff. In this case it could be the best to only to estimate the integral

    \int\limits_{-R}^R \log|u| e^{iux} du

    and try to solve some relevant behavior in the limit [itex]R\to\infty[/itex].
    Last edited: Aug 1, 2009
  5. Aug 1, 2009 #4
    Wolfram Integrator told that some integral function of [itex]\log(x)e^{Ax}[/itex] would be

    \frac{1}{A}\log(x)e^{Ax} - \frac{1}{A}\textrm{Ei}(Ax),

    where Ei is the exponential integral. (Wolfram, Wikipedia)

    It could be, that the problem can be solved by using some known asymptotic properties of the exponential integral.
    Last edited: Aug 1, 2009
  6. Sep 22, 2009 #5
    I think you can try to solve it using complex analysis. Consider the complex plan with z = x+iy. Now the integral can be solved in the complex domain, uisng Residue Theorem. Hope this helps.
  7. Sep 22, 2009 #6
  8. Sep 22, 2009 #7
    Thanks for the new method. It does solve most of the problem as the Fourier transform of 1/x was available in the table(entry 309). I had one problem, though. I am stuck at evaluating the limit for the expression exp(-i[tex]\omega[/tex]x)/x at the limits x approaching negative infinity. Can I use the Euler's identity and reason that as the sin and cos terms are always between -1 and 1, then as x approaches infinity on any side, it will always evaluate to zero. So, is the correct answer pi*w*sqn(w)
  9. Sep 23, 2009 #8
    Show the magnitude goes to zero. The magnitude of the numerator is one.
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