Fourier transform of Logarithm ?

  • #1
391
0

Main Question or Discussion Point

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]
 

Answers and Replies

  • #2
Cyosis
Homework Helper
1,495
0
I guess the absolute value is the problem?

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

Now you can split up the integral in a part that goes from -infinity to 0 and from 0 to infinity.
 
  • #3
2,111
16
Do I remember the definition of a distribution sense Fourier transform correctly, when I think that you want to learn something about the mapping

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

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?

update:

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

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

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

[tex]
\int\limits_{-R}^R \log|u| e^{iux} du
[/tex]

and try to solve some relevant behavior in the limit [itex]R\to\infty[/itex].
 
Last edited:
  • #4
2,111
16
Wolfram Integrator told that some integral function of [itex]\log(x)e^{Ax}[/itex] would be

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

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:
  • #5
18
1
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.
 
  • #6
  • #7
18
1
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)
 
  • #8
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)
Show the magnitude goes to zero. The magnitude of the numerator is one.
 

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