Fourier transform of Logarithm ?

[zetafunction]

does anyone know how to calculate (in the sense of distribution) the Fourier transform of

f(x)= ln|x|

that is to obtain the integral \int_{-\infty}^{\infty} dx ln|x|exp(iux)

 

[Cyosis]

I guess the absolute value is the problem?

<br /> \ln|x| = \begin{cases} \ln ( x), &amp; \mbox{if } x \ge 0 \\ \ln (-x), &amp; \mbox{if } x &lt; 0. \end{cases} <br />

Now you can split up the integral in a part that goes from -infinity to 0 and from 0 to infinity.

 

[jostpuur]

Do I remember the definition of a distribution sense Fourier transform correctly, when I think that you want to learn something about the mapping

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

where s:\mathbb{R}\to\mathbb{R} 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:

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

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

<br /> \int\limits_{-R}^R \log|u| e^{iux} du<br />

and try to solve some relevant behavior in the limit R\to\infty.

 

[jostpuur]

Wolfram Integrator told that some integral function of \log(x)e^{Ax} would be

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

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.

 

[Jivesh]

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.

 

[John Creighto]

How about using integration by parts. This will give you a new integral that is the Fourier Transform of 1/x instead.
Then use property (105)
http://en.wikipedia.org/wiki/Fourier_transform#Functional_relationships

That is if you know the inverse Fourier transform of a function then you can calculate the Fourier transform of the function.

 

[Jivesh]

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\omegax)/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)

 

[John Creighto]

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\omegax)/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.