# Fourier transform of Logarithm ?

## Main Question or Discussion Point

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
Homework Helper
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.

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

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

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 $R\to\infty$.

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

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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.

• Complexiologist
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$$\omega$$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)

• Complexiologist
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$$\omega$$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.

• Complexiologist