# 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 $R\to\infty$.

Last edited:
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.

Last edited:
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.

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)

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.