What is the Fourier Transform of $f'(x)/x$

Question

What is the Fourier Transform of $f'(x)/x$?

Is this even possible to find? It's deceptively simple looking.

Answer 1

We can derive a formula, but nothing close to the nice formulas we have for the Fourier transform of $f'(x)$ for example.

The reason we have a nice formula for the Fourier transform of $f'(x)$ is that we can easily do a integration by part on the integral

$$\mathcal{F}(f'(x)) = \int f'(x)e^{-ikx}dx = ik\int f(x) e^{-ikx} dx = ik\mathcal{F}(f(x))$$

giving us $\mathcal{F}(f'(x)) = ik \mathcal{F}(f(x))$. For your case the Fourier transform integral reads $$\mathcal{F}\left(\frac{f'(x)}{x}\right) = \int \frac{f'(x)}{x}e^{-ikx}dx$$ for which integration by parts does not simplify the expression as much:

$$\int \frac{f'(x)}{x}e^{-ikx}dx = ik\int \frac{f(x)}{x} e^{-ikx} dx + \int \frac{f(x)}{x^2}e^{-ikx} dx$$

But we do get a formula that avoids $f'(x)$ on the right hand side:

$$\mathcal{F}\left(\frac{f'(x)}{x}\right) = ik\mathcal{F}\left(\frac{f(x)}{x}\right) + \mathcal{F}\left(\frac{f(x)}{x^2}\right) = \mathcal{F}\left(\frac{f(x)(ikx+1)}{x^2}\right)$$

and this is really the best we can do. In general this is not very useful unless one knows the Fourier transform of $f(x)/x$ and $f(x)/x^2$ or if these transforms are simpler to evaluate.