What is the inverse Fourier transform of $1/|\omega|$?

Question

What is the inverse Fourier transform of $\frac{1}{|\omega|}$?

$$ \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{|\omega|}e^{i\omega t}d\omega=? $$

Answer 1

Your integral doesn't converge because $\frac{1}{|x|}$ is not integrable at $x=0$.

Also, even if we look at the Fourier transform of distributions (not an easy subject) then $\frac{1}{|x|}$ is not a distribution, $\frac{1}{x}$ neither.

Now $vp(\frac{1}{x})$ is well-defined as a distribution, by $$\forall \varphi \in S(\mathbb{R}), \quad \langle vp(\frac{1}{x}), \varphi \rangle = \langle \frac{1}{x}, \varphi-\varphi(0)e^{-x^2} \rangle = \int_{-\infty}^\infty \frac{\varphi(x)-\varphi(0)e^{-x^2}}{x}dx$$ where $S(\mathbb{R}$ is the Schwartz space).

Note that $T$ is a tempered distribution iff $T \ast e^{- x^2/a^2}$ is well-defined. $T \ast e^{- x^2/a^2}$ has of course a Fourier transform, and the Fourier transform of $T$ is by definition the limit of $\mathcal{F}[T \ast \frac{e^{- x^2/a^2}}{|a|\sqrt{\pi}}]$ as $a \to 0$.

From this, the Fourier transform of $vp(\frac{1}{x})$ is $\frac{i}{2} \, sign(\omega)$, because $i x \,vp(\frac{1}{x}) = i \implies \frac{d}{d\omega} \mathcal{F}[vp(\frac{1}{x})] = i\delta$

Answer 2

The answer in terms of distributions is given here. The given formula for the transform of $\ln(x^2+a^2)$ is still applicable for $a=0$, yielding $$\frac 1 {2\pi} \int_{-\infty}^\infty \frac 1 {|\omega|} e^{i t \omega} d\omega = -\frac {\ln |t| + \gamma} \pi.$$