Fourier transform of function $1/ \vert x \vert$

Question

What is the Fourier transform of function $$f(x) = \frac{1}{\vert x \vert}?$$ This is not a homework. I would also appreciate help for calculating it myself.

Answer 1

Solution in one dimension

We first note that $x \dfrac{1}{|x|} = \operatorname{sign}(x)$. Taking the Fourier transform of both sides clearly gives $$ \mathcal{F}\{x \dfrac{1}{|x|}\} = \mathcal{F}\{\operatorname{sign}(x)\}. $$ But the left hand side can now be rewritten as $i\frac{d}{d\xi}\mathcal{F}\{\dfrac{1}{|x|}\}$ and the right hand side can be calculated by $$ i\xi \, \mathcal{F}\{\operatorname{sign}(x)\} = \mathcal{F}\{\frac{d}{dx}\operatorname{sign}(x)\} = \mathcal{F}\{2\delta(x)\} = 2, $$ since $\mathcal{F}\{\delta(x)\} = 1.$ Therefore $$ \mathcal{F}\{\operatorname{sign}(x)\} = \operatorname{pv}\frac{2}{i\xi} + A\delta(\xi) $$ for some constant $A.$ But $\operatorname{sign}(x)$ is an odd function and so shall be the Fourier transform; therefore $A=0.$

Thus, $$ i\frac{d}{d\xi}\mathcal{F}\{\dfrac{1}{|x|}\} = \operatorname{pv}\frac{2}{i\xi} $$ so $$ \mathcal{F}\{\dfrac{1}{|x|}\} = -\ln|\xi| + B $$ for some constant $B.$

Determining the constant can be done, but is not trivial.