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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.

tohoyn
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  • You can use the radial and homogeneity properties of the Fourier Transform i.e. the FT of a radial function will be radial and the FT of a homogeneous function with deg $s$ is also a homogeneous function of deg $-n-s$ where $n$ is the dimension of the space. – Ninad Munshi Apr 04 '21 at 04:25
  • The function is in neither $L^1$ nor $L^2,$ so what do you mean, exactly> – Igor Rivin Apr 04 '21 at 04:25
  • Yes but is it a distribution? Fourier transform of $1/x$ is defined. See https://en.wikipedia.org/wiki/Fourier_transform#Functional_relationships,_one-dimensional. – tohoyn Apr 04 '21 at 04:26
  • Is it in one dimension or in several? – md2perpe Apr 04 '21 at 06:36
  • https://math.stackexchange.com/questions/1522986/what-is-the-fourier-transform-of-1-x – Svyatoslav Apr 04 '21 at 09:31
  • https://math.stackexchange.com/questions/3723136/the-fourier-transform-of-1-p3 – LL 3.14 Apr 21 '22 at 04:55

1 Answers1

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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.

md2perpe
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