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Fourier Transform of Norms

  1. Jul 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Compute the Fourier transform of a function of norm f(\norm{x}).


    2. Relevant equations
    \mathbb{F}{\frac{1}{1+\norm{x}}

    3. The attempt at a solution
    Attempt at using Cauchy theorem and the contour integral with the contour [(-R,R),(R,R+ip),(R+ip,-R+ip),(-R+ip,-R)] does not seem to work.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 26, 2013 #2

    Ray Vickson

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    Is x a scalar or a vector? If it is a scalar, do you mean |x| when you write norm(x)?
     
  4. Jul 26, 2013 #3
    Yes, you are correct. While I need to generalise it to x\in\mathbb{R}^n I would like to start with the case in which

    x\in \mathbb{R}

    and

    \norm{x}=\abs{x}
     
  5. Jul 26, 2013 #4

    LCKurtz

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    Emra1976, put ## or $$ tags around your tex (double characters on each end). You can check what it will look like by clicking the Go Advanced button and previewing your post.
     
  6. Jul 26, 2013 #5

    Mark44

    Staff: Mentor

    To use LaTeX here at PF you need to put the right tags around your expressions, as in the following.

    Inline LaTeX
    Code (Text):

    ##x\in \mathbb{R}##
    or
    [itex]x\in \mathbb{R}[/itex]
     
    Standalone LaTeX (for lack of a better name for it)
    Code (Text):

    $$x\in \mathbb{R}$$
    or
    [tex]x\in \mathbb{R}[/tex]
     
     
  7. Jul 26, 2013 #6
    Alright. Let's just crank it out. Let ##f(x)=|x|##. Then, ##\displaystyle \mathcal{F}\left[f(x)\right](\xi)=\int\limits_{(-\infty,+\infty)}|x|e^{-2\pi i x\xi}~dx##. You tried the residue theorem, you said?
     
  8. Jul 26, 2013 #7
    1) it should be $$f(x)=\frac{1}{1+\abs{x}}$$

    2) I tried Cauchy theorem using the fact that $$f(z)=\frac{1}{1+\abs{z}}$$ is analytical. I tried both the rectangular contour $$([-R,R],[R,R+ip],[R+ip,-R+ip],[-R+ip,-R])$$ and the the countour of the semidisk centered on the origin with radius $$R$$.
     
  9. Jul 26, 2013 #8

    Ray Vickson

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    In TeX, just write |z|, not \abs(z).

    The function z| is not an analytical function, so I don't think you can apply Cauchy directly.
     
  10. Jul 26, 2013 #9
    Do you mean that $$\frac{1}{1+|z|}$$ is not analytical (used as homomorphic)?
    If so, where are its poles?
     
  11. Jul 26, 2013 #10

    Ray Vickson

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    No poles, but it fails to satisfy the Cauchy-Riemann equations, so is not holomorphic.

    However, you can write
    [tex] F(w) = \int_{R} \frac{e^{-iwx}}{1+|x|} \, dx = 2\int_0^{\infty} \frac{\cos(w x)}{1+x} \, dx,[/tex]
    and then extend the latter integrand to
    [tex] \frac{\cos(w z)}{1+z},[/tex]
    which IS analytic.
     
    Last edited: Jul 26, 2013
  12. Jul 27, 2013 #11
    Now, given
    $$f(z)=\frac{cos(\omega z)}{1+\omega z}$$
    how would you choose the contour for the Cauchy Integral? My two choices would be

    1) the quarter disk in the top right quadrant of the complex plane
    2) a rectangular contour in the top right quadrant of the complex plane

    but I get stacked with both (1) and (2).
     
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