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Fourier transform problem

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data

    Compute the integral
    [tex] \int_{-\infty}^{\infty}\frac{\sin(a\xi)}{(1+i\xi)^3\xi}d \xi[/tex]

    by using Parseval's formula for Fourier transform
    [tex] <\overbrace{f}^{\wedge},\overbrace{g}^{\wedge}>=2 \pi<f,g>[/tex]
    where [tex] \wedge[/tex] means the Fourier transform of a function



    3. The attempt at a solution



    Using Parseval's formula we can rewrite the integral as

    [tex] \int_{-\infty}^{\infty}\frac{\sin(a\xi)}{(1+i\xi)^3\xi}d \xi = \Bigl<\overbrace{\frac{\sin(a\xi)}{\xi}}^{\mathcal{F}},\overbrace{\frac{1}{(1+i\xi)^3}}^{\mathcal{F}}\Bigr>[/tex]

    with their inversions as follows

    [tex] \Bigl(\frac{\sin(a \xi)}{\xi}\Bigr)^{\wedge \wedge }=\frac{1}{2} \chi_{[-a,a]}(x)[/tex] by the table 2 in Folland p.223

    and the other one
    [tex] \Bigl(\frac{2}{(i \xi+1)^3}\Bigr)^{\wedge \wedge}=\frac{1}{2}x^2e^{-x}[/tex]

    which I calculated myslef as follows

    we let [tex] g(x)= e^{-x}[/tex] and then the Fourier transform of g(x) is [tex]\mathcal{F}[g(x)]= \frac{1}{1+i \xi}[/tex]

    then using formula in Folland again table 2 [tex] \mathcal{F}[xf(x)]=i(\overbrace{f}^{\wedge})^{'})(\xi)[/tex]

    and applying it to our function [tex] x^2e^{-x}[/tex]
    we get
    [tex] \mathcal{F}[x^2e^{-x}]=i(\overbrace{f}^{\wedge})^{''})(\xi)=\frac{2}{(1+i \xi)^3}[/tex]


    1. The problem statement, all variables and given/known data

    going back to the integral and plugging in everything in the Parseval's formula gives me smth like this

    [tex] \int_{-\infty}^{\infty}\frac{\sin(a\xi)}{(1+i\xi)^3\xi}d \xi = \Bigl<\overbrace{\frac{\sin(a\xi)}{\xi}}^{\mathcal{F}},\overbrace{\frac{1}{(1+i\xi)^3}}^{\mathcal{F}}\Bigr>=2 \pi \Bigl<\frac{1}{2} \chi_{[-a,a]}, \frac{1}{2}x^2e^{-x}\Bigr>[/tex]


    in the solution given by our teacher we find

    [tex] \frac{1}{4}<f^{\wedge}, g^{\wedge}>= \frac{\pi}{2}<f,g>[/tex]
    how did he get this [tex]\frac{1}{4}[/tex] on the left hand side and then [tex] \frac{\pi}{2}[/tex] on the right hand side??
     
  2. jcsd
  3. Dec 9, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    He is "norming" the inner product so that the inner product of a "basis vector" with itself is 1.
     
  4. Dec 9, 2011 #3
    Could you elaborate it a little bit more...I am very new to this...
     
  5. Dec 9, 2011 #4
    Of course it is correct, now I see it. The only thing that is done here is


    [tex] \Bigl<\frac{\sin(a\xi}{\xi},\frac{1}{(1+i \xi)^3}\Bigr>=2 \pi \Bigl<\frac{1}{2}\chi_{[-a,a]}, \frac{1}{2}x^2e^{-x}\Bigr>[/tex]
    so multiplying everything we simply get
    [tex]*2 \pi \cdot \frac{1}{4}\int_{0}^{a}x^2e^{-x}dx[/tex] like in the solution
     
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