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Math Help - Fourier transform problems

  1. #1
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    Fourier transform problems

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

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







    Using Parseval's formula we can rewrite the integral as

     \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>

    with their inversions as follows

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

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

    which I calculated myslef as follows

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

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

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



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

     \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>


    in the solution given by our teacher we find

     \frac{1}{4}<f^{\wedge}, g^{\wedge}>= \frac{\pi}{2}<f,g>
    how did he get this \frac{1}{4} on the left hand side and then  \frac{\pi}{2} on the right hand side??
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  2. #2
    Grand Panjandrum
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    Re: Fourier transform problems

    Quote Originally Posted by rayman View Post
    Compute the integral
     \int_{-\infty}^{\infty}\frac{\sin(a\xi)}{(1+i\xi)^3\xi}d \xi

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

    <snip>

     \frac{1}{4}<f^{\wedge}, g^{\wedge}>= \frac{\pi}{2}<f,g>
    how did he get this \frac{1}{4} on the left hand side and then  \frac{\pi}{2} on the right hand side??
    Divide the first of these by 4 to get the second...

    CB
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