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Plancherel formula and integral computing

  1. Dec 9, 2011 #1
    My task is to
    1) compute the Fourier transform of the function [tex] \frac{x}{1+x^2}[/tex]

    2) compute the integral [tex] \int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx[/tex]

    1) I can write my function as [tex] x \cdot \frac{1}{1+x^2}[/tex] and by using the formula

    we let [tex] f(x)=\frac{1}{1+x^2}[/tex]

    [tex]\mathcal{F}[xf(x)]=i(f^{\wedge})^{'}(\xi)=-\pi ie^{-|x|}[/tex]


    which finally gives gives

    [tex] \Bigl(\frac{x}{1+x^2}\Bigr)^{\wedge}=\begin{cases} -\pi ie^{-|\xi|} \therefore \xi>0 \\ \pi ie^{-|\xi|} \therefore \xi<0\end{cases}[/tex]

    which agrees with the answer.

    2) this one is a bit tricker and somehow it seems like I am on the correct track except for the sign I get....

    I use Plancherel formula for Fourier transform to solve this integral, namely
    [tex] ||f^{\wedge}||^2=2 \pi ||f||^2[/tex]

    and we have

    [tex] \int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=||\frac{x}{1+x^2}||^2=\frac{1}{2 \pi}||\overbrace{\frac{x}{1+x^2}}^{\wedge}||^2=[/tex]

    and we know from the part 1) that

    [tex]\Bigl(\frac{x}{1+x^2}\Bigr)^{\wedge}=\begin{cases} -\pi ie^{-|\xi|} \therefore \xi>0 \\ \pi ie^{-|\xi|} \therefore \xi<0\end{cases}[/tex]

    then our integral will be

    [tex] \int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=\frac{1}{2\pi}\int_{-\infty}^{\infty}-\pi^2 e^{-2|\xi|}d\xi=-\frac{\pi}{2}\cdot 2 \int_{0}^{\infty}e^{-2\xi}d\xi=-\pi \int_{0}^{\infty}e^{-2\xi}d\xi=[/tex]

    [tex]\frac{\pi}{2}\Bigl[e^{-2\xi}\Bigr]_{0}^{\infty}=-\frac{\pi}{2}[/tex]

    the answer should be [tex] \frac{\pi}{2}[/tex] where do I make mistake?
    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
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 10, 2011 #2
    You already wrote down the solution.
    ||.||^2 means absolute value squared, so the integrand is positive.
     
  4. Dec 10, 2011 #3



    yeah, you are right :) thank you
     
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