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Math Help - fourier transform and integral computing

  1. #1
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    fourier transform and integral computing

    My task is to
    1) compute the Fourier transform of the function  \frac{x}{1+x^2}

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

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

    we let  f(x)=\frac{1}{1+x^2}

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


    which finally gives gives

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

    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
     ||f^\wedge}||^2=2\pi ||f||^2

    and we have

     \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=

    and we know from the part 1) that

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

    then our integral will be

     \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=

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

    the answer should be  \frac{\pi}{2} where do I make mistake?
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  2. #2
    Grand Panjandrum
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    Re: fourier transform and integral computing

    Quote Originally Posted by rayman View Post
    My task is to
    1) compute the Fourier transform of the function  \frac{x}{1+x^2}

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

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

    we let  f(x)=\frac{1}{1+x^2}

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


    which finally gives gives

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

    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
     ||f^\wedge}||^2=2\pi ||f||^2

    and we have

     \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=

    and we know from the part 1) that

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

    then our integral will be

     \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=

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

    the answer should be  \frac{\pi}{2} where do I make mistake?
    On the right hand side of the Plancherel formula you are integrating

     (-\pi i e^{-|\xi|})(+\pi i e^{-|\xi|})=\pi^2 e^{-2|\xi|} for \xi>0

    and:

     (\pi i e^{-|\xi|})(-\pi i e^{-|\xi|})=\pi^2 e^{-2|\xi|} for \xi<0

    CB
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  3. #3
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    Re: fourier transform and integral computing

    thank you!

    alternatively one might say that because the norm ||*||\geq 0 so the expression  ||-\pi ie^{-|\xi|}||^{2} is always positive and equals to  \pi^2 e^{-2|\xi|}
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