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Thread: 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 $\displaystyle \frac{x}{1+x^2}$

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

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

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

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


    which finally gives gives

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

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

    and we have

    $\displaystyle \int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=||\frac{x} {1+x^2}||^2=\frac{1}{2 \pi}||\overbrace{\frac{x}{1+x^2}}^{\wedge}||^2=$

    and we know from the part 1) that

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

    then our integral will be

    $\displaystyle \int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=\frac{1}{2 \pi}\int_{-\infty}^{\infty}-\pi^2 e^{-2|\xi|}d\xi=-\frac{\pi}{2}\cdot 2 \int_{0}^{\infty}e^{-2\xi}d\xi=-\pi \int_{0}^{\infty}e^{-2\xi}d\xi=$

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

    the answer should be $\displaystyle \frac{\pi}{2}$ 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 $\displaystyle \frac{x}{1+x^2}$

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

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

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

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


    which finally gives gives

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

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

    and we have

    $\displaystyle \int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=||\frac{x} {1+x^2}||^2=\frac{1}{2 \pi}||\overbrace{\frac{x}{1+x^2}}^{\wedge}||^2=$

    and we know from the part 1) that

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

    then our integral will be

    $\displaystyle \int_{-\infty}^{\infty}\frac{x^2}{(1+x^2)^2}dx=\frac{1}{2 \pi}\int_{-\infty}^{\infty}-\pi^2 e^{-2|\xi|}d\xi=-\frac{\pi}{2}\cdot 2 \int_{0}^{\infty}e^{-2\xi}d\xi=-\pi \int_{0}^{\infty}e^{-2\xi}d\xi=$

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

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

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

    and:

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

    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 $\displaystyle ||*||\geq 0$ so the expression $\displaystyle ||-\pi ie^{-|\xi|}||^{2}$ is always positive and equals to $\displaystyle \pi^2 e^{-2|\xi|}$
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