Plancherel formula and integral computing

[rayman123]

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?

 

[susskind_leon]

You already wrote down the solution.
||.||^2 means absolute value squared, so the integrand is positive.

 

[rayman123]

You already wrote down the solution.
||.||^2 means absolute value squared, so the integrand is positive.

yeah, you are right :) thank you