Plancherel formula and integral computing

[rayman123]

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?

 

[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