Thread: fourier transform and integral computing

1. fourier transform and integral computing

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

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

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?

2. Re: fourier transform and integral computing

Originally Posted by rayman
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

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

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

3. 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 $\displaystyle \pi^2 e^{-2|\xi|}$