Find the Fourier Transform of $2x/(1+x^2)$ I tried doing this the same way you would find the Fourier transform for $1/(1+x^2)$ but I guess I'm having some trouble dealing with the 2x on top and I could really use some help here.
 A: Hint: Taking the derivative with respect to $k$ of $$F(k)=\int_{-\infty}^{\infty}\frac{1}{1+x^2}e^{ikx}dx$$
yields
$$F'(k)=i\int_{-\infty}^{\infty}\frac{x}{1+x^2}e^{ikx}dx$$
Thus, the Fourier Transform of $\frac{2x}{1+x^2}$ is $-2i$ times the derivative with respect to $k$ of the Fourier Transform of $\frac{1}{1+x^2}$.
$$\mathscr{F}\left(\frac{2x}{1+x^2}\right)(k)=-2i \mathscr{F}\left(\frac{1}{1+x^2}\right)(k)$$
A: Your function is square integrable. So the Fourier transform will be square integrable, and expressed as the Cauchy principal value
$$
    \lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}e^{-isx}\frac{2x}{1+x^{2}}dx \\
  = \lim_{R\rightarrow\infty}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}e^{-isx}\left(\frac{1}{x+i}+\frac{1}{x-i}\right)dx.
$$
Integration by parts gives evaluation terms that vanish as $R\rightarrow\infty$ so that the above becomes
$$
  = \frac{i}{s}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-isx}\left(\frac{1}{(x+i)^{2}}+\frac{1}{(x-i)^{2}}\right)dx.
$$
The Cauchy principal value limit was removed because this last integral expression is absolutely convergent.
A: Another way to solve this. As you know, $$-ix \mathcal{F}[g](x) = \mathcal{F}[g'](x)$$ and $$\mathcal{F}[\mathcal{F}[g]] = cg$$
You have (with some coefficients)
$$ \mathcal{F}[ x \mathcal{F}[g](x) ] (\xi)= \mathcal{F}[i\mathcal{F}[g'] ] (\xi) = i g'(\xi)$$
Now call $f = \mathcal{F}[g]$, and you got
$$ \mathcal{F}[ x f(x) ] (\xi)= i \mathcal{F}[f ]'(\xi)$$
So if you know $\mathcal{F}[f ]$, you can conclude
(I left the exaxct value of the coefficients in the wind, there may be some nasty $2\pi$ and other values here)
