Calculate $\left(\frac{x}{(1+x^2)^2}\ast \frac{1}{1+x^2}\right)(y)$ using Fourier transformations.
I have found a solution, but my method was very long. How could I shorten the solution?
Global idea: If I know $\mathcal{F}^\pm\left(\frac{x}{(1+x^2)^2}\right) (y) $ and $\mathcal{F}^\pm\left(\frac{1}{1+x^2}\right)(y) $ then I can use the following theorem
$$\text{If } f, g \in L^1(\mathbb{R})\text{ then } \mathcal{F}^\pm (f\ast g) = \sqrt{2\pi} \mathcal{F}^\pm f \cdot \mathcal{F}^\pm g$$
Then I will use the inverse Fourier transform $\mathcal{F}^\mp$ to find the solution.
Since $\mathcal{F}^\pm \left(\frac{1}{1+x^2}\right)(y) = \sqrt{\frac{\pi}{2}}e^{-|y|}$ all I need to calculate is $\mathcal{F}^\pm \left(\frac{x}{(1+x^2)^2}\right)(y)$.
This proved rather difficult, I use complex contour integration, can this be calculated any other way?
I found $\mathcal{F}^\pm \left(\frac{x}{(1+x^2)^2}\right)(y) = \pm \frac{i}{2}\sqrt{\frac{\pi}{2}} y e^{-|y|}$.
I the final step I needed to calculate
$$C\cdot \mathcal{F}^\mp\left(xe^{-2|x|}\right)(y) \qquad\text{for some constant }C$$ Once again a bit tedious. Could anyone point towards a shortcut?
You can simplify a lot of the calculations by using the fact that (under suitable hypotheses) the Fourier transform of $f'$ is $\xi\hat f(\xi)$ (times some constant that depends on where the $\pi$'s are in your particular definition of the Fourier transform). "Suitable hypotheses": Assuming $f$ is integrable and absolutely continuous is enough, for example.
First, $x/(1+x^2)^2$ is (up to a constant) the derivative of $1/(1+x^2)$, so you can calculate that transform that way.
And then at the end when you need the inverse transform of $\xi e^{-2|\xi|}$ you can use this again; this inverse transform is some mysterious constant times the derivative of the inverse transform of $e^{-2|\xi|}$. (And this mysterious constant exactly cancels the previous one.)