Hilbert transform of $ \frac{x}{(1+x^2)^2} $.

Question

I want to calculate the Hilbert transform of the function $ \frac{x}{(1+x^2)^2} $ but I find the way by direct computations that $$ Hf(x)=\frac{1}{\pi}\lim_{\epsilon\to 0^+}\int_{|y|>\epsilon}\frac{f(x-y)}{y}dy $$ is too complicated. How can I simplify the calculations?

Answer 1

You may use complex analysis for this integral. Below is a sketch with missing details for the reader to fill in.

For a given $x\in\mathbb{R}$, define $g:\mathbb{C}\to\mathbb{C}$ by $$g(z) = \frac{f(x-z)}{z} = \frac{x-z}{z(1+(x-z)^2)^2}$$ Take a closed contour $C_{r,R}$ as in https://i.stack.imgur.com/oEpgl.png . The only singularity of $g$ that lies inside of this contour is $x+i$, for $r>0$ small and $R>0$ large enough. Other singularities do not fall inside this countour for any $0<r<R$. Then, by the residue theorem $$\oint_{C_{r,R}} g(z)dz = 2\pi i\ \textrm{Res}(g,x+i) = 2\pi i \lim_{z\to x+i} \frac{d}{dz} \frac{x-z}{z(z-x+i)^2}$$ As $R\to\infty$ and $r\to 0$, we get $$\pi\ Hf(x)-i\pi\frac{x}{(1+x^2)^2} = \lim_{r\to 0}\lim_{R\to\infty}\oint_{C_{r,R}} g(z)dz = 2\pi i \lim_{z\to x+i} \frac{d}{dz} \frac{x-z}{z(z-x+i)^2}$$