Hilbert transformation

There is a very important case of discontinuity: 1/t. It does not belong to our set of standard discontinuities. It will be useful to expand our set of discontinuities by introducing the Hilbert transformation:

$$R_{q,1}(t)={\bf H} R_q(t).$$

We see that $R_{-1,1}={\bf H}\delta(t)={1\over \pi t}$. But what about R_q,1 for q>-1? Unfortunately, the integral:

$${\bf H}R_{q,1}(t)={1\over \pi q!}\int_{0}^{\infty}{\tau^q\over (t-\tau)}d\tau$$

does not exist because of the non-integrable singularity at the point $\tau$=0. But we may use the following trick: if q>-1, we apply the operator ${\bf D}^{q+1}$ and obtain the discontinuity R_-1. After that we may apply the Hilbert transformation ${\bf H}$ and then the inverse operator ${\bf D}^{-(q+1)}$.

Remembering the equation (14), we get:

$$R_{q,1}(t)\stackrel{\infty}{\sim}{\bf D}^{-(q+1)}{\bf H}{\bf D}^{q+1}R_q(t).$$

This is a definition of the Hilbert transformation for discontinuities. We define also:

$$R_{q,\nu}(t)=\left[\cos({\pi\nu\over 2}){\bf E} + \sin({\pi\nu\over 2}){\bf H} \right]R_q(t) \equiv {\bf H}^{\nu}R_q(t)$$

where ${\bf E}$ is the unit operator.