I haven't found a duplicate question, but apologies in advance if this is a dup.
At the DSP SE we were asked about the Hilbert Transform of the unit step (you math guys call it the Heaviside step function which is the sum of a constant and the signum function). So I attempted to answer the question thus:
To integrate the convolution integral of the Hilbert transform, one must make use of the concept of the Cauchy principal value (p.v.) of an integral with some kinda nasty infinity or singularity in it.
Hilbert Transform:
$$\begin{align} \hat{x}(t) &\triangleq \mathscr{H}\Big\{ x(t) \Big\} \triangleq \frac{1}{\pi t} \ \circledast x(t) \\ \\ &= \mathrm{p.v.} \int\limits_{-\infty}^{\infty} \frac{1}{\pi v} \ x(t-v) \mathrm{d}v \\ \\ &= \lim_{\epsilon \to 0^+} \int\limits_{-1/\epsilon}^{-\epsilon} \frac{1}{\pi v} \ x(t-v) \mathrm{d}v \ + \ \int\limits_{\epsilon}^{1/\epsilon} \frac{1}{\pi v} \ x(t-v) \mathrm{d}v \\ \end{align}$$
That's in general. Now let $x(t)$ be the unit step function:
$$ x(t) = \tfrac{1}{2}\big(1 + \operatorname{sgn}(t)\big) = \begin{cases} 0 \qquad & t<0 \\ \tfrac12 \qquad & t=0 \\ 1 \qquad & t>0 \\ \end{cases} $$
and the sign function $\operatorname{sgn}(t)$ is
$$ \operatorname{sgn}(t) \triangleq \begin{cases} -1 \qquad & t<0 \\ \ \ 0 \qquad & t=0 \\ \ \ 1 \qquad & t>0 \\ \end{cases} $$
Because of the linearity of convolution and of the fact that the integral of a sum is the sum of integrals, we can think of this as the Hilbert transform of $\tfrac12$ added to the Hilbert transform of the sign function $\tfrac12 \operatorname{sgn}(t)$.
The Hilbert transform of $\tfrac12$ or of any constant is zero because the two integrals expressing the principal value above will be exact negatives for any value of $\epsilon$ in the limit. So the problem is just that of the Hilbert transform of $\tfrac12 \operatorname{sgn}(t)$.
In the integral above, let's say $t>0$. The math will be identical with flipped signs for the case $t<0$. For $t=0$ there will simply be the problem of infinity.
$$\begin{align} \mathscr{H}\Big\{ \tfrac12 \operatorname{sgn}(t) \Big\} & = \frac12 \lim_{\epsilon \to 0^+} \left( \int\limits_{-1/\epsilon}^{-\epsilon} \frac{1}{\pi v} \ \operatorname{sgn}(t-v) \mathrm{d}v \ + \ \int\limits_{\epsilon}^{1/\epsilon} \frac{1}{\pi v} \ \operatorname{sgn}(t-v) \mathrm{d}v \right) \\ \\ & = \frac{1}{2\pi} \lim_{\epsilon \to 0^+} \left( \int\limits_{-1/\epsilon}^{-t} \frac{1}{v} (+1) \mathrm{d}v \ + \ \int\limits_{-t}^{-\epsilon} \frac{1}{v} (+1) \mathrm{d}v \ + \ \int\limits_{\epsilon}^{t} \frac{1}{v} (+1) \mathrm{d}v \ + \ \int\limits_{t}^{1/\epsilon} \frac{1}{v} (-1) \mathrm{d}v \right) \\ \end{align}$$
Wow! This is problematic. The two integrals in the interior cancel each other, but the two integrals on the outside team up
$$\begin{align} \mathscr{H}\Big\{ \tfrac12 \operatorname{sgn}(t) \Big\} &= \frac{1}{2\pi} \lim_{\epsilon \to 0^+} \left( \int\limits_{-1/\epsilon}^{-t} \frac{+1}{v} \mathrm{d}v \ + \ \int\limits_{t}^{1/\epsilon} \frac{-1}{v} \mathrm{d}v \right) \\ \\ &= \frac{1}{2\pi} \lim_{\epsilon \to 0^+} 2\int\limits_{t}^{1/\epsilon} \frac{-1}{v} \mathrm{d}v \\ \\ &= \frac{1}{\pi} \lim_{\epsilon \to 0^+} \ -\big(\log(1/\epsilon) - \log(t)\big) \\ \\ &= \frac{1}{\pi} \big( \lim_{\epsilon \to 0^+} \ \log(\epsilon) + \log(t) \big) \\ \end{align}$$
I don't see how that converges to a finite value.
But another person, using indirect means, appears to show that the answer is simply $\tfrac1\pi \log(|t|)$ but I cannot see how that can be true since I thought I was pretty careful with this principal value and the integrals with finite limits were all well-defined and exist.
