Hilbert transform of unit step function
Absolute integrability is not a necessary condition for the Hilbert transform to exist. Simple counterexamples are the Hilbert transforms of $\cos(\omega t)$ and $\sin(\omega t)$, given by $\sin(\omega t)$ and $-\cos(\omega t)$, respectively, if $\omega&gt;0$.

Since the functions $u(t)$ and $\textrm{sgn}(t)$ are related by scaling and an additive constant, and since the Hilbert transform of a constant is zero, the Hilbert transforms of the two functions - if they exist - must be scaled versions of each other:

$$\textrm{sgn}(t)=2u(t)-1\\\Rightarrow \mathcal{H}\{\textrm{sgn}(t)\}=2\mathcal{H}\{u(t)\}\tag{1}$$

I believe that the Hilbert transform of these two functions can be computed in the following way. [Edit: this turned out to be true. See below and the comments to RBJ's answer.] We know that the Hilbert transform can be written as a convolution:

$$\mathcal{H}\{x(t)\}=\frac{1}{\pi t}\star x(t)\tag{2}$$

This can be rewritten as

$$\mathcal{H}\{x(t)\}=\frac{1}{\pi}\left(\log(|t|)\right)'\star x(t)=\frac{1}{\pi}\log(|t|)\star x'(t)\tag{3}$$

where $'$ denotes differentiation w.r.t. $t$. The rightmost equality in $(3)$ is true because we generally have

$$(f\star g)'=f'\star g=f\star g'\tag{4}$$

With $x(t)=u(t)$ we get $x'(t)=\delta(t)$, and from $(3)$ we obtain

$$\mathcal{H}\{u(t)\}=\frac{1}{\pi}\log(|t|)\star \delta(t)=\frac{1}{\pi}\log(|t|)\tag{5}$$

From $(1)$ and $(5)$ we get

$$\mathcal{H}\{\textrm{sgn}(t)\}=\frac{2}{\pi}\log(|t|)\tag{6}$$



The results given by Eqs $(5)$ and $(6)$ can also be derived in the frequency domain. The Fourier transform of the convolution

$$\mathcal{H}\{u(t)\}=\frac{1}{\pi t}\star u(t)$$

is given by

$$\mathcal{F}\left\{\frac{1}{\pi t}\right\}\mathcal{F}\{u(t)\}=-j\;\textrm{sgn}\{\omega\}\left(\pi\delta(\omega)+\frac{1}{j\omega}\right)=-\frac{\textrm{sgn}(\omega)}{\omega}=-\frac{1}{|\omega|}$$

So we have

$$\mathcal{H}\{u(t)\}=\mathcal{F}^{-1}\left\{-\frac{1}{|\omega|}\right\}$$

From this document (p. 27, Ex. 19.4), we know that

$$\mathcal{F}^{-1}\left\{-\frac{1}{|\omega|}\right\}=\frac{1}{\pi}\log(|t|)$$

confirming Eq. $(5)$.

(Note the different scaling in the cited document due to the use of the unitary Fourier transform.)


EDIT:

The problem can also be solved using the definition of the Hilbert transform of bounded functions:

$$\mathcal{H}\{f(t)\}=\frac{1}{\pi}\left[\mathrm{p.v.}\int_{|\tau|&lt;1}\frac{f(\tau)}{t-\tau}d\tau+\mathrm{p.v.}\int_{|\tau|&gt;1}f(\tau)\left(\frac{1}{t-\tau}+\frac{1}{\tau}\right)d\tau\right]$$

where $\mathrm{p.v.}$ denotes the Cauchy principal value.

For $f(t)=u(t)$ we have

$$\mathcal{H}\{u(t)\}=\frac{1}{\pi}\left[\underbrace{\mathrm{p.v.}\int_{0}^{1}\frac{d\tau}{t-\tau}}_{\displaystyle I_1}+\underbrace{\mathrm{p.v.}\int_{1}^{\infty}\left(\frac{1}{t-\tau}+\frac{1}{\tau}\right)d\tau}_{\displaystyle I_2}\right]$$

If $t$ is outside the interval $(0,1)$, the first integral is simply

$$I_1=\int_{0}^{1}\frac{d\tau}{t-\tau}=\log\left(\frac{t}{t-1}\right)$$

If $t\in (0,1)$ we get

$$I_1=\lim_{\epsilon\to 0^+}\int_0^{t-\epsilon}\frac{d\tau}{t-\tau}+\int_{t+\epsilon}^1\frac{d\tau}{t-\tau}=\log\left(\frac{t}{\epsilon}\right)+\log\left(\frac{-\epsilon}{t-1}\right)=\log\left(\frac{t}{1-t}\right)$$

Since in both cases the argument of the logarithm is positive, both results can be summarized as

$$I_1=\log\left|\frac{t}{1-t}\right|$$

If $t&lt;1$ the second integral is

$$I_2=\lim_{\epsilon\to 0^+}\left[\int_1^{1/\epsilon}\frac{d\tau}{t-\tau}+\int_1^{1/\epsilon}\frac{d\tau}{\tau}\right]=\lim_{\epsilon\to 0^+}\left[\log\left(\frac{t-1}{t-1/\epsilon}\right)+\log\left(\frac{1}{\epsilon}\right)\right]=\\=\lim_{\epsilon\to 0^+}\log\left(\frac{t-1}{\epsilon t-1}\right)=\log\left(1-t\right)$$

If $t&gt;1$ we have

$$I_2=\lim_{\epsilon\to 0^+}\left[\int_1^{t-\epsilon}\frac{d\tau}{t-\tau}+\int_{t+\epsilon}^{1/\epsilon}\frac{d\tau}{t-\tau}+\int_1^{1/\epsilon}\frac{d\tau}{\tau}\right]=\\=\lim_{\epsilon\to 0^+}\left[\log\left(\frac{t-1}{\epsilon}\right)+\log\left(\frac{-\epsilon}{t-1/\epsilon}\right)+\log\left(\frac{1}{\epsilon}\right)\right]=\\=\lim_{\epsilon\to 0^+}\log\left(\frac{1-t}{\epsilon t-1}\right)=\log(t-1)$$

Again, both results can be summarized as

$$I_2=\log|1-t|$$

Combining these results, we obtain the Hilbert transform of $u(t)$:

$$\mathcal{H}\{u(t)\}=\frac{1}{\pi}(I_1+I_2)=\frac{1}{\pi}\left(\log\left|\frac{t}{1-t}\right|+\log|1-t|\right)=\frac{1}{\pi}\log|t|$$

which agrees with our previous result given by Eq. $(5)$.