Concerning functions in question are not integrable on the line, the Fourier transform has to be considered in the sense of distributions. Particularly for the logarithm, it is known that (Vladimirov, Equations of Mathematical Physics, $\S2.5$)
$$
F\left[{\cal P}\frac1{|x|}\right]=-2\gamma-2\log|\xi|,
$$
where $\gamma$ is the Euler constant and ${\cal P}\frac1{|x|}$ is a distribution defined by
$$
({\cal P}\frac1{|x|},\varphi)=
\int_{|x|\le 1}\frac{\varphi(x)-\varphi(0)}{|x|}\,dx+
\int_{|x|> 1}\frac{\varphi(x)}{|x|}\,dx.
$$
With inverse FT one can get from here the FT of $\log|x|$:
$$
F[\log|x|](\xi)=-2\pi\gamma\delta(\xi)-\frac\pi{|\xi|},
$$
taking into account that FT is defined in this book as
$$
F[f](\xi)=\int_{\mathbb R}f(x)e^{ix\xi}\,dx.
$$
There is a good reason to use $z$ instead of $e^{sT}$. Before starting with any analysis, let me remind you that in analysis of signals and systems we are interested in analyzing the frequency spectrum of the signal, i.e. the Laplace transform on the imaginary line $s = jw$. And since your signal $x[n]$ is discrete, then its frequency spectrum its periodic, so its more general define $s=jwT$.
Now, let $X(z) = \mathcal{Z}\{x[n]\}$ of a causal or non-causal discrete signal $x[n]$, i.e.
$$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}. $$
Since $z\in\mathbb{C}$ we have $z = |z| e^{j\arg z}$. Without loss of generality we rewrite $|z| = r$ and $\arg z = wT$, i.e. $z=r e^{jwT}$ (note that not necessarily $r=1$). Then
$$ \begin{aligned}
X(z) &= \sum_{n=-\infty}^{\infty} x[n] z^{-n}\\
&= \sum_{n=-\infty}^{\infty} x[n] (r e^{jwT})^{-n}\\
% &= \sum_{n=-\infty}^{\infty} (x[n] r^{-n}) e^{-njwT}\\
&= \sum_{n=-\infty}^{\infty} (x[n] r^{-n}) (e^{jwT})^{-n},
\end{aligned} $$
which implies $X(z) = \left. \mathcal{L}\{x[n]r^{-n}\} \right\rvert_{s=jwT} = \mathcal{F}\{x[n]r^{-n}\}$. As a consequence, $X(z)$ is a Fourier transform more generic than the Fourier transform $X(e^{jwT}) = \mathcal{F}\{x[n]\}$ of our signal of interest.
So, if the convergence radius of $X(z)$ is less than unity then $X(e^{jwT})$ does not exist and therefore its Fourier transform does not either, which represents a problem because there are many signals with this problem of convergence, e.g. non-causal signals such as a digital image filter. Therefore, it is convenient (and even necessary in non-causal signals) to use the Z-transform.
Or informally, use $z$ instead of $\left. e^{sT} \right\rvert_{s=jwT} = e^{jwT}$ whenever you can.
We also recommend to see this link about radius convergence.
At this point, it is clear that the Z-transform has the same objective as the Laplace transform: ensure the convergence of the transform in some region of $\mathbb{C}$, where the Z-transform does it for discrete signals and Laplace transform for continuous signals.
Best Answer
The Hilbert transform is anti-self-adjoint. Therefore, it is natural to define it on distribution by passing $\mathcal H$ to the test functions, similar to "pass the hat" definition of the Fourier transform. In fact, the Wikipedia article already says this.
Since the stated relation between $\mathcal{F}$ and $\mathcal H$ holds for test functions, the duality-based definition implies that it holds for distributions as well.