Nice question! The Hilbert transform $\mathcal{H}\left\{f(\omega)\right\}$ with
$$f(\omega)=-\frac12\log(1+\omega^2)\tag{1}$$
can be calculated in the following way. First, note that
$$\frac{df(\omega)}{d\omega}=-\frac{\omega}{1+\omega^2}\tag{2}$$
From this table we know that
$$\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}=\frac{\omega}{1+\omega^2}\tag{3}$$
We also know that
$$\mathcal{H}\{\mathcal{H}\{f\}\}=-f\tag{4}$$
Combining $(3)$ and $(4)$ we get
$$\mathcal{H}\left\{\frac{\omega}{1+\omega^2}\right\}=\mathcal{H}\left\{\mathcal{H}\left\{\frac{1}{1+\omega^2}\right\}\right\}=-\frac{1}{1+\omega^2}\tag{5}$$
So
$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{1}{1+\omega^2}\tag{6}$$
Now we also know that the Hilbert transform operator and the differentiation operator commute:
$$\mathcal{H}\left\{\frac{df(\omega)}{d\omega}\right\}=\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}\tag{7}$$
which yields
$$\frac{d}{d\omega}\mathcal{H}\{f(\omega)\}=\frac{1}{1+\omega^2}\tag{8}$$
Integrating $(8)$ finally gives
$$\mathcal{H}\{f(\omega)\}=\arctan(\omega)\tag{9}$$