Phase spectrum: $\mathrm{arg}\mathcal{F}(\omega)$ or $\mathrm{-arg}\mathcal{F}(\omega)$? Do we need minus or not?
I need to extract phase spectrum from that thing

$$f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega x}d\omega\int_{-\infty}^{\infty}f(t)e^{-i\omega t}dt$$

I can do it by taking Fourier Transform of $f(x)$, right? And phase spectrum will be:

$$S(\xi)=\int_{-\infty}^{\infty}f(x)e^{-i\xi x}dx $$

$$\Phi (\xi)=-\mathrm{arg}\,S(\xi)$$

I found that in book, but don't understand do we need minus
 before arg or not? I'm looking on different plots and some sources plotting with minus and some without. Help me please
Both definitions are common. However, if you talk about the phase, then what is usually meant is the (frequency dependent) phase angle $\phi(\omega)$ of the Fourier transform $F(\omega)$:

$$F(\omega)=|F(\omega)|e^{j\phi(\omega)}\tag{1}$$

The negative phase $\beta(\omega)=-\phi(\omega)$ is often called the phase shift or phase lag. It is commonly used together with the attenuation $\alpha(\omega)$, which is the negative natural logarithm of $|F(\omega)|$:

$$\alpha(\omega)=-\log (|F(\omega)|)\tag{2}$$

$$F(\omega)=e^{-(\alpha(\omega)+j\beta(\omega))}\tag{3}$$