quantum field theory - Sokhotski–Plemelj theorem

In quantum field theory we always deal with complex integrals. They arise, e.g. when we calculate Feynman diagrams in either $T=0$, $T \neq 0$ and non-equilibrium formalism (or just Fourier transforms in any branch of physics). If we have free propagators in, e.g. self-energy diagrams, we often have expressions like $1/(x \pm i0)$ due to analytical properties of Green's functions.

In all QFT books, we learn about Sokhotski formula, which reads $$\frac{1}{x + i0} = \text{P}\frac 1 x - i \pi\delta(x) $$ If we apply it to $$\int\limits_{-\infty}^{\infty} \frac{dx}{x+i0}$$ we therefore get $- i \pi$.

On the other hand, we can seemingly compute this integral using the residue theorem. We can regularise it by adding $$\lim_{\varepsilon \rightarrow 0} e^{- i \varepsilon x} \, ,$$ so we can close the contour in the lower half plane and get $- 2 \pi i$.

Integral regularization is certainly quite artificial trick. However, it is commonly and successfully used in quantum mechanics (and in Fourier transforms, e.g. of Coulomb potential) as well as Sokhotski formula. If I just put $$\int\limits_{-\infty}^{\infty} \frac{dx e^{- i 0 x}}{x+i0}$$ into Mathematica, I get $- 2 \pi i$. So, I can't see how to resolve this contradiction.

Just to let you know - I also posted this question here https://math.stackexchange.com/questions/2189979/sokhotski-plemelj-theorem