# Sokhotski–Plemelj theorem [closed]

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

• Would Mathematics be a better home for this question? – Qmechanic Mar 16 '17 at 20:53
• @Qmechanic, I also posted this question there. However, this is a so common integral in QFT, so I think it's worth posting it here. – user2891657 Mar 16 '17 at 20:54
• I'm voting to close this question as it has been crossposted. – Qmechanic Mar 16 '17 at 21:16
• @Qmechanic Could you give an idea how to answer such a question rather than vote here? – user2891657 Mar 16 '17 at 21:21
• You need to calculate the large circular boundary. The boundary contribution is $-i \pi$. The total circle is the residue which is $-2i \pi$. So use the contour the result is still the $-i\pi$. – user153663 May 25 '17 at 15:15