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Is the following statement true? $$ \lim\limits_{\epsilon \to 0^{+}} \ln( x \pm i \epsilon ) = \mathscr{P}\ln|x| \pm i \pi \Theta(-x) $$

where $\mathscr{P}$ is the Cauchy principal value. The above is a statement about distributions, so I guess it only really means anything under an integral sign, meaning: $$ \lim\limits_{\epsilon \to 0^{+}} \int_a^b dx\ f(x) \ln( x \pm i \epsilon ) = \mathscr{P} \int_a^b dx\ f(x) \ln|x| \pm i \pi \int_a^b dx\ f(x) \Theta(-x) $$

I've been encountering expressions in which it seems the above should be true while doing loop integrals in Quantum Field Theory. If you differentiate the above with respect to $x$ you also seem to return: $$ \lim\limits_{\epsilon \to 0^{+}} \frac{1}{x \pm i \epsilon} = \mathscr{P}\frac{1}{x} \mp i \pi \delta(x) $$

which is a rather famous identity (see pg. 113 in Chapter 3.1 of Weinberg's Quantum Theory of Fields Volume I, for example).

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  • $\begingroup$ I’m not sure that the principal value makes sense in this case, and you need to specify the branch cut for the logarithm. But otherwise this is correct, and quite commonly used when doing contour integrals, I think. $\endgroup$ – Stephen Powell Mar 26 '18 at 6:17
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    $\begingroup$ Yes the formula is evidently true. The principal part symbol is not necessary in front of $\ln |x|$ since it is a locally integrable function. $\endgroup$ – Valter Moretti Mar 26 '18 at 6:30

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