I am starting to learn about QFT and something that I noticed is that integrals who would diverge otherwise are assigned a value if we do it by contour integration using the residues theorem and the Feynman prescription, e.g.

$$ \begin{alignat}{7} i\int\dfrac{\mathrm{d}^4p}{\left(2\pi\right)^4}\dfrac{1}{p^2-M^2} & ~~=~~ & i & \int \dfrac{\mathrm{d}^3\textbf{p}}{(2\pi)^4} && ~~ \int_{-\infty}^{\infty} \mathrm{d}p^0\dfrac{1}{\left(p^0\right)^2-E^2_{\textbf{p}}} \\ & ~~=~~ & \dfrac{1}{2} &\int \dfrac{\mathrm{d}^3\textbf{p}}{\left(2\pi\right)^3} && ~~ \dfrac{1}{E_{\textbf{p}}}. \end{alignat} $$

The integral at the most right in the first line diverges ($p^0$ is real as is $E_{\textbf{p}}$), but if we use the feynman prescription, i.e., changing the integrand$$ \dfrac{1}{\left(p^0\right)^2-E^2_{\textbf{p}}} ~~ \rightarrow ~~ \dfrac{1}{\left(p^0\right)^2-E^2_{\textbf{p}} + i\epsilon} \,,$$we can evaluate this to a finite value (which was used to go from the first to the second line) using the residues theorem. Incidentally, the integral on the second line still diverges and we need to regularize it.

But my question concerns the first step, why is it valid to assign a value to this integral just by shifting its poles? The integral isn't convergent, when we shift its poles we are evaluating another integral, not the same we started with.

  • $\begingroup$ While not the same, a related question (and very good answer to the shifting contour part) physics.stackexchange.com/q/138217. Mostly a side note on why you can add the imaginary part in the denominator $\endgroup$ – Triatticus May 19 '18 at 22:39

This is a common (and good) question, which is the result of taking introductory texts more seriously than one should.

Reading such texts, one gets the impression that the philosophy is

We proceed as naively as possible and, when we find a divergence, we regulate it.

On the other hand, the correct (and much more useful) attitude is

We take things more seriously from the start, and make sure all expressions are always finite. We don't regulate things on the go, but take a finite theory as a starting point.

The second attitude is perhaps slightly more involved to formulate, and that's the reason introductory texts don't follow it. It is easier to fix things on the go rather than to anticipate problems that the reader is unfamiliar with. But once the global picture is more or less clear, one should change the philosophy to the more correct one, where everything is finite from the beginning.

In this sense, OP's question becomes meaningless when thought of in the context of the second attitude: we are not introducing $+i\epsilon$ and changing the integral; rather, the $+i\epsilon$ was always there, since square one. OP asks why is it valid to assign a finite value to an integral by changing the integrand. They are right to be skeptical: integrals are what they are, and if you modify them, you are computing something else. You are not allowed to change an integrand, because if you do so, you are not computing what you wanted to compute.

The resolution is that if you do things right from the beginning, the integral you actually want to compute is the modified one, not the divergent one.

Now, formulating QFT correctly from the beginning is beyond the scope of this answer, but let me mention that the $+i\epsilon$ prescription and its origin within the context of the path-integral formulation of QFT is discussed in this PSE post. In the operator formalism, the origin is slightly different (but the philosophy is the same). All in all, the answer to OP's question is: the $+i\epsilon$ is not introduced by hand, but it was always there. If you didn't see it before, it is because the text you are following was trying to keep things as simple as possible.

  • $\begingroup$ I don't have the background to go into the math you described in your other post yet. But to answer my question, the origin to the $i\epsilon$ prescription in QFT comes from the boundary conditions on the fields in the path-integral formalism, right? If not the same origin, the boundary conditions are applied to ''what'' in the operator formalism? $\endgroup$ – Slayer147 May 20 '18 at 14:52
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    $\begingroup$ @Slayer147 1) Yes, the $i\epsilon$ comes from the path-integral boundary conditions. 2) In the operator formalism, there are no boundary conditions (well, not at least in the standard case; in some more complicated cases there are). There are different ways to understand the origin of $i\epsilon$ in the operator formalism. The one I like the most is that it ensures causality (i.e., it comes from the time-ordering symbol; more precisely, from the Fourier transform of the step function, cf. Weinberg's QFT, Vol.1, §6.2). $\endgroup$ – AccidentalFourierTransform May 20 '18 at 15:43

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