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In Feynman diagram calculations, we seem to invariably Euclideanise loop integrals in order to exploit the resulting spherical symmetry. This Wick rotation is simply a deformation of the contour; providing we avoid all poles and providing our integrand falls off at infinity sufficiently fast, this is legitimate. See, for instance, Figure 6.1 of Peskin & Schroeder.

However, when looking at the integral at face value, in Minkowski space, it sometimes appears that the integral is divergent (or at least, not absolutely convergent), even if the above analysis suggests everything is kosher. Consider, for instance, the 2D integral

$$ \int \mathrm{d}^2k \frac{1}{(k^2 + i \epsilon)^2} \,. $$

The integrand does not fall off along the line $k^0 = k^1$, which stretches out to infinity. Indeed, if we demand that the absolute value of the integrand be larger than some numer $\delta > 0$, we find an infinite area region that extends along this line. So the integral is not absolutely convergent. On the other hand, the Euclideanised integral is absolutely convergent.

Question: is there some analytic subtelty in performing this Wick rotation that is typically overlooked? And if so, why doesn't it impact the calculations we go on to perform? If the Wick rotation is implicitly serving as some form of regularisation, is it obvious that the physical observables we compute are independent of this regularisation?

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    $\begingroup$ Strictly speaking, you should only Wick rotate absolutely convergent integrals, so you must introduce a regulator first. In dimReg, the Wick rotation is part of the definition of the regularisation procedure and, as of today, no inconsistency has been found in this prescription. But yeah, it's not perfectly justified. $\endgroup$ Commented Apr 24, 2018 at 18:08
  • $\begingroup$ related: Wick rotation in field theory - rigorous justification? and links therein. $\endgroup$ Commented Apr 24, 2018 at 18:10
  • $\begingroup$ We also require that the integral has no singularities in the complex $k^0$ plane in first and third quadrants. This important requirement can be proven for arbitrary loop integrals with some work. $\endgroup$
    – QuantumDot
    Commented Apr 24, 2018 at 20:23

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