I thought that for momentum integrals in Minkowski space, the Wick rotation to Euclidean space $k_0 \to ik_0$ allows one to write (let's say $f$ comes with an $i\epsilon$ prescription):
$$\int_{\mathbb{M}^4} d^4k \ f(k^2) = i \int_{\mathbb{E}^4} d^4k \ f(-k^2) = 2\pi^2i\int_0^\infty dk \ k^3 f(-k^2).$$
Sorry if the notation is weird, I wasn't sure how to denote the difference between Minkowski space and Euclidean space.
But I've come across a problem (calculating a loop integral) where doing this doesn't give the right answer, namely for
$$f(k^2)= \frac{1}{\left( k^2 - \Delta +i \epsilon \right)^2}.$$
Doing a Wick rotation results in:
$$I = \int_0^\infty \frac{k^3 dk}{\left( k^2 + \Delta \right)^2}$$
But the correct answer should be:
$$I = \int_0^\infty \frac{k^2 dk}{\left( k^2 + \Delta \right)^{3/2}}$$
According to Aitchison, Hey - 'Gauge Theories in Particle Physics', eq. (10.42). I uploaded it here.
Of course, both of these integrals are formally divergent, but suppose there's an energy cutoff. What went wrong here? What conditions must be met for a Wick rotation to work?
One suspicion I have is that maybe these two turn out to be equivalent, up to a choice of the cutoff energy. Especially since there is still a $\int_0^1 dx$ integration to be done to obtain the actual observable quantity, and $\Delta$ is a quadratic function of $x$:
$$\Delta = m_1^2 (1-x) + m_2^2 x -p^2 x(1-x) \equiv Ax^2 +Bx + C.$$
Could this be the reason? Or is the Wick rotation a mistake for some mathematical reason?
EDIT: I checked and the difference between the two integrals exists and equals $\log 2 - 1/2 \approx 0.2$, independent of $\Delta$. So they're not exactly equal even in the limit, but does it matter...?
