# Why doesn't Wick rotation work for this integral?

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...?

As noted in the edit, I also believe the two results will be the same. Let's do an example using a convergent integral, taking $I\equiv \int d^4k \, f(k^2),$ where $f(k^2)=(k^2-\Delta+i\epsilon)^{-3}$. Then performing the Wick rotation yields $I = -2\pi^2i\int_0^\infty \frac{k^3dk}{(k^2+\Delta)^3}=-\frac{i}{2}\frac{\pi^2}{\Delta}.$
On the other hand, direct integration over $k^0$ yields $I = -\frac{3\pi i}{8}4\pi\int_0^\infty \frac{k^2dk}{(k^2+\Delta)^{5/2}} = -\frac{i}{2}\frac{\pi^2}{\Delta}.$