Why can I not asymptotically expand a Feynman integral this way?

Question

I would like to asymptotically expand a series of Feynman diagrams in Euclidean space, and as a toy I started with the following integral, for which I know the full solution in $4d$ ($\omega \to 2$):

$$\int_{-\infty}^\infty d\tau_3\ X_{1123} := \int_{-\infty}^\infty d\tau_3\int d^{2\omega} x_5 \frac{1}{\left(x_{15}^2 x_{15}^2 x_{25}^2 x_{35}^2\right)^{\omega-1}} \tag{1}$$

with $x_1 := (1,0,0,0)$, $x_2 := (x_2,0,0,0)$, $x_3 := (0,0,0,\tau_3)$, $x_{ij} := x_i - x_j$.

For my question it suffices to look at the divergent part of the integral, which occurs at $x_5 \sim x_1$:

$$\left. \int_{-\infty}^\infty d\tau_3\ X_{1123} \right|_\text{div} = \int_{-\infty}^\infty d\tau_3 \frac{1}{\left(x_{21}^2 x_{31}^2\right)^{\omega - 1}} \int d^{2\omega} x_5 \frac{1}{\left(x_5^4\right)^{\omega-1}} \propto \frac{1}{\left|x_1 \right| x_{12}^2} \log \epsilon^2 \tag{2}$$

This result is robust, since $X_{1123}$ is known analytically in point-splitting regularization (see Drukker, Plefka, eq. $21$) and the remaining integral over $\tau_3$ is simple. I want to do an asymptotic expansion for $x_2 \to \infty$. In that case the divergence becomes:

$$\left. \int_{-\infty}^\infty d\tau_3\ X_{1123} \right|_{\text{div}, x_2 \to \infty} \propto \frac{1}{\left|x_1 \right| x_2^2} \log \epsilon^2 \tag{3}$$

Now I try to do the same expansion, but assuming that I don't know the result in $(2)$, i.e. I start by using $x_2 \sim \infty$ in $(1)$ and look at the divergence after. This is what I am really interested in, since I would like to do similar operations on integrals for which I do not know the analytical solution. I find:

$$\begin{align} \left. \int_{-\infty}^\infty d\tau_3\ X_{1123} \right|_{x_2 \to \infty} &= \int_{-\infty}^\infty d\tau_3\ \int d^{2\omega} x_5 \frac{1}{\left[ \left(x_{12} - x_5 \right)^4 x_5^2 \left( x_{32} - x_5 \right) \right]^{\omega-1}} \\ & \sim \int_{-\infty}^\infty d\tau_3\ \int d^{2\omega} x_5 \frac{1}{\left[ \left(x_2 - x_5 \right)^4 x_5^2 \left( x_{32} - x_5 \right) \right]^{\omega-1}} \\ & = \int_{-\infty}^\infty d\tau_3\ \int d^{2\omega} x_5 \frac{1}{\left( x_5^4 x_{25}^2 x_{35}^2 \right)^{\omega-1}} \tag{4} \end{align}$$

in which I shifted the integral accordingly in the first and last steps.

The problem is clear: when I look at the divergence of $(4)$, it is not located at $x_5 \sim x_1$ anymore and computing the integral further logically gives me something promotional to $1/(x_2^2)^{3/2}$ (it had to be the case for dimensional reasons anyway), which contradicts $(3)$.

Now my questions are:

1. Why can I not shift the integral the way I did in $(4)$ ? Surely it is related to the fact that there is a divergence at $x_1$, but I do not see clearly what prevents me concretely to do so.

2. Does the step in $(4)$ affect only the divergent part or also the finite part (I also wish to compute that)? I would think that the finite part is also affected, since the dependency $1/(x_2^2)^{3/2}$ also contradicts what I know about the finite part of $(1)$.

3. What would be the right way to go? I am mostly interested about computing the finite part, since I already checked that the divergences cancel with other diagrams.

Many thanks in advance!