quantum field theory - The relation of two integrals

In this post about the integral

$$\int\frac{d^4k}{(2\pi)^4}\,\frac{1}{(k^2)^2}e^{ik\cdot\epsilon}=\frac{i}{(4\pi)^2}\log\frac{1}{\epsilon^2},\quad \epsilon\rightarrow 0, \tag{19.43}$$ which is (19.43) on page 660 of Peskin and Shroeder,
Luboš answered as follows.

That's equivalent simply to $c\int dx/x$. Switch to the Euclidean spacetime, $k_0=ik_4$ where $(k_1,\dots k_4)$ is $k_E$; i.e. analytically continue in $k_0$ (Wick rotation). The integral is

$$\int \frac{i\cdot d^4 k_E}{(2\pi)^4} \frac{1}{(k_E^2)^2} \exp(-ik_E\cdot \epsilon_E)$$ So it's proportional to the Fourier transform of $1/k_E^4$. The original function is $SO(4)$ symmetric, so the Fourier transform must be symmetric as well and depend on $\epsilon^2$ only. Dimensional analysis implies that the result is dimensionless i.e. it must be a combination of a constant and $\ln(\epsilon^2)$. The logarithm is there with a nonzero coefficient so the constant only determines how to take the logarithm: it should properly be written as $\ln(\epsilon^2/\epsilon_0^2)$ for some constant $\epsilon_0$ with the same dimension.

Indeed when

$$f(x_E)=\int\frac{d^4p_E}{(2\pi)^4} \tilde{f}(p_E)\,e^{ip_E \cdot x_E}$$ and $\tilde{f}(p_E)=\tilde{f}(M_1 p_E)$, we can show $f(M_2 x_E)=f(x_E)$ by a short calculation, where $M_1,\,M_2$ are orthogonal matrices.
But I'm not sure why the result is proportional to a combination of a constant and $\ln(\epsilon^2)$.
Though using dimensionlessness, $\sum_n c_n (\epsilon^2/\epsilon_0^2)^n$ is also dimensionless.
How can we relate $$\displaystyle \int \frac{i\cdot d^4 k_E}{(2\pi)^4} \frac{1}{(k_E^2)^2} \exp(-ik_E\cdot \epsilon_E)$$ with $\int dx/x$ rigorously?