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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?

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    $\begingroup$ For spherical symmetry in 4D Euclidean space you can replace $d^4k_E$ by $2 \pi^2 k^3 dk$ $\endgroup$ – jim Jan 29 '17 at 22:01
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    $\begingroup$ After replacing $d^4k_E$ by $2 \pi^2 k^3 dk$, we get$\frac{i}{4\pi^2} \int_0^{\infty} ds\frac{1}{s^2} J_1(s)$. And then ...? $\endgroup$ – GotchaP Jan 29 '17 at 23:24
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    $\begingroup$ You may calculate these things if you learn the Schwinger and Feynman parameterizations: en.wikipedia.org/wiki/Schwinger_parametrization and en.wikipedia.org/wiki/Feynman_parametrization - All the power laws in the integrand disappear, and you can turn the things to an integral over the Schwinger parameters of an integral of a Gaussian. The Gaussian integrals may always be calculated etc. My comment is that it's equivalent to $dx/x$ means that you encounter $dx/x$ as the hardest part in the calculation, or that the process applied to $\int dx/x$ is analogous. $\endgroup$ – Luboš Motl Jan 31 '17 at 15:40
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    $\begingroup$ Just to be sure, my previous comment isn't meant to say that the Schwinger and Feynman parameterizations are the only way how to calculate the integrals. But they're a systemic, industrialized, "straightforward" method to calculate lots of such integrals. $\endgroup$ – Luboš Motl Jan 31 '17 at 15:52
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    $\begingroup$ I upvoted your calculation as a good one. Note that you did use the Schwinger parameters in your calculation so that's indeed close to the hint recommended in my comments above. But there are other calculations, even by other people on the page you linked to. There are also general arguments why it has to end up depend on $\ln\epsilon$ etc. $\endgroup$ – Luboš Motl Feb 3 '17 at 7:17

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