homework and exercises - A four-dimensional integral in Peskin & Schroeder

The following identity is used in Peskin & Schroeder's book Eq.(19.43), page 660:

$$\int\frac{d^4k}{(2\pi)^4}\,\frac{1}{(k^2)^2}e^{ik\cdot\epsilon}=\frac{i}{(4\pi)^2}\log\frac{1}{\epsilon^2},\quad \epsilon\rightarrow 0$$

I can't figure out why it holds. Could someone provide a method to prove this? Many thanks in advace.

Answer

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

$$\int \frac{i\cdot d^4 k_E}{(2\pi)^4} \frac{1}{(k_E^2)^2} \exp(ik\cdot \epsilon)$$ 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.

The only remaining unknown is the coefficient and one gets $4\pi^2$ from the remaining integral. It's a sort of waste of resources to compute this special integral; it's better to compute the more general integrals in appendix A.4, see especially formulae (A.44)-(A.49) on page 807, which I won't copy here because that's why Peskin and Schroeder wrote the textbook.