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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.

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I haven't attempted the integral, but with these sort of things, spherical polars in k space are sometimes a useful approach. Have you tried that? –  twistor59 Apr 18 '13 at 11:16
I have tried this approach, but I can't get it right. The trouble is how to take the integration with respect to $k^0$. –  soliton Apr 18 '13 at 11:24
It's just a loop integral which can be evaluated using the formulae given in the appendix of P&S. Then Taylor-expand the result in $\epsilon$ and you have the result. Edit: oh, just saw that Lubos already said that... –  A friendly helper Apr 18 '13 at 12:20
It looks to me like this can be evaluated using the integral representation of the Dirac delta: $$\delta(\epsilon) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ik\epsilon} dk.$$ This would mean that you just have to understand the above identity, which is a standard and historical problem. –  Douglas B. Staple Apr 18 '13 at 16:14
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2 Answers

up vote 5 down vote accepted

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.

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Why can we say that "it's proportional to the Fourier transform of $1/k^4_E$" since $(k^2_E)^2=(k_1^2+k_2^2+k_3^2+k_4^2)^2$ and $e^{ik\cdot\epsilon}$ is not $SO(4)$ symmetric? –  soliton Apr 18 '13 at 13:12
Dear Soliton, the $\exp(ik\cdot \epsilon)$ factor is the phase that is a part of the definition of the Fourier transform! It's nothing we have added to the function we're Fourier-transforming. –  Luboš Motl Apr 18 '13 at 16:08
Thanks a lot. Dimensional analysis is a shortcut to obtain the result. –  soliton Apr 19 '13 at 12:01
I have found another approach by using the result of Eq.(5.2.9) in Weinberg's book (vol. 1, page 202) and the asymptotic expansion of Bessel function $K_1(x)=\frac{1}{x}+\frac{x}{2}\log\frac{x}{2}$ as $x\rightarrow 0$. –  soliton Apr 19 '13 at 12:10
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I will give another approach to this identity. First, we notice that

$$\int\frac{d^4k}{(2\pi)^4}\,\frac{1}{(k^2-m^2)^2}e^{ik\cdot\epsilon}=-i\frac{\partial}{\partial m^2}D_F(x)\big|_{x=\epsilon}$$

For space-like vector $\epsilon^2=-r^2<0$, we have


whose derivation refers to Weinberg's book vol. 1, page 202. For $r\rightarrow 0$, the following expansion holds

$$ K_1(mr)=\frac{1}{mr}+\frac{mr}{2}\log\frac{mr}{2}$$

With this conditions, we finally obtain


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