$$\int \frac{d^4k}{(2\pi)^4} \frac{1}{k^2}\frac{1}{k^2}\frac{1}{k^2}.\tag{1}$$

We can Wick rotate $k_0 \to i k_0$:

$$ i \int \frac{d^4k_E}{(2\pi)^4} \frac{1}{k_E^2}\frac{1}{k_E^2}\frac{1}{k_E^2}\tag{2}$$ and switch to spherical coordinates

$$ = 2\pi^2i \int \frac{dk_E}{(2\pi)^4} \frac{k_E^3}{k_E^6} \tag{3}$$ $$ = 2\pi^2i \int \frac{dk_E}{(2\pi)^4} \frac{1}{k_E^3}\tag{4}$$

which doesn't converge according to Mathematica. I'm basically confused because $\int dk \frac{1}{k^n}$ doesn't converge for any $n$ using Mathematica.

I may be missing an obvious point. Why isn't this finite?

Does it change significantly if the denominator involves a mass?


2 Answers 2


The integral is infra-red divergent. Power counting is about ultra-violet divergences. As you can check for yourself, the singular behaviour is in the lower limit: $$ \int_\epsilon^\infty \frac{1}{x^n}\mathrm dx=(1-n)^{-1}x^{1-n}\bigg|^\infty_\epsilon\sim\epsilon^{1-n} $$ which blows up if $n>1$.

The superficial degree of divergence tells you that the integral is superficially convergent as far as the $k\to\infty$ integration region is concerned. It tells nothing about possible singularities at finite values of $k$. In other words, the degree of divergence is all about the UV, but it knows nothing about the IR. In this example, the singularity is of the second type.

To regulate this integral you have to introduce a non-zero mass in the propagators, or use the dimensional regularisation trick $$ \int x^n\mathrm dx\equiv 0\quad \forall n $$ which is known as Veltman's formula. It can be argued that this formula is consistent as far as perturbation theory is concerned. Pragmatically speaking, this can be accomplished by introducing a non-zero mass and taking the massless limit carefully. In more formal terms, the consistency of the prescription can be proven by means of analytical continuation to complex $n$ (cf. 1.1666512).

For more details, see Massless integrals in dim-reg and Divergence of the tree level scattering amplitude in quantum field theory. The latter contains a list of useful references on IR divergences, their physical meaning and regularisation techniques.

  • $\begingroup$ In renormalized perturbation theory do our counterterms have to cancel both UV and IR divergences? @AccidentalFourierTransform $\endgroup$
    – Dwagg
    Jan 4, 2018 at 18:21
  • $\begingroup$ @Dwagg The don't have to. Counter-terms are only supposed to absorb UV divergences. These signal irrelevant physics at higher, undetectable scales. The counter-terms parametrise that irrelevant physics. IR divergences, OTOH, signal relevant physics at detectable scales. Using dimReg you can regulate IR divergences, although this obscures the physics. Dealing with IR divergences and extracting real-life phenomenons from them is a rather subtle but extremely interesting and rich subject. You may want to have a look at the references listed in the second link above. $\endgroup$ Jan 5, 2018 at 14:29

This is because the integral is UV convergent while IR divergent. If the integration is from the whole space, both UV and IR behaviors are taken into account so it is certainly divergent. If you only focus on the range $k\rightarrow\infty$, the integrand decays fast enough that the infinite integration measure does not lead to any divergences.

To summarize, (suppose we are working in momentum space) UV divergence is from the infinity of integrand measure while IR divergence is from the singularities of the integrand.

If you add an extra nonzero mass $m$ to the propagator(s), then actually you are regulating the IR divergence.


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