Consider the massless divergent integral
$$
\int dk^4 \frac{1}{k^2},
$$
which occurs in QFT. We can't regularize this integral with dim-reg; the continuation from the massive to the massless case is ill-defined. It can be shown, however, that no "inconsistencies" occur if
$$
\int dk^4 \frac{1}{k^2} "=" 0.
$$
I think this is the now proven 't Hooft-Veltman conjecture. I don't understand this "equation". The integral is certainly not zero (although inconsistencies might occur in dim-reg if it weren't treated as zero in some contexts.)
Suppose I think about $\lambda\phi^4$ theory, with no bare mass for the field. Is it reasonable to claim that I will chose dim-reg and calculate the one-loop correction to the mass as
$$
\delta m^2 \propto \int dk^4 \frac{1}{k^2} = 0 \qquad ?
$$
Such that the particle stays massless, with no fine-tuning. I think this is simply incorrect – an unreasonable application of the 't Hooft-Veltman conjecture (I've no reason to worry about the consistency of dim-reg because I'm not regulating any integrals with it). Surely the 't Hooft-Veltman conjecture can only be applied in particular contexts? I can't start making any calculation in QFT, see an integral such as
$$
\int dk^n \frac{1}{k^a} = 0 \text{ for $n>a$},
$$
and set it zero, citing dim-reg and 't Hooft-Veltman?
P.S. This is not a straw man. I read people saying such things in the context of classical scale invariance.
Best Answer
The problem with such an integral is that it is both UV and IR divergent. We therefore need to introduce two regulators (to regulate both divergences). To regulate the UV divergence, we use dimreg. To regulate the IR divergence, we give the particle a mass and then take the massless limit. Doing both of them, the integral becomes $$ I = \lim_{m^2 \to 0} \int d^d k \frac{1}{k^2 - m^2 + i \varepsilon} $$ We can now Wick rotate and we get (I won't keep track of overall constants) $$ I \sim \int d^d k_E \frac{1}{k_E^2 + m^2} = \Omega_{d-2} \int_{0}^{\infty} \frac{k_E^{d-1} d k_E}{k_E^2 + m^2} \sim m^2 \left[\frac{2}{\epsilon} + \log \frac{m^2}{\mu^2} + {\cal O}(\epsilon) \right] $$ Now, we take the massless limit to make the integral well-defined and we get $$ I = 0 $$
More generally, it is easy to show (just using dimensional analysis) that if we introduce the mass to regulate the IR divergence $$ \int d^n k \frac{1}{k^a} = \lim_{m^2 \to 0} \int d^n k \frac{1}{\left( k^2 - m^2 \right)^{a/2}} \propto \lim_{m^2 \to 0} m^{n-a} $$ Thus, as long as $n > a$, this limit gives us 0.