I am confused about how to interpret the apparently discontinuity which can arise in the massless limit of dimensionally regulated integrals.
My questions are illustrated by the following example:
Consider the three-point amplitude ${\cal A}_3$ for the Lorentz-violating, effective field theory $${\cal L}=-\frac{1}{2}(\partial\phi)^2-\frac{m^2}{2}\phi^2-\frac{\lambda_3}{\Lambda^2} \dot{\phi}(\vec{\nabla}\phi)^2 \quad ({\rm mostly \ plus \ signature})$$ where $\dot{\phi}=\partial_t\phi$ and $(\vec{\nabla}\phi)^2=\delta^{ij}\partial_{i}\phi\partial_{j}\phi$.
The tree-level contribution to the amplitude is ${\cal A}_3^{\rm tree}\sim \lambda_3 \frac{\omega_1\omega_2\omega_3}{\Lambda^2}$ while the one-loop contribution is of the schematic form $${\cal A}_{3}^{\rm one-loop}\sim \frac{\lambda_3^3}{\Lambda^6}\int\frac{{\rm d}^4k}{(2\pi)^4} \frac{F_9[p_i,k]}{(k^2+m^2)((k+p_1)^2+m^2)((k-p_2)^2+m^2)}$$ where $p^{\mu}_i=(\omega_i, \vec{p}_i)$ are the momenta of the on-shell states and $F_6[p_i,k]$ is some 9th-order function of momenta. The result of the integral includes, among other terms,
$${\cal A}_{3}^{\rm one-loop}\supset \lambda_3^3\frac{F_7[\omega_i]}{\Lambda^6}\times (\frac{c_1}{\epsilon}+c_2\ln\left( m^2/\mu^2)\right) $$
as computed in dimensional regularization, schematically. $F_7[\omega_i]$ is some 7-th order function of the $\omega_i$. Only $m^2$ can can appear in the logarithm since 3-pt kinematics of the free action fix all momenta contractions to be $\propto m^2$.
The divergence corresponds to some ${\cal L}_{\rm ct}\sim \frac{\lambda_7}{\Lambda^6}(\partial^7 \phi^3)$ counterterm whose coefficient runs due to the above logarithm:
$$\beta(\lambda_7)\sim \lambda_3^3$$
Question: what happens in the (technically natural) massless limit of the theory? The naive massless limit of the above is divergent. However, as documented on this site (e.g. in this post) and elsewhere, the massless limit of dimensionally regulated integrals is not a continuous limit, generally. In particular, since the integral is scale-less in the massless limit, one would expect that the above contribution to ${\cal A}_{3}^{\rm one-loop}$ strictly vanishes when $m=0$, meaning that
$$\beta(\lambda_7)\sim \theta(m^2)\lambda_3^3$$
with $\theta$ the usual Heaviside function.
Is this the right result? I am not familiar with other examples where such discontinuities in $\beta$-functions arise. It feels physically incorrect.
