3 to 3 scattering in massless $\phi^4$ theory

During my QFT study I faced a problem of calculating amplitude of 3 to 3 scattering in massless $\phi^4$ theory in zero momenta limit at tree level. One of topologically distinct diagrams corresponding to this process gives a contribution of $\sim \frac{1}{(p_1+p_2+p_3)^2}$ (up to symmetry coefficients and coupling constant), where $p_1,p_2,p_3$ are momenta of incident particles. I have no idea how to take a limit $p_i \rightarrow 0$. Can anyone help?

The process is indeed divergent in the IR (low energy) limit. There is no "trick" to take the limit as it is infinite. This is an idealization as in practice the amplitude would be cutoff by the mass of the particle and the diagram would go as $1/m^2$.
• In this theory we don't have any mass parameter because the theory is massless. The Lagrangian density is $\mathcal L = \frac{1}{2}(\partial_\mu \phi)^2 - \frac{\lambda}{4!}\phi^4$ – user43283 Mar 26 '14 at 15:02
• Actually this problem arose from finding an effective low energy potential in theory with spontaneous symmetry breaking with Lagrangian density $\mathcal L = \vert \partial_\mu \xi \vert^2 - \frac{\lambda}{4} (\vert \xi \vert^2 -v^2)^2$. $\phi$ in the problem above is Goldstone field. – user43283 Mar 26 '14 at 15:10