I have to find the order of 1/ε poles in dimensional regularization of $\dfrac{\lambda}{4!}\phi^4$ theory. The Feynman integral is the following:
\begin{equation}
I(p)=-λ^6\int \frac{d^4p_1}{(2\pi)^4}\frac{d^4p_3}{(2\pi)^4}\frac{d^4p_5}{(2\pi)^4}\frac{d^4p_7}{(2\pi)^4}\frac{d^4p_8}{(2\pi)^4}\frac{d^4p_{10}}{(2\pi)^4}\frac{1}{p_1^2+m^2}(\frac{1}{p_7^2+m^2})^2(\frac{1}{(p-p_1-p_7)^2-m^2})^2 \\ \frac{1}{p_3^2+m^2}\frac{1}{p_5^2+m^2}\frac{1}{(p-p_1-p_7-p_3-p_5)^2+m^2}\frac{1}{p_8^2+m^2}\frac{1}{p_{10}^2+m^2}\frac{1}{(p_7-p_8-p_{10})^2+m^2}
\end{equation}
My question is whether I can tell what the order of $\dfrac{1}{\epsilon}$ poles will be, without calculating the Feynman integral.
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$\begingroup$ On general grounds, you should expect at worst a pole 1/epsilon^2L, L being the loop order, because each loop integration can provide a UV 1/epsilon and a IR 1/epsilon. If all propagators are massive, as in your case, then the IR are not expected and you should get at worst 1/epsilon^L. In this case, a closed formula due to BPHZ is known: generate all forests, i.e. collection of non-overlapping nested subgraphs. Each forest give rise to a pole 1/epsilon^N, where N is the number of graphs \gamma of the forest with E_\gamma - L_\gamma D/2 < 0. $\endgroup$– giulio bullsaverCommented Jan 12 at 8:46
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$\begingroup$ In your case, assuming D=4, (p3,p4,p5), (p8,p9,p10), and the whole graph form a forest of graphs all with E_\gamma - L_\gamma D/2 < 0 (this is called the superficial degree of divergence of the graph gamma), therefore at least a pole 1/epsilon^3. There may be other forests giving a higher pole, although I can't see them. $\endgroup$– giulio bullsaverCommented Jan 12 at 8:48
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