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I have a question with a perhaps well-known answer. Consider a two-loop sunset (log divergent) integral in two dimensions:

$$ I_S = \int \frac{d^2k d^2l}{(2\pi)^4} \frac{ k^2}{(k^2-m^2)(l^2-m^2)((p-k-l)^2-m^2)} $$

for some positive masses $m$ and external momenta $p$. If I evaluate this integral in standard Feynman parameterization and dim-reg, I find that the answer goes like'

$$ I_S= \frac{1}{\epsilon}f_1(p)+\textrm{finite-terms} $$ Where, of course, the finite terms are sensitive to the regularization scheme (dim-reg). However, using trivial algebraic identities I can write $I_S$ as $$ I_S=\int \frac{d^2k d^2l}{(2\pi)^4}\frac{1}{(l^2-m^2)((p-k-l)^2-m^2)}+\textrm{finite sunset-integral} $$ The leading-order singularity in the divergent integral above should be insensitive to shifts in $k_\mu$ so we can write $$ \int \frac{d^2k d^2l}{(2\pi)^4}\frac{1}{(l^2-m^2)((p-k-l)^2-m^2)}=\int \frac{d^2k d^2l}{(2\pi)^4}\frac{1}{(l^2-m^2)(k^2-m^2)}+\textrm{possible finite or $1/\epsilon$-terms} $$ Since the first term in the second expression is a double tadpole it will at leading order go like $1/\epsilon^2$. Thus, the leading-order divergence is regularization dependent. I didn't think this could happen, hence, my question: what's going on?

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  • $\begingroup$ While not necessary for the calculations, it would be nice to know which theory you are looking at. $\phi^4$ in 2D? $\endgroup$ – ACuriousMind Nov 7 '14 at 14:44
  • $\begingroup$ Hi, no. Its a supersymmetric sigma-model. For 2D $\phi^4$ we wouldn't get a $k^2$-term in the numerator (no derivatives are acting on the vertexes). $\endgroup$ – Nid Nov 7 '14 at 16:26

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