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During the renormalization procedure of a massive $\phi^4$ theory at two loop level, one finds that the quantity $K_{\Lambda}(k^2,m^2)-K_{\Lambda}(0,m^2)$ that appears in the bare $\Gamma^{(2)}$ function is divergent, where $$ K(k^2,m^2)=\int_{\Lambda} \frac{d^4 q}{(2 \pi)^4}\frac{d^4 q'}{(2 \pi)^4} \frac{1}{(q^2+m^2)(q'^2+m^2) [(k-q-q')^2+m^2]}. $$

I computated explicitly $K_{\Lambda}(k^2,m^2)-K_{\Lambda}(0,m^2)$ and found $$ K_{\Lambda}(k^2,m^2)-K_{\Lambda}(0,m^2) = \int_{\Lambda} \frac{d^4 q}{(2 \pi)^4}\frac{d^4 q'}{(2 \pi)^4} \frac{-k^2 +2k(q+q')}{(q^2+m^2)(q'^2+m^2) [(k-(q+q'))^2+m^2][(q+q')^2+m^2]}, $$ but I can't see why this integral is UV divergent; is there an easy way to prove it?

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In the UV, i.e. either $|q|\gg |k|,m$ or $|q'|\gg |k|,m$, the integration measure goes like $|q|^3 dq\cdot |q'|^3 dq'$.

The denominator goes like $|q|^2 |q'|^2 |q+q'|^4$, one may neglect all the smaller terms in the factors. The numerator goes like $2k\cdot (q+q')$, I suppose that the capitalized $K$ is the same thing as $k$.

So just by counting the powers for a generic large value of $q,q'$, treated to be of the same order, you see that there is $q^7 dq$ from the integration measure (I merged $dq\cdot dq' = q\cdot d\phi\cdot dq$), $q$ from the numerator, and $1/q^8$ from the denominator, which means $dq$ in total, so the leading term in the integral is linearly divergent.

In reality, the leading linear divergence almost certainly cancels (linear divergences only tend to survive in the presence of anomalies) but the term suppressed by one power of $q$ is the logarithmic divergence $dq/q$ that survives.

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  • $\begingroup$ I edited the capitalized K mistake in the integrand. Anyway, thanks for your answer; I was counting powers in a wrong way because I wasn't sure how to cope with terms like $(q+q')$. Thanks a lot for your clarification! $\endgroup$ – Alex A Aug 23 '13 at 14:49

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