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I calculated the correction to the self energy of the W boson due to a fermion doublet (below I have n(e) for neutrino (electron), but it could be up and down quarks or just any 4th generation contribution). My result checks with the results from section 21.3 in Peskin, so I am fairly confident in the form below. I think of this as a fairly straight-forward calculation, but I am getting inconsistent results. The interaction comes from the term

$$\mathcal{L}= \frac{g}{\sqrt{2}}W^+_\mu \bar{n}P_R \gamma^\mu P_L e + \frac{g}{\sqrt{2}}W^-_\mu \bar{e}P_R \gamma^\mu P_L n $$

so we get a single diagram with a electron (neutrino) running in the top(bottom) leg of the loop. For the self energy correction I get (with $\bar{MS}$ scheme):

$$\Pi^{\mu \nu}(q^2) \sim \eta^{\mu \nu} \int^1_0 dx \left( x(1-x)q^2 - \frac{1}{2} ( x m_e^2 +(1-x)m_n^2)\right) \log \left[ \frac{(1-x)m_n^2 + x m_e^2 +(x-1)x q^2}{\mu^2} \right] $$

where I have dropped the $q^\mu q^\nu $ term.This seems innocuous enough however if I perform the x integration and then expand in the external momentum $q^2$ I get weird terms like

$$\Pi(q^2) \approx(m_n^3-m_e^2)^3 \log \frac{m_e}{m_n}\frac{1}{q^2}$$

which seems incorrect since this should correspond to a term in the effective action, but there shouldn't be $\frac{1}{q^2}$ terms in the action. Furthermore the original integrand is not singular in $q^2$ so I could have just as well expanded in $q^2$ first and then performed the Feynman parameter integral and no such $\frac{1}{q^2}$ terms would be present. Any idea why the Feynman parameter integral and the momentum expansion don't seem to commute, or why I get these weird terms if I integrate first? Thanks a lot.

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Right now this seems to be semi-resolved as an artifact of careless expansions in Mathematica ($\epsilon$ for dim-reg and then in k to get the effective action terms). I'll post a more complete answer if I get a handle on the subtleties of what went wrong in the nuts-and-bolts of the expansion(s). – DJBunk Jun 7 '12 at 14:56

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