# Computing integrals for divergent loop amplitudes?

I am trying to compute the cross-section for the diagram below with a divergent triangle loop: $$\qquad\qquad\qquad\qquad\qquad$$

where $$X^0$$ and $$X^-$$ are some fermions with zero and negative charge respectively. I am interested in low energy limits, so you can consider W-propagator as $$\frac{i\eta_{\mu\nu}}{M_w^2}$$.

When computing the amplitude, ignoring the external wave functions, you end of with an integration of the form: $$\int \frac {k_\mu \gamma^\mu +m_-} {k^2 -m_-^2 +i\epsilon} \frac {d^4 k} {(2\pi)^4}$$ where $$m_-$$ is mass of $$X^-$$.

Any ideas how to solve this integral in terms of kinematic parameters, masses etc?

• Well, you notice yourself that the expression diverges. And I am not sure I understand how you get rid of the dependence of the W propagators on the loop momentum. The integrals are standard, look up Passarino-Veltman functions.
– user178876
Commented Oct 3, 2018 at 3:23
• @marmot, for simplicity, I just assumed that the loop momentum is less than W boson mass. Maybe, I oversimplified the integral as that is an internal momentum. Commented Oct 3, 2018 at 6:36
• You cannot make this assumption since the loop momentum gets integrated all the way up to infinity, as your formula shows.
– user178876
Commented Oct 3, 2018 at 14:08

Agree with marnot that you can't get rid of W propagators in the loop.

That being said, the integral per se $$\int \frac {k_\mu \gamma^\mu +m_-} {k^2 -m_-^2 +i\epsilon} \frac {d^4 k} {(2\pi)^4}$$ is a typical single fermion loop (e.g. a vacuum bubble diagram), where the first term related to $$k_\mu \gamma^\mu$$ drops out since it's odd in $$k_\mu$$.

The second term related to $$m_-$$ is quadratic divergent, which amounts to zero however in dimensional regularization (one of the peculiarities of DR).

• Why do you say the second term is zero?
– CAF
Commented Oct 5, 2018 at 20:42
• It's one of the quirks of DR. Commented Oct 5, 2018 at 20:45
• It’s well understood and only vanishing for certain $d$ and if there’s no scale. Here there is a scale $m$_ present so you’re not dealing with a scaleless integral.
– CAF
Commented Oct 5, 2018 at 20:50