# Tag Info

## New answers tagged regularization

1

Your statement that the integral "is actually" $\propto g^{\mu\nu}\Pi$ is incorrect because you can clearly see the $q^\mu q^\nu$ in the numerator of the integrand. The correct expression that you have in your final equation is not a choice, it is the result of calculating the loop using dimensional regularization. Finally, indeed $q_\mu \Pi_2^{\mu\nu}$ is ...

0

I don't think it is exactly the same regulator: In the first method, you integrate $\int_{-\infty}^\infty dk^0 \int^\Lambda d^3k$, but in the second calculation you integrate $\int^\Lambda d^4 k_E$.

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