2
$\begingroup$

On page 43 of Manohar notes on effective field theories, he argues that since all the integrals in EFT are scale less when expanded in terms of the IR parameter, they all vanish. To me it seems absolutely correct for the examples he's given on the same page, but if one goes to higher order terms in the IR expansion of the EFT one faces integrals that are either only UV divergent or only IR divergent, and they don't seem to be zero but rather infinity! Can someone help me with this problem please?

How cam I deal with scale less integrals of higher orders!

Or how can I rewrite all the scale less integrals in the (5.21) format? Like how to write $$\int 1/k^8 \, d^4k$$ or $$\int k^2\, d^4k$$ in the (5.41) format?

The paper: https://arxiv.org/abs/1804.05863

$\endgroup$
7
  • 1
    $\begingroup$ In dim reg d=4+eps these Integrals vanish $\endgroup$
    – Thomas
    Jul 6, 2020 at 1:16
  • $\begingroup$ Unfortunately I can't grasp how. I can only visualize how $$\int 1/k^4 \, d^4k$$ vanishes! Could you please help me with this. $\endgroup$ Jul 6, 2020 at 10:09
  • $\begingroup$ I don't think that you can "visualize" integrals in dimensional reglarization. You take the master formula for integrals in dim reg and set M=0 before taking $\epsilon\to 0$ $\endgroup$
    – Thomas
    Jul 6, 2020 at 15:34
  • $\begingroup$ I'm sorry but why do we really take the $m=0$ limit prior to the $\epsilon$ limit?!! I mean for the case of $ k^-4 $ one can very reasonably introduce a scale in the integrand namely $m$ and divide the integral to a UV-divergent and an IR-divergent integral and the mass dependence cancels out by itself and the only parameter that remains is $\epsilon$. I'm just curious if such a less "biased" method exist or not?! $\endgroup$ Jul 6, 2020 at 18:08
  • $\begingroup$ Your question was: What is \int d^4k/k^8 in dim reg? Answer: Take master formula for $\epsilon\neq 0$ and set M=0. $\endgroup$
    – Thomas
    Jul 6, 2020 at 18:18

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.