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Consider the Feynman diagram below: enter image description here

in the case of $\phi^4$ theory where there is no bare mass: $$\mathcal{L}=\frac{1}{2} \partial_\mu \phi\partial^\mu\phi-\frac{\lambda}{4!} \phi^4$$ the contribution of this diagram is given by: $$I=\frac{-i\lambda}{2} \int \frac{d^d k}{(2\pi)^d}\frac{1}{k^2+i\varepsilon}$$ typically such integrals are done using Feynman and Schwinger Parameterizations. In this case, however Schwinger parameterization won't work as you will have a completely imaginary exponential. It also appears to be the case that $$\lim_{m\rightarrow 0}\frac{-i\lambda}{2} \int \frac{d^d k}{(2\pi)^d}\frac{1}{k^2-m^2+i\varepsilon}$$ which can be done using Schwinger parameterization does not converge onto the same value as $I$. Thus my question is: what is the most common way to deal with integrals of the form $I$ in QFT ideally using minimal subtraction.

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  • $\begingroup$ As an amateur, I think it doesn’t matter if it converge to the same thing as long as you use the same regularization scheme everywhere $\endgroup$ – Darkseid Dec 11 '17 at 9:02
  • $\begingroup$ possible duplicate: Massless integrals in dim-reg and links therein. $\endgroup$ – AccidentalFourierTransform Dec 12 '17 at 16:28
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In dimensional regularization, loop integrals which do not depend on physical external momenta are automatically regularized to be zero. This is so because, in absence of a mass scale, the integral is both UV and IR divergent and the divergences are regulated to zero.

Then, in dimensional regularization, you can regulate any integral of this sort

$$ I_a =\int d^d k\, k^a = 0 $$

for any value of $a$. You can see this result is really obvious since the mass dimension of $I_a$ is $[I_a] = d+a$ but since you don't have any mass scale in the problem, what dimensionful result different from zero would you expect?

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  • $\begingroup$ "In dimensional regularization, loop integrals which do not depend on physical external momenta are automatically regularized to be zero." It's just a peculiar artifact of DR. It's a convenient trick for most QFT calculations. However, when you do want a non-zero value for OP's quadratic integral (for instance dynamical chiral symmetry breaking and mass generation), this DR 'convenience' becomes a liability and thus DR is not a proper regularization. $\endgroup$ – MadMax Jun 5 at 14:02

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