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I need to determine Feynman one-loop integrals to work out some Feynman diagrams, in particular $I_{2,1}$. Starting from the general formula: $$I_{n,m}=\frac{1}{(4\pi)^2}\frac{\Gamma(m+2-\frac{\epsilon}{2})}{\Gamma(2-\frac{\epsilon}{2})\Gamma(n)}\frac{1}{\Delta^{n-m-2}}(\frac{4\pi M^2}{\Delta})^{\frac{\epsilon}{2}}\Gamma(n-m-2+\frac{\epsilon}{2})$$ I arrived to the following: $$I_{2,1}=\frac{\Delta}{(4\pi)^2}\frac{(2-\frac{\epsilon}{2})}{(\epsilon-1)}[\frac{2}{\epsilon}-\gamma+ln(\frac{4\pi M^2}{\Delta})-\gamma\frac{\epsilon}{2}ln(\frac{4\pi M^2}{\Delta})+O(\epsilon)]$$ The term $\frac{1}{\epsilon-1}$ is giving me some trouble and I'm wondering if I can just set $\epsilon=0$ like I'll do to compute all the terms in the numerator.

Edit: I got to the answer, using the suggestion from the comments:

$$I_{2,1}=-2\Delta I_{20}-\frac{\Delta}{(4\pi)^2}$$

Can someone confirm this?

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    $\begingroup$ Since it is multiplying a pole in $\epsilon$ you should be careful and expand that term (and the others) if you want the part that is finite for $\epsilon \rightarrow 0$. For example, $1/(\epsilon(1-\epsilon)) = 1/\epsilon + 1 + O(\epsilon)$. See also the answer and comments on this post: physics.stackexchange.com/q/506744 $\endgroup$ Nov 5, 2019 at 2:02
  • $\begingroup$ Thank you very much! I'll expand $\frac{1}{\epsilon -1}$ as $1+\frac{\epsilon}{2}$ and then set $\epsilon\rightarrow 0$. $\endgroup$
    – RicardoP
    Nov 5, 2019 at 13:46
  • $\begingroup$ Actually as $-(1+\frac{\epsilon}{2})$, if i'm not mistaken. $\endgroup$
    – RicardoP
    Nov 5, 2019 at 14:10

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