In lectures on effective field theory the professor wanted to find the correction to the four point vertex in massless $\phi^4$ theory by calculating the diagram,
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We consider the zero external momentum limit and denote $p$ as the momentum in the loop. Then we get, \begin{align} \int \frac{ d ^d p }{ (2\pi)^4}\frac{1}{p ^4 } & = \frac{ - i }{ 16 \pi ^2 } ( 4\pi ) \Gamma ( \epsilon ) \mu ^\epsilon \\ & = - \frac{ i }{ 16 \pi ^2 } \left( \frac{1}{ \epsilon _{ UV}} - \gamma + \log 4\pi - \log \mu ^2 \right) \\ & = \frac{ i }{ 16 \pi ^2 } \left( \frac{1}{ \epsilon _{ UV}} - \frac{1}{ \epsilon _{ IR}} \right) \end{align} where we introduced $\mu$ as an IR cut-off and then take $\log \mu ^2 $ as a $\frac{1}{\epsilon_{IR}}$.
This is fine, however the professor then goes on to say that this diagram is zero since the two divergences cancel. Why would this be the case? The two divergences arise for completely different reasons. The UV divergence is due to a UV cutoff (possibly from new high energy particles arising at some high up scale) and the second is a consequence of studying a massless theory.
For more context the lecture notes are available here under Effective Field Theory (Eq. 4.17)