In https://arxiv.org/abs/1708.01256 on p.8, it says:
> The contribution corresponding to the rightmost graph yields zero in the ¯MS scheme since we consider massless fermions, since at this order in collinear factorisation the
photon is implicitly on shell and so the vacuum polarisation is evaluated at zero virtuality.
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The process in question is depicted in the Feynman diagram above. Can somebody explain to me why the vacuum polarisation (also called self-energy of the photon in other literature) should vanish in this case?
The polarisation function in the ¯MS scheme reads
Πμν(p,μ2)=e22π2(pμpν−gμνp2)∫10dxx(1−x)lnμ2m2−p2x(1−x).
If the photon is on-shell and the fermions massless, then 0=p2=m2. But then the logarithm is not well-defined anymore since
lnμ2p2x(x−1)=lnμ2x(x−1)−lnp2p2→0→−∞.
Can somebody show me **by calculation** how Πμν vanishes for an on-shell photon?