In https://arxiv.org/abs/1708.01256 on p.8, it says:
> The contribution corresponding to the rightmost graph yields zero in the $\overline{\mathrm{MS}}$ 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.
![](https://www.physicsoverflow.org/?qa=blob&qa_blobid=13958304483618283180)
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 $\overline{\mathrm{MS}}$ scheme reads
$$\Pi^{\mu\nu}(p,\mu^2)=\frac{e^2}{2\pi^2}(p^\mu p^\nu-g^{\mu\nu}p^2)\int^1_0dxx(1-x)\ln\frac{\mu^2}{m^2-p^2x(1-x)}.$$
If the photon is on-shell and the fermions massless, then $0=p^2=m^2$. But then the logarithm is not well-defined anymore since
$$\ln\frac{\mu^2}{p^2x(x-1)}=\ln\frac{\mu^2}{x(x-1)}-\ln p^2\stackrel{p^2\rightarrow0}{\rightarrow}-\infty.$$
Can somebody show me **by calculation** how $\Pi^{\mu\nu}$ vanishes for an on-shell photon?