# Polarization at zero virtuality

In this paper 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. 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?