# Why do the Euler-Mascheroni constant $\gamma$ and $\ln 4\pi$ not show up in observables (renormalisation of electric charge)?

The one-loop contribution of the vacuum polarisation of the photon after using dimreg is given by

$$\Pi_2^{\mu\nu}= e^2 J(q) \left(\eta^{\mu\nu} - \frac{q^\mu q^\nu}{q^2}\right),$$

with the metric tensor $$\eta$$, coupling constant $$e$$ and momentum $$q$$. Also

$$J(q) = -\frac{q^2}{2\pi^2} \int_0^1 dx x(1-x)\left(\frac{2}{\epsilon}-\ln\Delta-\gamma+\ln(4\pi) + O(\epsilon)\right)$$

with $$\epsilon=d-4$$ and $$\Delta=m^2-x(1-x)q^2$$.

The renormalised electric charge can be expressed as

$$e^2 = e_0^2\left(1-\frac{1}{2}J^{\prime\prime}(0) + O(e_0^4)\right)$$

Dimreg then yields

$$e^2 = e^2_0\left(1-\frac{e_0^2}{6\pi^2\epsilon}+O(e_0^4)\right)$$

Where did the $$\gamma$$ and $$\ln 4\pi$$ go?

It's implicitly performing the modified minimal subtraction $$\bar{MS}$$ renormalization scheme, only in a sloppy way.
One should retain the finite terms at the intermediate stage: $$e^2 = e^2_0\left(1-\frac{e_0^2}{12\pi^2}[\frac{2}{\epsilon} - \ln(m^2) - \gamma+\ln(4\pi)]+O(e_0^4)\right)$$
The usual minimal subtraction $$MS$$ renormalization scheme will only subtract the divergent term (the $$O(\frac{1}{\epsilon})$$ part), while the modified minimal subtraction $$\bar{MS}$$ renormalization scheme will subtract the finite terms as well ($$\ln(m^2)$$, $$\gamma$$, and $$\ln(4\pi)$$). The difference between $$MS$$ and $$\bar{MS}$$ all boils down to whether the counterterm should include only a divergent term ($$MS$$) or both divergent and finite terms ($$\bar{MS}$$).
The OP's formula is implicitly performing $$\bar{MS}$$ at different stages, causing unnecessary confusion. At any rate, the final physical results should be the same, being it $$MS$$, $$\bar{MS}$$, or sloppy $$\bar{MS}$$.