Confution on UV cut-off in the calculation of effective action and Beta function

Question

I am reading David Tong's gauge theory notes and meet some difficulties.

In section 2.4.2, he uses background field to calculate effective action $S_{eff}$ and Beta function. Simply like follows: Writting gauge field $A_{\mu}$ as $A_{\mu}=\bar{A}_{\mu}+\delta A_{\mu}$ for $\bar{A}_{\mu}$ being a fixed field, $\delta A_{\mu}$ being fluctuation. Also we introduce Faddeev-Popov ghost fields $c$ and $c^{\dagger}$. The path integral of gauge field $A_{\mu}$ goes like \begin{equation} e^{-S_{eff}}=Z=\int \mathcal{D}A \mathcal{D}c \mathcal{D}c^{\dagger}e^{S(A,c,c^{\dagger})}=(\det[\Delta_{gauge}])^{-1/2} \det[\Delta_{ghost}]e^{(-1/2g^2) S_{YM}(\bar{A})} \end{equation} where \begin{equation} S_{YM}(A)=\int d^4x tr(F^{\mu\nu}F_{\mu \nu}) . \end{equation} So the effective action $S_{eff}$ is then \begin{equation} S_{eff}=\frac{1}{2g^2}S_{YM}(\bar{A})+\frac{1}{2}Tr\log\Delta_{gauge}-Tr\log\Delta_{ghost} \end{equation} See eq(2.63) in the notes.

Now he gets the contribution of $-Tr\log\Delta_{ghost}$ (see page 71 in the notes) as \begin{equation} -Tr\log\Delta_{ghost}=constant \times \int \frac{d^4k}{(2\pi)^4}tr(\bar{A}_{\mu}(k)\bar{A}_{\nu}(-k))(k^{\mu}k^{\nu}-k^2\delta^{\mu\nu})\log\bigg(\frac{\Lambda^2}{k^2}\bigg). \end{equation}

Here is what I don't understand: how does this $\log\bigg(\frac{\Lambda^2}{k^2}\bigg)$ appear in our integral in terms of $\log$? Is it a regularization? If it is, then the integrand cannot turn back to $tr(\bar{A}_{\mu}(k)\bar{A}_{\nu}(-k))(k^{\mu}k^{\nu}-k^2\delta^{\mu\nu})$ under limit $\Lambda=\infty$.

Answer 1

Tong is using a cutoff regularization in this calculation, the cutoff being this $\Lambda$. Specifically this factor appears because he has just completed the loop momenta integral (integral over the momenta $p$). This integral is divergent without a regularization scheme, so of course taking $\Lambda\rightarrow\infty$ (removing the cutoff) will cause the what you have written to diverge.