# Usage of the cutoff momentum in QFT integral

I am trying to calculate the following divergent integral, I cite directly from the book

\begin{align} V\left(\phi_{c}\right) &=\frac{1}{2} \mu^{2} \phi_{c}^{2}+\frac{\lambda}{4 !} \phi_{c}^{4}-\mathrm{i} \int \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \sum_{n=1}^{\infty} \frac{1}{2 n}\left[\frac{(\lambda / 2) \phi_{c}^{2}}{k^{2}-\mu^{2}+\mathrm{i} \varepsilon}\right]^{n} \\ &=\frac{1}{2} \mu^{2} \phi_{c}^{2}+\frac{\lambda}{4 !} \phi_{c}^{4}-\frac{\mathrm{i}}{2} \int \frac{\mathrm{d}^{4} k}{(2 \pi)^{4}} \ln \left[1-\frac{\lambda \phi_{\mathrm{c}}^{2} / 2}{k^{2}-\mu^{2}+\mathrm{i} \varepsilon}\right] \end{align}

and then proceeds

The integral is divergent. If it is cut off at some large momentum, we obtain \begin{align} V\left(\phi_{\mathrm{c}}\right)=& \frac{1}{2} \mu^{2} \phi_{\mathrm{c}}^{2}+\frac{\lambda}{4 !} \phi_{\mathrm{c}}^{4}+\frac{\Lambda^{2}}{32 \pi^{2}}\left(\mu^{2}+\frac{\lambda}{2} \phi_{\mathrm{c}}^{2}\right) \\ &+\frac{1}{64 \pi^{2}}\left(\mu^{2}+\frac{\lambda}{2} \phi_{\mathrm{c}}^{2}\right)^{2}\left[\ln \left(\frac{\mu^{2}+\lambda \phi_{\mathrm{c}}^{2} / 2+\mathrm{i} \varepsilon}{\Lambda^{2}}\right)-\frac{1}{2}\right] \end{align}

I understand that the cutoff is a upper limit of some quantity. But how to use it? Does it mean every component of $$k$$ $$\leq\Lambda$$?

As @knzhou noted, we first Wick-rotate so $$\color{blue}{k}\in\Bbb R^4$$ is Euclidean. Then $$\int_{\Bbb R^4}f(\color{blue}{k}^2)d^4\color{blue}{k}=2\pi^2\int_0^\infty f(\color{red}{k}^2)\color{red}{k}^3d\color{red}{k}$$, where $$\color{red}{k}\in[0,\,\infty)$$ is the radius of $$\color{blue}{k}$$, and the proportionality constant is the solid angle in $$4$$ dimensions. The cutoff momentum lets us instead consider $$2\pi^2\int_0^\Lambda f(\color{red}{k}^2)\color{red}{k}^3d\color{red}{k}$$, since any theory that mandates the original integral cannot be trusted to accurately compute contributions beyond the cutoff.
In general, such cutoffs only have meaning when you Wick rotate to Euclidean signature. In that context, $$k$$ is a vector in $$\mathbb{R}^4$$, and the condition is that its magnitude is less than $$\Lambda$$.