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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 $k$ $\leq\Lambda$?

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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.

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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 $\Lambda$.

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