Suppose we have the Lagrangian in 3 dimensions:
$$
\mathcal{L} = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2-\frac{g_1}{4!}\phi^4-\frac{g_2}{6!}\phi^6
$$
The superficial degree of divergence could be found as $\omega = 3-(1/2)n-n_4$, where $n$ is the number of external lines, and $n_4$ is the number of 4-point vertices.
I'm not quite sure how to analyze the divergence of these 2-point functions:
At high momenta, diagram (a) goes like
$$
\int^\Lambda\frac{d^3k}{k^2}\sim \Lambda
$$
diagram (b) goes like
$$
\int^\Lambda\frac{d^6k}{k^6} \sim \log\Lambda
$$
diagram (c) goes like
$$
\int^\Lambda\frac{d^{12}k}{k^{10}} \sim \Lambda^2
$$
I actually don't quite understand how those approximation work, I was just told that the divergence of each diagram sometimes could be found by finding the superficial degree of divergence $\omega$, and then the approximation works like $\Lambda^\omega$ for $\omega>0$, and $\log\Lambda$ for $\omega = 0$. How can I justify the approximations above? For $(b)$, I remember I saw some reference that shows me
$$
\int^\Lambda\frac{d^6k}{k^6}\sim \frac{|k|^5d|k|d\Omega}{k^6} \sim \log\Lambda
$$
But I don't know how the $d^6k$ is separated, and how to evaluate the integral of this form in general.
Thanks for the help!
Best Answer
In this post we are going to consider all the integrals already in Euclidean signature, so that Wick rotation has already been performed.
We want to evaluate integrals of the form $$\int_{|k|<\Lambda} \dfrac{d^nk}{k^{2m}}$$
Since we are in an $n$-dimensional Euclidean space we can introduce hyperspherical coordinates. For our purposes, all we need to know is that the volume element is of the form
$$d^n k = |k|^{n-1} d|k| d^{n-1}\Omega,$$
where $|k|$ is the radial coordinate in $n$-dimensional $k$-space and $d^{n-1}\Omega$ is the volume element on $S^{n-1}$. In that case the integral becomes $$\int_{|k|<\Lambda}\dfrac{d^nk}{k^{2m}}=\int_0^\Lambda \dfrac{|k|^{n-1}}{|k|^{2m}}d|k|\int_{S^{n-1}}d^{n-1}\Omega={\rm Vol}(S^{n-1})\int_0^\Lambda |k|^{n-1-2m}d|k|$$
Now there are two cases: $n=2m$ in which case we get $\log \Lambda$ and $n\neq 2m$ in which case we get $\Lambda^{n-2m}$. The complete result is $$\int_{|k|<\Lambda}\dfrac{d^nk}{k^{2m}}=\begin{cases}{\rm Vol}(S^{n-1})\log \Lambda,& n=2m,\\{\rm Vol}(S^{n-1})\dfrac{\Lambda^{n-2m}}{n-2m},& n\neq 2m.\end{cases}$$