I am studying the degrees of divergence of Feynman diagrams. I feel that I miss something but I don't really understand what. Please apologize if this question is silly. Anyway.
As an introduction to systematics of renormalization, Peskin and Schroeder (chapter 10) introduce the counting of ultraviolet divergences. They say that an expression for any diagram typically looks like
$$\sim \int \dfrac{d^4 k_1 d^4 k_2 \dots d^4 k_L}{(\bar{k}_i-m) \dots k^2_j \dots }$$ where $\bar{k}$ is $k$ slash (i.e. $\gamma^{\mu} k_{\mu}$). The number of $d^4 k$ will depend on the number of loops, while the number of $(\bar{k}-m)$ and $k^2$ in the denominator will depend on the number of fermion/boson propagators. Ok.
They then say :
For each loop there is a potentially divergent 4-momentum integral, but each propagator aids the convergence of this integral by putting one or two powers of momentum into the denominator.
I don't understand that. To be more clear, I'll develop a little bit more...
They then introduce the superficial degree of divergence $D$ as $$ D \equiv (\text{power of } k \text{ in numerator}) - (\text{power of } k \text{ in denominator}) $$
and they say that we can expect a diagram to have a divergence proportional to $\Lambda^D$ ($\Lambda$ is a cut-off) when $D>0$, proportional to $\log(\Lambda)$ when $D=0$ and no divergence when $D<0$.
I know that it's a tool and not always exact and so on to evaluate the divergence of a Feynman diagram. That's not the point. What I don't get is why is a diagram with $D<0$ not divergent ? Integrals like
$$ \int \dfrac{d^4 k}{k^7} $$ don't converge, so the associated diagram should be divergent too... Or is it because we are only dealing with UV divergences, and that diagrams with $D<0$ diverge because of IR divergences ?