Are these IR and UV divergences equal let be 2 divergent integrals
$$ \int_{0}^{\infty}\frac{p^{3}dp}{(p^{2}+m^{2})^{2}}= A $$
$$ \int_{0}^{\infty}\frac{dp}{p(p+q)^{2}}=B $$
B has a divergent as $ p \to 0 $ however i can use a change of variable $ xu=1 $ so for B we have
$$ \int_{0}^{\infty} \frac{udu}{(1+uq)^{2}}=B $$
so now this integral B has a logarithmic divergence, in this case is enough if we add and substract the integral $ \int_{0}^{\infty}\frac{dx}{x+1} $
to regularize it ?¿is my argument correct ??
 A: No. One can always find a substitution to map similar divergent integrals to each other. However, that doesn't mean that they have the same "degree of divergence" in the sense of quantum field theory. In particular, one must always distinguish UV and IR divergences. The former appear for high momenta or short distances; the latter come from low momenta or long distances. By making a substitution $p\to p´= 1/p$, the new primed $p'$ isn't a real momentum anymore – it has different units – so you can't call $p'to \infty$ the UV region. In fact, $p'\to \infty$ is $p\to 0$ which is the IR region.
Similarly, the power-law type of a divergence can't be changed by a substitution. For example, $\int p^n dp$ diverges as $p_{max}^{n+1}$ at high $p$. By writing $p=q^k$ it becomes $\int q^{kn+k-1} dq$ and the exponent changes to $kn+k-1$ but it is the exponent above a different base. The original divergence was quadratic/cubic etc. according to the value of the power above the momentum $p$ and if there is a power above something else, one can't directly translate it to the name of the divergence.
One must be extremely careful about subtracting or comparing infinities. Substitutions like yours violate various symmetries. Note that $\infty-\infty$ is an indeterminate form which can be regulated (in a limit) to yield any result. But only some result is the physically right one and most procedures how to subtract infinities would be wrong.
