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 ??


closed as off topic by Waffle's Crazy Peanut, Brandon Enright, Michael Brown, Emilio Pisanty, Manishearth Jun 7 '13 at 5:33

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    $\begingroup$ Instead of manipulating objects that don't exist, you should introduce a cutoff explicitly, so instead of $B$ you write $B_\Lambda$ and cut off the integral at $p = 1/\Lambda.$ Then you can massage the integrand, change variables etc. and see exactly what happens. Otherwise it's voodoo. $\endgroup$ – Vibert May 30 '13 at 21:29
  • $\begingroup$ @JoseJavierGarcia This looks like an interesting issue, but the title seems not quite to fit together with what you are asking in the body of the question. Maybe you could clarify this a bit since the question has attracted some close votes :-/ $\endgroup$ – Dilaton May 31 '13 at 9:19

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.

  • $\begingroup$ however if i make $ p= 1/u $ then 'u' should be a wavelenght since $ \lambda = \frac{h}{p} $ $\endgroup$ – Jose Javier Garcia May 30 '13 at 16:45
  • $\begingroup$ anyway in dimensional regularization (to name one) can one simply make a change of variables inside the d-regularized integrals ?? $\endgroup$ – Jose Javier Garcia May 31 '13 at 10:27
  • $\begingroup$ Yes, that's possible. $\endgroup$ – Vibert May 31 '13 at 16:08

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