I know that if the integral is convergent we can always make a change of variable to make it better, however what happens with DIVERGENT integrals? can we make a change of variable into a divergent integral after having it regularized?
I mean we insert the regulator $(q+a)^{-s}$ or similar on each variable and then make the change of variable to polar, cilindrical or other
Here's some elementary stuff (so that we don't get bogged down in unnecessary technicalities) to think about. Suppose we'd like to integrate something like $$\int_0^{\infty} {1 \over x^2} {\rm d}x.$$ This of course diverges but we can insert some regularizator $M$ that captures the divergence in the $M \to 0$ limit $$I(M) := \int_0^{\infty} {1 \over (x + M)^2} {\rm d}x = \left[- {1 \over x + M} \right]_0^{\infty} = {1 \over M}.$$
We obviously can't change the value of $I(M)$ by using any (smooth) substitution. so, suppose we changed our original integral using $x = y^2$. Then we need to evaluate
(Edit: there was a mistake in this calculation that changes the result of the discussion).
$$\int_0^{\infty} {2 \over y^3} {\rm d}y$$ and using the same regularization procedure $x \to x + M$ or $y \to \sqrt{y^2 + M}$ we get $$I(M) = \int_0^{\infty} {2 \over (y^2 + M)^{3/2}} {\rm d}y = \int_M^{\infty} {1 \over z^2} {\rm d}y = {1 \over M} .$$
On the other hand if we naively performed the regularization $y \to y + M$ we would obtain a different result $$I(M) = \int_0^{\infty} {2 \over (y + M)^3} {\rm d}y = {1 \over M^2} .$$
So I guess the point is that one needs to make a regularization that respects the substitution to obtain consistent results.
Important point: a change of variables cannot improve convergence! When you regulate an divergent integral $I$ ($=\infty$), you're actually calculating another integral $J$ (that might depend on some parameters $s,t,\ldots$) and you try to 'recover' $I$ by exchanging the order of integration and taking some limits. However, since you know that $I$ doesn't exist and $J$ does, it's obvious that the two aren't equal.
The same goes for you current question: by construction, a change of variables leaves the value of the integral unchanged and cannot change the convergence of an integral. Of course, you can change the integrand to turn an existing integral $I$ into something divergent, but that doesn't make a lot of sense, mathematically. You're probably interested in common regularization methods (like the one you cite in your problem). They can actually be tested on convergent integrals to see if they work: suppose you regularize the already convergent integral $I$ by $J(\varepsilon)$ and $$\lim_{\varepsilon \rightarrow 0} \, J(\varepsilon) \neq I,$$ then you know something is wrong with your regularization.
Yes. The answer for the specific type of regularistation you mention, adding complex analytic convergence factors, yes. the answer is yes for zeta-regularisation also. Because the original version and the change of variable version both agree on an open set in the region of convergence, their analytic continuations and other limits have to agree everywhere (up to the usual choose-a-branch issues).
but « regularisation » is a loose term, and there may be other methods, such as lattice regularisation, for which it is not quite true.
If an integral is convergent, nothing can make it better ;-)
If an integral is divergent, every means is good nowadays to make it convergent and equal to what you want, unfortunately.
