# Can we make a change of variables (for example to polar coordinates) into a divergent integral?

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

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I am not sure what you mean. Are you perhaps asking whether the correct regularized result can be recovered also after changing variables in the divergent integral? My guess is that the answer is yes but this is of course in a purely physical way; mathematically it probably doesn't make much sense. – Marek Jul 24 '11 at 10:04
You seem to mean first insert the terms you mention, then change variable. You seem to be assuming that for some values of $s$ the integral then converges, so you are changing variables in a convergent integral, even though you earlier said « make a change of variables into a divergent integral », and when you say « after having it regularised » that also implies you make the change of variables after regularisation. – joseph f. johnson Jan 7 '12 at 23:32

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.

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I believe there's a factor of 2 missing in your 4th line (i.e. $\int_{y > 0} 2/(y^2 + M)^{3/2} = 2/M$) - maybe you can check your calculation. – Gerben Jul 25 '11 at 1:04
@Gerben: it seems you are correct. I just threw this out as random ideas and I was wondering myself where it would lead (since it doesn't make much mathematical sense). So are we back at version 1 where I stated that most of the time any substitution will spoil the result? I'll have to check this more carefully later. – Marek Jul 25 '11 at 7:41

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.

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of course that if the integral is DIVERGENT no substitution will make it convergent however there may be some variables that make the inetgral to look nicer than others – Jose Javier Garcia Jul 24 '11 at 13:05

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.

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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.

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