My teachers say this trick (regularization) is not valid

  • Context: Graduate 
  • Thread starter Thread starter zetafunction
  • Start date Start date
  • Tags Tags
    Regularization Teachers
Click For Summary

Discussion Overview

The discussion revolves around the validity of using regularization techniques to compute divergent integrals, specifically in the context of a rational function integral. Participants explore the implications of regularization methods, such as introducing a regulator and using analytic continuation, while addressing concerns raised by the original poster's teachers about the legitimacy of these approaches.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • The original poster proposes a method to regularize a divergent integral by introducing a regulator and changing to polar coordinates, leading to a new integral that can be analytically continued.
  • Some participants argue that regularization is a way to assign meaning to divergent expressions, but emphasize that such assignments are not standard and require careful interpretation.
  • One participant suggests that the integral may diverge at a specific value of the regulator, which could undermine the proposed regularization strategy.
  • Another participant draws a parallel to the Riemann zeta function, noting that while certain divergent sums can be assigned values through analytic continuation, this does not imply that the original divergent expressions have standard meanings.
  • There is a discussion about the rules of calculus and whether they apply to divergent integrals, with participants expressing uncertainty about the implications of using standard calculus rules in these contexts.

Areas of Agreement / Disagreement

Participants express differing views on the validity and interpretation of regularization techniques for divergent integrals. There is no consensus on whether the methods proposed by the original poster are acceptable or if they lead to meaningful results.

Contextual Notes

Participants highlight the need for nonstandard definitions to make sense of divergent integrals or series, indicating that standard mathematical definitions may not apply in these cases. The discussion remains open-ended regarding the implications of regularization methods.

zetafunction
Messages
371
Reaction score
0
i want to compute the integral

[tex]\iint_{D} f(x,y)dxdy[/tex] here f(x,y) is a Rational function and the integral is DIVERGENT.

in order to regularize i had the following idea , i introduce a regulator[tex]\int_{D} \frac{f(x,y)}{(x^{2}+y^{2}+1)^{s}}dxdy[/tex] so for big 's' the integral is convergent

i make now a change of variable to polar coordinates so

[tex]g(r)= \int_{\theta}f(r, \theta)[/tex]

then the initial intengral becomes (after integration over the angular variable, this integraiton can be made by either numerical or exact analytic methods)

[tex]\int_{0}^{\infty} \frac{f(r ,\theta)}{(r^{2}+1)^{s}}rdr = I(s)[/tex]

then this integral is easier to handle, i use tables to compute it and then to analytically continue with external parameter 's' to s =0 in order to botain a regularization of the initial integral

however my teachers say this can not be done how is it possible ¿¿ are they cheating me ?
 
Physics news on Phys.org
Regularisation like you are doing here can be seen as a trick to make some sense of an otherwise meaningless expression. Strictly speaking, if an integral is divergent, that is the end of the story; the integral cannot be assigned any particular value.

Inspired by your namesake, let's take a slightly simpler example that should be closely analogous. Consider the sum

[tex]S=\sum_{n=1}^\infty n = 1+2+3+\cdots[/tex]

which is divergent in a strict mathematical sense, and I'm sure you'll agree that it can't be assigned any sensible value. But it can be regarded as a special case of the series

[tex]\zeta(s)=\sum_{n=1}^\infty n^{-s}[/tex]

for s=-1. This sum does make sense, but only for certain values of s (with real part greater than 1). It is in fact the famous Riemann zeta function. But this function can naturally be extended to every s except s=1. At s=-1, it turns out that this extension gives you a value of -1/12. So our original sum can in some sense be assigned the value -1/12*. But this doesn't really mean that 1+2+3+...=-1/12; our trick hasn't suddenly given a divergent sum a value in the ordinary sense, and to have any strict mathematical truth, the sum must be interpreted in some non-standard way (check out Hardy's "Divergent Series" if you can get hold of a copy).


* Since I like theoretical physics, I'll note that this assignment can be used to 'prove' that bosonic string theory must live in 26 dimensions!
 
zetafunction said:
i want to compute the integral

[tex]\iint_{D} f(x,y)dxdy[/tex] here f(x,y) is a Rational function and the integral is DIVERGENT.

in order to regularize i had the following idea , i introduce a regulator


[tex]\int_{D} \frac{f(x,y)}{(x^{2}+y^{2}+1)^{s}}dxdy[/tex] so for big 's' the integral is convergent

i make now a change of variable to polar coordinates so

[tex]g(r)= \int_{\theta}f(r, \theta)[/tex]

then the initial intengral becomes (after integration over the angular variable, this integraiton can be made by either numerical or exact analytic methods)

[tex]\int_{0}^{\infty} \frac{f(r ,\theta)}{(r^{2}+1)^{s}}rdr = I(s)[/tex]

then this integral is easier to handle, i use tables to compute it and then to analytically continue with external parameter 's' to s =0 in order to botain a regularization of the initial integral

however my teachers say this can not be done how is it possible ¿¿ are they cheating me ?
Just a thought:
Your integral is most likely to become divergent already at the value s=1/2, since then the r's in the numerator and denominator have roughly equal exponent.

Thus, your strategy would seem to break down prior to the limit of I(s) at s=0.
 
use analytic continuation arildno, consider 's' big so the integral is convergent the final integral will depend on 's' so we can analytically continue to s -->0

just like the zeta function regularization [tex]\sum _{n \ge 1} n^{k}= \zeta (-k)[/tex] that has a pole at s=1 but it can be analytically continue to s <0 , this case is the same

by the way , the rules of calculus remain the SAME even for divergent integrals ??
 
Sorry, perhaps my first post wasn't that clear.

The point I was trying to make is that though we might try to make sense of a divergent integral or series, any answer we come up with would lie beyond the standard definitions of such objects. To make any rigorous sense of a statement like 1+2+3+...=-1/12 requires more work. I have shown just one possible way we might try to give it meaning, but who is to say someone else might come up with an equally convincing argument and yet a different answer?

From a modern point of view, to interpret such a statement requires a series (or integral) to be defined differently from the standard way, in order to catch a broader class of objects. But no such definition is mainstream mathematics.

zetafunction said:
by the way , the rules of calculus remain the SAME even for divergent integrals ??

Once we've come up with a decent definition, it would definitely be a very attractive property if we could apply the ordinary rules of arithmetic, calculus etc. But this sort of thing would require proof from our new definition, just as it requires proof in the ordinary definition.

In summary: talking about assigning value to a divergent series/integral is nonsense, at least until you specify some nonstandard meaning for your series/integral.

If you're interested in this kind of thing, I really do recommend Hardy's book I mentioned above, and in fact you can read it for free on the http://www.archive.org/details/divergentseries033523mbp" At least skim through the first few pages, it explains things much more lucidly than I ever could.
 
Last edited by a moderator:

Similar threads

  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K