My teachers say this trick (regularization) is not valid

  • #1
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Main Question or Discussion Point

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 ?
 

Answers and Replies

  • #2
160
1
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!
 
  • #3
arildno
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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.
 
  • #4
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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 ??
 
  • #5
160
1
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.

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" [Broken] At least skim through the first few pages, it explains things much more lucidly than I ever could.
 
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