1. Sep 4, 2006

### lokofer

Hello..compared to most people of this forum i'm just a "newbie"... but once i read (or i think) that to deal with perturbation theory it would be a good idea if the divergences of the type:

$$I(m)= \int_{0}^{\infty}dpp^{n}$$ n>1,n=0 or n<0 could be expressed in a "recursive" form for example if we could write:

$$I(m)=aI(m-1) +bI(m-2) +............+zI(0)$$

where a,b,c,d,e,.....,z are "finite" and real numbers..is that true?..i think in other forums heard a similar idea but i don't know if it worth working on it.

2. Sep 5, 2006

### mathman

The integral that you presented doesn't exist, i.e. it is infinite. You will have to clarify your question.

3. Sep 6, 2006

### Chronos

It's a mathematical formalism I'm not familiar with.

4. Sep 6, 2006

### Careful

Hi, what you propose seems meaningful since the divergences on the right hand side are less harmful than the ones on the left hand side. To make it accurate, I guess you might want to substitute for :
$$I(n)= \int_{0}^{\infty}dpp^{n} e^{- \epsilon p}$$ for n > 0, for n < 0 you have to remove the pole at p=0 and take the limit for epsilon to zero at the end of your calculation.
Anyway, this is just one particular naive regularization procedure.

Cheers,

Careful

Last edited: Sep 6, 2006
5. Sep 6, 2006

### mathman

Careful's integral is simply n!(eps)-n-1, which becomes infinite as eps ->0.

6. Sep 7, 2006

### Chronos

This stuff hurts my head, but, I am fairly confident that summing up an infinite series of non-zero fractional entities does not necessarily result in infinity.

7. Sep 7, 2006

### Careful

Sure, I typed that in, but the latex didn't get through. In physics such divergent expressions occur all the time, and we can regularize them in such ways by putting in a cutoff of some kind.