Can Recursive Formulas Help Solve Divergences in Perturbation Theory?

[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.

 

[mathman]

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

 

[Chronos]

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

 

[Careful]

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.

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

 

[mathman]

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

 

[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.

 

[Careful]

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

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.