Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Renormalization and divergent integrals.

  1. Aug 12, 2007 #1
    If the problem of renormalization is that there are divergent integrals for x-->oo couldn't we make the change.

    [tex] \int_{0}^{\infty}dx f(x) \approx \sum_{n=0}^{\infty}f(nj) [/tex]

    using rectangles with base 'j' small , and approximating the divergent integral by a divergent series and 'summing' by Borel or other kind of resummation, to solve the problem

    for Infrared divergences [tex] f(x)= \frac{C}{x^{n}} [/tex] n >0 we could apply some kind of 'Hadamard finite-part integral' or Cauchy Principal Value to get finite results
  2. jcsd
  3. Aug 12, 2007 #2
    If the integral is divergent, it is divergent. Period. No "resummation" can change that. Divergent integrals in QED are signs of a serious problem. The problem is that the original Hamiltonian is wrong and cannot be used beyond the first non-vanishing order. Tomonaga-Schwinger-Feynman renormalization theory corrects this problem: (infinite) counterterms are added to the Hamiltonian. Then all infinities in S-matrix elements cancel out, and very accurate results are obtained.

  4. Aug 12, 2007 #3


    User Avatar
    Science Advisor

    Sometimes you can borel resum the integral in that way, there are examples of field theories like that (typically in d = 3). Unfortunately it doesnt work in the general case.
  5. Aug 13, 2007 #4
    Why not 'Haelfix' ?? if a integral is divergent a series is also divergent.

    the easiest example is the finite value attached to divergent integral

    [tex] \int_{0}^{\infty}dp(1+ p)^{n} \approx h^{n+1} \zeta(1/h,-n) [/tex]

    where 'h' is the step and we have used the 'Hurwitz zeta function' since for a function we can make the expansion (Laurent series)

    [tex] \sum_{n=-\infty}^{\infty}a_{n}z^{n} [/tex]

    the rest of the problem should be a trivial question, for a power-law divergent integral you must find the Borel sum of a certain sequence...
  6. Aug 13, 2007 #5


    User Avatar
    Science Advisor

    Eh? I never said the contrary.

    Borel resummation is precisely the process whereby you try to make sense of certain classes of divergent series (or integrals, its a straightforward generalization).

    Like I said, it works for some field theories, but not all of them (most do not). Actually, when you can find such an example, you tend to become rather famous, there are preciously few analytic results in QFT.

    Try to Borel resum some of the textbook cases of field theories as an exercise, you will find there are obstructions to doing this. (In fact a lot was learned about general nonperturbative contributions by comparing perturbation theory and the analytic results in the few cases where we could resum)
  7. Aug 14, 2007 #6
    You mean that the integrals are not Borel summable or that you can't find the Borel transform of the sequence a(n) mainly:

    [tex] \sum_{n=0}^{\infty}\frac{a_{n}}{n!}x^{n} [/tex]

    or that there is a theorem that says ..that series with factor [tex] e^{iax} [/tex] are not Borel summable.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook