Given a 2-loop divergent integral $\int F(q,p)\,\mathrm{d}p\mathrm{d}q$, can it be solved iteratively? I mean

  1. I integrate over $p$ keeping $q$ constant
  2. Then I integrate over $q$

In both iterated integrals I use dimensional regularization.

Can it be solved iteratively? I presented a paper to a teacher of mine about regularization of integrals using the Zeta regularization. He told me that for one dimensional integrals (or one loop integrals) it was fine, but that my method could not handle multi loop integrals. I argue back that you could apply the regularization method by introducing a regulator of the form


I could make the integral on each variable by iterated regularization, that is applying the algorithm iteratively.

EDIT: i think they have cheated me :) making up excuses not to put me atention

by the way , how can i insert math codes on my posts ?? is just setting $ at the beginning and the end of the equation ??

Edit: 19th of July

thanks :) your answer was quite useful :)

however can I not insert a term $(qi+a)^{-s} $ on each variable and then apply the regularization iteratively ??

I say so because I made a paper http://vixra.org/abs/1009.0047 to regularize integrals using the regulator $(q+1)^{-1}$ and tried to extend it to several variables, however that was my doubt, if i could apply my regularization scheme to every variable :) thanks again

I mean for a one dimensional integral $\int dx (x+a)^{m-s}$, I know how to regularize it by using the Euler-Maclaurin summation formula plus the Riemann Zeta function $ \zeta (s-m)$

Then for multiloop integrals, I had thought that i could introduce the $s$-regulators (x+a)^{-s}(y+a)^{-s}.. and so on on each variable, and then apply iterated integration... :)

so this is a resume of my method..

a) i know how to use Zeta regularization to get finite values for the integral $ \int dx(x+a)^{m-s} $ in terms of the Riemann Zeta function $ \zeta (s-k) $ with k=-1,0,1,....,m

b) for a more general 1-dimensional integral $ \int dx f(x)(x+a)^{-s}$ i add substract a Polynomial $K(x+a)(x+a)^{-s} $ to get a finite part and then regularize the divergent integrals $dx(x+a)^{m-s} $

c) for a more complicate 2-loop integral $ \iint dpdq F(q,p)(p+a)^{-s}(q+a)^{-s} $ to obtain a regularization of it i do the sema method , first on 'p' considering 'q' a constant and i treat it as a one dimensional integral over 'p' and then over 'q' by substracting a Polynoamials $ K(q,p+a)(p+a)^{-s} $ and so on

  • $\begingroup$ I don't know the answer, but I can't really see why it wouldn't work. Hm. Good question. $\endgroup$
    – wsc
    Commented Jul 14, 2011 at 1:26
  • $\begingroup$ What do You do with the partial results (first over p, then over q?). Add them? Multiply? $\endgroup$
    – Georg
    Commented Jul 17, 2011 at 12:59
  • 1
    $\begingroup$ @wsc: I know it doesn't work and also that two (not to mention higher higher) loop diagrams can get very tricky (e.g. a friend of mine very recently published a result on some two-loop radiative corrections to $\pi^0 \to e^+ e^-$). But unfortunately that's all I know as this isn't my field. $\endgroup$
    – Marek
    Commented Jul 17, 2011 at 18:35
  • $\begingroup$ Response to Jul19 edit: Of course you can apply your regularization like that. But remember that there's no canonical way of routing momentum through a Feynman diagram, so you regularized integral will be dependent on this arbitrary choice. This makes it more difficult (but not impossible) to get a consistent renormalization scheme. That is, to consistently subtract subdivergences so that you always have local counterterms. I also imagine that your regularization will break gauge invariance / Ward identites, which will need to be fixed with finite counterterms. $\endgroup$
    – Simon
    Commented Jul 19, 2011 at 11:36
  • $\begingroup$ (Continued) Anyway, try to use your regularization to produce something like 2-loop propagator results in QED and see what happens. Can you always get local counterterms? What about the Ward identities? $\endgroup$
    – Simon
    Commented Jul 19, 2011 at 11:37

1 Answer 1


Assuming that you can actually do the dimensionally regularized integrals, then yes, you can integrate iteratively. Normally, two-loop and higher-loop integrals are quite difficult and you need good tricks like turning them into differential equations or using Mellin-Barnes parametrization of the propagators (or even just Feynman or Schwinger parametrizations). The new standard book on evaluating dimensionally regularized multiloop integrals is Feynman Integral Calculus by Vladimir Smirnov (2006). Also see the notes for the Father & Son lectures at Durham a couple of years ago.

Not only are multiloop integrals hard, but the regularization scheme becomes important once you leave one-loop. This is because you have to make a consistent renormalization / subtraction of counter-terms. Regularization schemes that modify the diagrams/integrals and not the original Lagrangian can lead to messy renormalization analysis. The simplest renormalization analyses are associated with regularizations that change the Lagrangian (or dimension) combined with a mass-independent minimal subtraction scheme.

Inserting terms like $(q^2)^s$ is a type of analytic regularization, which (like most regularizations) is easy at one-loop, but tricky at higher loops. Analytic regularization was extensively examined by Eugene Speer in the 60s and 70s.

Zeta regularization does only work at one-loop, but it has a generalization/extension called Operator Regularization that was studied by McKeon and friends. From memory, the calculations get a bit messy, although there is a new(ish) paper on using operator regularization and Feynman diagrams that is on my reading list (previous work used a more Schwinger-like, functional approach to loop calculations).

I hope that something I said in the above is useful and not too much is wrong. ;)

  • $\begingroup$ i managed to obtain finite results for one dimensional and multi-dimensional integrals in my paper vixra.org/abs/1009.0047 however they reject my saying it is WRONG but without giving any proof, they say that for one loop it may be okey but that it would not work on multi-loo`ps however i believe that by iterated integral i could apply zeta regularization to multiple integrals :( but since i can not get a grant i can not publish my papers $\endgroup$ Commented Jul 27, 2011 at 16:21
  • $\begingroup$ @Jose: Getting a finite result from a divergent integral is easy. Doing it in a sensible way and getting consistent physical results is hard. For example, your method breaks momentum routing invariance - not that that is necessarily fatal. Try reading Consistency Conditions for 4-D Regularizations and the papers by Bonneau [1, 2] and the intro to Breitenlohner, Maison. $\endgroup$
    – Simon
    Commented Jul 29, 2011 at 12:14
  • $\begingroup$ @Jose: A grant and a uni position is helpful for getting papers published, but not essential. Your paper might have some useful stuff in it, but it's not very clear. Try to really clarify what your new idea is and how it fits into the ecosystem of other regularizations. How does your generalization of zeta function regularization (if it is that) compare to that of McKeon et al? How does it work in the two-loop $\phi^4$ calculation? How about gauge invariance? $\endgroup$
    – Simon
    Commented Jul 29, 2011 at 12:17
  • $\begingroup$ here it is explained with some examples i gave no renormalization so i do not know how to work for the $ \phi ^4 $ theory, however i have checked that for finite integrals the results are the same..for multi-loop or multiple integrals we can do iterated integration over one of each variable to get a regularization for them , by introducing the regulator $ (q+1)^{-s}$ we can make the integral convergent $\endgroup$ Commented Jul 29, 2011 at 14:02
  • $\begingroup$ why my method will not work ?? zeta regularization WORKS well for series why not for integrals :( , sorry i would like to know before giving up , why DImensional regularization works but not zeta regularization ? , why can not we make iterated integration over every variable and then apply my method of zeta regularization for integrals ?, why can not this method be applied to define the product of delta function with itself?, i think that my methdo would respect ALL the rules of calculus and that can be used to get finite correction for the UV divergences and IR divergences $\endgroup$ Commented Jul 29, 2011 at 14:21

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