Regularization of an Non-conergent Integral

In summary, the regularization process is supposed to change the problem such that the integral becomes convergent, as long as the cut-off value of \Lambda is larger than any of the masses in the theory.
  • #1
topsquark
Science Advisor
Insights Author
Gold Member
MHB
2,021
804
The equations here come from calculating the amplitude of a Feynman diagram. I can set up the problem if you really want me to but here I am just interested in why and how the regularization process is supposed to work Mathematically.

The generalized meaning of this is if we are given a divergent integral, say \(\displaystyle \int _0^{\infty} f(x) ~ dx\), we can (supposedly) change the problem to \(\displaystyle \lim_{ \Lambda \to \infty } \int_0^{ \Lambda } R( \Lambda ) f(x) ~ dx\) such that the integral will become convergent so long as \(\displaystyle \lim_{ \Lambda \to \infty } R (\Lambda ) = 1\). The cut-off value of \(\displaystyle \Lambda\) is taken as \(\displaystyle x << \Lambda\). (I know not just any function R will work and I don't know how to find one.)

But that changes the nature of the integral! I don't see how this is supposed to work. My Physics text says that certain properties of the integral don't change so we can use the regularized integral for those. It doesn't directly say what those properties are that are kept, just that presumably finite properties, like mass, can be split up into finite and infinite parts. I can give you more on the Physics but I can't seem to find a good source on this lower level. My QFT text starts talking about the renormalization group and Ward identities and suddenly the problem goes over my head. I am just working Mathematically on the cut-off technique for specific integrals for right now.

Specifically I am working on the integral
\(\displaystyle \int _0^{ \infty } \left ( \dfrac{1}{q^2 - r^2} \right ) \left ( \dfrac{1}{(q - p)^2 - s^2} \right ) ~ q^3 ~ dq\)
where p, r, and s are constants. This integral is logarithmically divergent.

The regularization used in my text is
\(\displaystyle \lim_{ \Lambda \to \infty} \int _0^{ \Lambda } \left ( \dfrac{- \Lambda ^2 }{ q^2 - \Lambda ^2} \right ) \left ( \dfrac{1}{q^2 - r^2} \right ) \left ( \dfrac{1}{(q - p)^2 - s^2} \right ) ~ q^3 ~ dq\)
which does converge.

-Dan
 
Physics news on Phys.org
  • #2
topsquark said:
The cut-off value of \(\displaystyle \Lambda\) is taken as \(\displaystyle x << \Lambda\).
Sorry, I mispoke here. \(\displaystyle \Lambda = Mc\) where M is much larger than the other masses in the theory.

-Dan
 

Related to Regularization of an Non-conergent Integral

1. What is regularization of a non-convergent integral?

Regularization of a non-convergent integral is a mathematical technique used to assign a finite value to an integral that would otherwise be infinite or undefined. It involves manipulating the original integral to make it convergent and then taking the limit as the manipulated parameter approaches its original value.

2. Why is regularization necessary for non-convergent integrals?

Non-convergent integrals arise when the function being integrated grows too quickly or has a singularity at the limits of integration. Regularization is necessary to assign a meaningful value to these integrals and make them useful in mathematical calculations.

3. What are some common methods of regularization for non-convergent integrals?

Some common methods of regularization include dimensional regularization, zeta function regularization, and cutoff regularization. These methods involve manipulating the integral in different ways to make it converge and then taking the limit as the manipulated parameter approaches its original value.

4. How does regularization affect the accuracy of the integral's value?

Regularization can affect the accuracy of the integral's value depending on the method used. Some methods may introduce errors, while others may provide more accurate results. It is important to carefully choose the appropriate regularization method for a given integral to ensure accurate results.

5. Can regularization be applied to all non-convergent integrals?

No, not all non-convergent integrals can be regularized. Some integrals may be too complex or have no meaningful interpretation, making it impossible to assign a finite value to them. In these cases, other mathematical techniques may need to be used to approximate the integral's value.

Similar threads

Replies
12
Views
2K
Replies
3
Views
1K
  • Calculus
Replies
4
Views
2K
Replies
1
Views
2K
Replies
15
Views
3K
Replies
16
Views
1K
Replies
1
Views
1K
Replies
2
Views
808
Replies
16
Views
3K
Back
Top