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    #1
    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

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  3. MHB Master
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    topsquark's Avatar
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    #2 Thread Author
    Quote Originally Posted by topsquark View Post
    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

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