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What is principal value integral?

  1. Feb 27, 2009 #1
    In one paper (PRL 89, 144101 (2002)),
    [tex]k=<Tr\sigma>_{p.v.}[/tex], (1)
    where p.v. stipulates a principal-value evaluation and

    [tex]\sigma_{n+1}=(\sigma_{n}^{-1}+T)^{-1}-\nabla\nabla f(q_{n+1})[/tex], (2)

    then the author deduces the following equation:
    [tex]k=lim_{N\rightarrow\infty}\sum_{n=0}^{N-1}ln|det(1+\sigma_{n}T)|[/tex] (3).
    Can you show me how to deduce the equation (3)?
    Thank you!
    Last edited: Feb 27, 2009
  2. jcsd
  3. Feb 27, 2009 #2
    I have known how to deduce the equation (3)?
    There is another equation:
    [tex]\sigma (t)=(t+\sigma_{n}^{-1})^{-1}[/tex],
    [tex]\int_{0}^{T}\sigma (t)=ln|det(1+\sigma_{n}T)|[/tex]
  4. Feb 27, 2009 #3
    If [tex]x = x_0[/tex] is contained in the interval (a,b), then the Principal value of the Integral



    [tex]P.V. \int_{a}^{b}f(x)dx\ =\ lim_{\epsilon\rightarrow 0^+} [\int_{a}^{x_0-\epsilon}f(x)dx\ +\ \int_{x_0+\epsilon}^{b}f(x)dx]

    Note that the two separated limits or their sum can not exist, but the limit of their sum can, as in the case

    [tex] f(x) = 1/x;\ x_0\ =\ 0[/tex]:

    in this case, the two separated limits are infinite: one -oo and the other +oo, so their sum doesn't exist, but the limit of the sum (the principal value) is zero.

    See also:
    Last edited: Feb 27, 2009
  5. Feb 27, 2009 #4

    Thank you! you have talked about the principal value integral very clearly.
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