What is principal value integral?

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In one paper (PRL 89, 144101 (2002)),
k=<Tr\sigma>_{p.v.}, (1)
where p.v. stipulates a principal-value evaluation and
<f>=^{def}lim_{t\rightarrow\infty}t^{-1}\int_{0}^{t}f(\bar{t})d\bar{t}.

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

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

<br /> \int_{a}^{b}f(x)dx

is:

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]<br />

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

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

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:
http://en.wikipedia.org/wiki/Cauchy_principal_value
 
Last edited:
lightarrow said:
If x = x_0 is contained in the interval (a,b), then the Principal value of the Integral

<br /> \int_{a}^{b}f(x)dx

is:

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]<br />

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

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

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:
http://en.wikipedia.org/wiki/Cauchy_principal_value


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