# What is principal value integral?

1. Feb 27, 2009

### xylai

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!

Last edited: Feb 27, 2009
2. Feb 27, 2009

### xylai

I have known how to deduce the equation (3)?
There is another equation:
$$\sigma (t)=(t+\sigma_{n}^{-1})^{-1}$$,
then
$$\int_{0}^{T}\sigma (t)=ln|det(1+\sigma_{n}T)|$$

3. Feb 27, 2009

### lightarrow

If $$x = x_0$$ is contained in the interval (a,b), then the Principal value of the Integral

$$\int_{a}^{b}f(x)dx$$

is:

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

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.