# Talk:Cauchy principal value

WikiProject Mathematics (Rated Start-class, Mid-priority)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 Start Class
 Mid Priority
Field:  Analysis

## Notation

I would rewrite the article, using notation

${\displaystyle -\!\!\!\!\!\!\int _{a}^{c}f(x){\rm {d}}x}$


In the current version, the "definitions" are a little but fussy. dima (talk) 02:29, 28 July 2008 (UTC)

I would only use this notation if you think it is worth introducing to readers who have never seen it before. The notation is not common, from what I can tell.Njerseyguy (talk) 07:53, 11 March 2009 (UTC)

## Confusing sentence in "Formulation"

"In the case of Lebesgue-integrable functions, that is, functions which are integrable in absolute value, these definitions coincide with the standard definition of the integral."

I don´t understand the intened meaning of this sentence, since Lebesgue-integrable functions ≠ functions which are integrable in absolute value. The class of Lebesgue-integrable functions is broader than that.--79.235.151.134 (talk) 09:56, 20 April 2010 (UTC)

## Almost the only?

"It is the inverse distribution of function x and is almost the only distribution with this property" What does this mean? 78.91.83.39 (talk) 16:51, 22 May 2013 (UTC)

## Definition

I think the requirement that the lateral limits exist as extended reals is not necessary. Considere the example:

${\displaystyle f(x)={\frac {1}{x^{2}}}\sin \left({\frac {1}{x}}\right)}$

We see that

${\displaystyle \int _{\varepsilon }^{a}f(x)dx=\cos \left({\frac {1}{a}}\right)-\cos \left({\frac {1}{\varepsilon }}\right),~~\forall \varepsilon >0}$

which implies the lateral limit does not exist. Nonetheless, I would say that

${\displaystyle -\!\!\!\!\!\!\int _{a}^{b}f(x)\,\mathrm {d} x=\lim _{\varepsilon \rightarrow 0+}\left[\int _{a}^{-\varepsilon }f(x)\,\mathrm {d} x+\int _{\varepsilon }^{b}f(x)\,\mathrm {d} x\right]=\int _{-a}^{b}f(x)dx=\cos \left({\frac {1}{b}}\right)-\cos \left({\frac {1}{a}}\right),~~a<0

where I have used that f is odd so that

${\displaystyle \int _{a}^{-\varepsilon }f(x)\,\mathrm {d} x+\int _{\varepsilon }^{-a}f(x)\,\mathrm {d} x=0,~\forall a<0.}$

By the way, I see no problem in saying that

${\displaystyle -\!\!\!\!\!\!\int _{a}^{b}\mathrm {d} x=b-a}$.

Do you agree with me? Lechatjaune (talk) 17:27, 26 June 2014 (UTC)