Talk:Cauchy principal value

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I would rewrite the article, using notation

-\!\!\!\!\!\!\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"[edit]

"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.-- (talk) 09:56, 20 April 2010 (UTC)

Almost the only?[edit]

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


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


We see that

\int_\varepsilon^af(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

-\!\!\!\!\!\!\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}^bf(x)dx = \cos\left(\frac{1}{b}\right)-\cos\left(\frac{1}{a}\right),~~a<0<b

where I have used that f is odd so that

\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

-\!\!\!\!\!\!\int_a^b \mathrm{d}x=b-a.

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