Why is ε→0⁺ used in both terms of the Cauchy principal value formula?

[Jhenrique]

The cauchy principal value formula is:

But why ε→0⁺ in both terms? The correct wouldn't be ε→0⁻ in 1st term and ε→0⁺ in 2nd term? Like:

\lim_{\varepsilon \to 0^-}\int_{a}^{c-\varepsilon}f(x)dx + \lim_{\varepsilon \to 0^+}\int_{c+\varepsilon}^{b}f(x)dx

?

 

[lurflurf]

no if you do that c is in the interval
we want to exclude c

 

[Jhenrique]

but if I define the superior limit in first integral like c+ε, so the expression below will be correct now?

\lim_{\varepsilon \to 0^-}\int_{a}^{c+\varepsilon}f(x)dx + \lim_{\varepsilon \to 0^+}\int_{c+\varepsilon}^{b}f(x)dx

 

[PeroK]

but if I define the superior limit in first integral like c+ε, so the expression below will be correct now?

\lim_{\varepsilon \to 0^-}\int_{a}^{c+\varepsilon}f(x)dx + \lim_{\varepsilon \to 0^+}\int_{c+\varepsilon}^{b}f(x)dx

There's no difference between adding a small -ve ε and subtracting a small +ve ε.

... although that definition is not equivalent as now you have two separate limits, so what you've defined is the improper integral as both limits must exist independently.

 

[vanhees71]

This is definitely wrong! The correct definition has been given in the posting by Jhenrique! The important point of the definition of the Cauchy PV is to leave out a tiny SYMMETRICAL "window" around the singularity and then make this window arbitrarily small.

The difference can be demonstrated by a simple example. E.g., take the Cauchy principle value
I=\text{PV} \int_{-1}^{1} \mathrm{d} x \frac{1}{x}.
Now the correct definition is
I=\lim_{\epsilon \rightarrow 0^+} \left (\int_{-1}^{-\epsilon} \mathrm{d} x \frac{1}{x}+\int_{\epsilon}^1 \mathrm{d x} \frac{1}{x} \right ) = \ln \epsilon-\ln \epsilon=0.
If you try to take the limits of the two integrals separately, these limits do not even exist in this way!

 

[PeroK]

This is definitely wrong! The correct definition has been given in the posting by Jhenrique!

With all due respect, I think you've got confused about who posted what. JHenrique posted an alternative defn of the CPV, which I pointed out was in fact the defn of an Improper Integral.

You've really muddied the waters if you're saying JH is correct with his alternative definition.