# Cauchy principal value

1. Feb 15, 2014

### 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:

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

?

2. Feb 15, 2014

### lurflurf

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

3. Feb 16, 2014

### Jhenrique

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

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

4. Feb 16, 2014

### PeroK

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.

5. Feb 17, 2014

### 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!

6. Feb 17, 2014

### PeroK

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.