# Cauchy Principal Value

1. Apr 20, 2014

### Zag

Hello everyone,

I have recently bumped into the Kramers Kronig Relations while reviewing some of my Eletromagnetism notes, and as you may know those relations are written in terms of the Cauchy Principal Value (CPV) of certain integrals. Well, I've never been very familiar with with the concept of a CPV, so I decided to read a little more about it hoping to achieve a better understanding of that idea.

After some reading, as far as I understand, the CPV seems to be just a definition to assign values to integrals that in the first place would be indeterminate - which sounds a bit weird for me. So I was hoping to ask if this is the correct way of thinking about it. Is it really just a funny way of expressing certain values that would be otherwise ill-defined or is there something more to it?

Thank you very much!
Zag

2. Apr 20, 2014

### micromass

Staff Emeritus
Yes, that's exactly it! I know it's weird, but it's very useful nonetheless.

3. Apr 20, 2014

### lurflurf

Often it is convenient to group together a family of limits. This is often done in integration. We may do this because we want to replace one limit with another, find all of them at once, or require that all are equal. Sometimes we find that in such a grouping all the limits do not exist or different limits have different values. This does not mean none of the limits are useful. So the situation is an integral is a family of limits that we treat as one. Sometimes we find that we cannot treat them as one so we handle each limit separately. The Cauchy Principal Value is one particular such limit that we find useful at times. Indeterminate often and in this case means a question does not have the same answer in all cases and we must consider each case individually.

A typical example is
$$\int_{-\infty}^\infty \! x \, \mathrm{d}x$$
this represents a family of limits that do not agree
the cpv is
$$\operatorname{P \! V}\int_{-\infty}^\infty \! x \, \mathrm{d}x=\lim_{a\rightarrow \infty}\int_{-a}^a \! x \, \mathrm{d}x=0$$
it is one particular limit that we find of use

4. Apr 20, 2014

### Zag

Nice! Thank you for replying micromass and lurflurf. I will look more into the usefulness of the Cauchy Principal Value. But knowing that there is nothing deeper to it already makes it look less weird. :D

Thanks again!

5. Apr 21, 2014

### disregardthat

Does anyone have an example where this definition is particularly useful?

6. Apr 21, 2014

### jbunniii

7. Apr 21, 2014

### disregardthat

This article makes an interesting note, any PV can be expressed as improper integrals:

$$PV \int^{\infty}_{-\infty} f(x)dx = \lim_{a \to \infty} \int^{a}_{-a} f(x)dx = \lim_{a \to \infty} \left( \int^{a}_{0} f(x)dx+\int^{0}_{-a} f(x)dx \right)$$
$$= \lim_{a \to \infty} \int^{a}_{0} f(x)+f(-x)dx = \int^{\infty}_{0} f(x)+f(-x)dx$$

And for the other case:

$$PV \int^b_a f(x)dx = \lim_{\epsilon \to 0^+} \left( \int^{c-\epsilon}_{a}f(x)dx + \int^{b}_{c+\epsilon}f(x)dx \right)= \lim_{\epsilon \to 0^+} \left( \int^{c-\epsilon}_{a}f(x)dx + \int^{c-\epsilon}_{2c-b}f(2c-x)dx \right)$$
$$= \lim_{\epsilon \to 0^+} \left( \int^{c-\epsilon}_{2c-b}f(x)dx + \int^{2c-b}_af(x)dx+ \int^{c-\epsilon}_{2c-b}f(2c-x)dx \right) = \int^{2c-b}_af(x)dx + \int^{c}_{2c-b}f(x)+f(2c-x)dx$$

Last edited: Apr 21, 2014
8. Apr 21, 2014

### jbunniii

Yes, I think of it as being somewhat analogous to how we define the Fourier series as $\lim_{N \rightarrow \infty} \sum_{n=-N}^{N} c_n e^{2\pi i n t/T}$ instead of $\lim_{M,N \rightarrow \infty}\sum_{n=-M}^{N} c_n e^{2\pi i n t/T}$. The former converges in some cases where the latter would not. For example, a square wave has coefficients $c_n$ whose magnitudes decay on the order of $1/n$, so for $t=0$ the second limit fails to exist.

In the case of the Fourier series, it's natural to use symmetric upper and lower endpoints because we want it to pair each exponential with its conjugate so we are summing sines and cosines. In the case of the Hilbert transform, I don't have the same intuition other than "we have to define it that way so the integral will converge."