What is the role of the principal value in the Cauchy principal value integral?

[McLaren Rulez]

Hi,

I came across this for the first time today.

$\int_0^\infty e^{i\omega t}dt = \pi\delta(\omega)+iP(\frac{1}{\omega})$

Here $P(\frac{1}{\omega})$ is the so called principal value. I haven't seen this term normally so can I ask where we get it from?

Googling principal value showed me a very different thing. It said that the principal value of an integral of a function is to take a sum of integrals such that we skip over those values where the function is not well defined. But here, the complex exponential is always well defined so what exactly is the reason for the second term?

Thank you :)

 

[McLaren Rulez]

Anyone at all?

 

[jasonRF]

these types of relations are usually referred to as Plemelj formulas. See the end of the article:

http://en.wikipedia.org/wiki/Sokhotski–Plemelj_theorem

to understand what it means. Your equation is symbolic, and is missing the fact that you are taking the limit as the imaginary part of $\omega$ goes to zero. These types of relations are used frequently in plasma physics. A more usual form is found by integrating the exponential first to yield:
$$\lim_{\epsilon \rightarrow 0^{+}} \frac{1}{x \pm i \epsilon} = P\left(\frac{1}{x}\right) - \pm i \pi \delta(x)$$

Again, this is symbolic and what it really means is found in the equation under "Version for the real line" in the article I linked.

hope that helped.

jason

 

[Anthony]

The equality is meant in the sense of distributions. For $\varphi \in C^\infty_c(\mathbf{R})$ the equality should be read as
$$\int_0^\infty \left(\int_{-\infty}^\infty e^{\mathrm{i}\omega t}\varphi(\omega)\, \mathrm{d} \omega \right)\mathrm{d} t = \lim_{\epsilon\rightarrow 0} \int_{|\omega|>\epsilon} \mathrm{i}\varphi(\omega)\, \mathrm{d}\omega + \pi \varphi(0).$$
The result follows from a simple calcuation:
$$\int_0^\infty \left(\int_{-\infty}^\infty e^{\mathrm{i}\omega t}\varphi(\omega)\, \mathrm{d} \omega \right)\mathrm{d} t \overset{1}{=} \lim_{\epsilon\rightarrow 0} \int_0^\infty \left(\int_{-\infty}^\infty e^{\mathrm{i}\omega t-\epsilon t}\varphi(\omega)\, \mathrm{d} \omega \right)\mathrm{d} t \overset{2}{=} \lim_{\epsilon\rightarrow 0} \int_{-\infty}^\infty \left( \int_0^\infty e^{\mathrm{i}\omega t-\epsilon t}\, \mathrm{d} t \right) \varphi(\omega)\, \mathrm{d}\omega = \lim_{\epsilon\rightarrow 0} \int_{-\infty}^\infty \frac{\mathrm{i}\varphi(\omega)}{\omega+ \mathrm{i} \epsilon}\, \mathrm{d}\omega.$$
Now apply Plemelj. In 1 we used a regularisation, which can be justified by dominated convergence (the Fourier transform of a smooth function of compact support is Schwartz, so integrable). In 2 we used Fubini, justified since the double integrable is absolutely integrable for each $\epsilon>0$.

 

[McLaren Rulez]

Thank you jasonRF and Anthony. Your replies have been very helpful.

 

[McLaren Rulez]

Hi,

I wanted to bring this thread back up because I encountered this again. Okay, here's the full problem.

Suppose we have $$\int^{\infty}_{-\infty}d\omega \int^{\infty}_{0}dt\ e^{-i(\omega-\omega_{a})t}$$


Here, $\omega_{a}$ is a constant. We write this as $$\lim_{\epsilon\to 0} \int^{\infty}_{-\infty}d\omega \int^{\infty}_{0}dt\ e^{-i(\omega-\omega_{a}-i\epsilon)t}$$


Now the time integral yields, $$\lim_{\epsilon\to 0}\frac{\epsilon}{\epsilon^{2}+(\omega - \omega_{a})^{2}} - i\frac{(\omega-\omega_{a})}{(\omega-\omega_{a})^{2}+\epsilon^{2}}$$


The first part is a delta function and that part I have no trouble with. The imaginary part however is still troubling me. I still have to do the omega integral. The main problem is that it blows up. One author has even expressed it as $$\int^{\infty}_{-\infty}d\omega\frac{1}{\omega-\omega_{a}}$$

Doesn't this principal part blow up? Yet, there is an expression (and a physical Lamb shift) associated with this part of the integral. How is the imaginary integral solved?

Thank you for your help!