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

In summary, 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. However, the exponential is always well defined, so the term in the second equation is needed. This 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.
  • #1
McLaren Rulez
292
3
Hi,

I came across this for the first time today.

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

Here [itex]P(\frac{1}{\omega})[/itex] 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 :)
 
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  • #2
Anyone at all?
 
  • #3
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 [itex]\omega[/itex] 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:
[tex]
\lim_{\epsilon \rightarrow 0^{+}} \frac{1}{x \pm i \epsilon} = P\left(\frac{1}{x}\right) - \pm i \pi \delta(x)
[/tex]

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
 
  • #4
The equality is meant in the sense of distributions. For [itex]\varphi \in C^\infty_c(\mathbf{R})[/itex] the equality should be read as
[tex] \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). [/tex]
The result follows from a simple calcuation:
[tex] \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.[/tex]
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 [itex]\epsilon>0[/itex].
 
  • #5
Thank you jasonRF and Anthony. Your replies have been very helpful.
 
  • #6
Hi,

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

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


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


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


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 [tex]\int^{\infty}_{-\infty}d\omega\frac{1}{\omega-\omega_{a}}[/tex]

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!
 

What is Cauchy Principal Value?

Cauchy Principal Value is a mathematical concept used to evaluate improper integrals that do not converge in the traditional sense. It takes into account the limiting behavior of the function as it approaches the singular point.

Why is Cauchy Principal Value necessary?

Cauchy Principal Value is necessary because some integrals do not have a well-defined value using traditional methods. These integrals have singularities or infinite limits, and Cauchy Principal Value provides a way to assign a meaningful value to them.

How is Cauchy Principal Value calculated?

Cauchy Principal Value is calculated by taking the average of the limit of the function as it approaches the singularity from both sides. This ensures that the value assigned is symmetric with respect to the singularity.

What are some real-world applications of Cauchy Principal Value?

Cauchy Principal Value has applications in physics, particularly in quantum mechanics and electromagnetism. It is also used in engineering and numerical analysis to approximate integrals that do not converge using traditional methods.

What are the limitations of Cauchy Principal Value?

Cauchy Principal Value can only be used for integrals that have specific types of singularities and infinite limits. It is not applicable to all improper integrals and cannot be used to evaluate integrals with infinitely oscillating behavior.

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