Emergence of Cauchy Principal value I have a problem

Caerswso1pc

Caerswso1pc

Answered question

2022-04-04

Emergence of Cauchy Principal value
I have a problem solving differential equation, where I think there should be a Cauchy Principal value involved, but I do not see how it it should emerge. Let's say we have a differential equation:
iC˙1(t)=0dων(ω)ei(ω10-ω)t,
where all the variables are real and positive. If we now solve the differential equation, it needs only to carry out the time integral
iC1(t)=dt0dων(ω)ei(ω10-ω)t=0dων(ω)i(ω10-ω)ei(ω10-ω)t.
So the problem is that if the last integral is not taken as principal value, it will give , so I would rather need it to be
P0dων(ω)i(ω10-ω)ei(ω10-ω)t
where P is the principal value. Would anyone have any idea where the principal value would emerge from, if at all? I was thinking maybe it has something to do when swapping the order of integrals in order to carry out the time integral first, but I am not sure.

Answer & Explanation

Ashton Conrad

Ashton Conrad

Beginner2022-04-05Added 11 answers

Explanation:
Yes, your observation is precisely that Fubini does not apply, because one of the iterated integrals is (sometimes) infinity. But you can cut off a small ball around ω10>0 before interchanging the integrals:
PSK\begin{align} \int_0^t \int_0^\infty v(\omega)e^{i(\omega_{10}-\omega)t}d\omega dt&= \lim_{\epsilon\to 0}\int_0^t\int_{\omega\in[0,\infty),|\omega-\omega_{10}|>\epsilon} v(\omega)e^{i(\omega_{10}-\omega)t}d\omega dt\&= \lim_{\epsilon\to 0}\int_{\omega\in[0,\infty),|\omega-\ \omega_{10}|>\epsilon} \int_0^tv(\omega)e^{i(\omega_{10}-\omega)t}d\omega dt\&= \lim_{\epsilon\to 0}\int_{\omega\in[0,\infty),|\omega-\omega|>\epsilon} \frac{v(\omega)e^{i(\omega_{10}-\omega)t}}{i(\omega_{10}-\omega)}d\omega \&=:\mathcal P\!\!\!\int_0^\infty \frac{v(\omega)e^{i(\omega_{10}-\omega)t}}{i(\omega_{10}-\omega)}d\omegaZSK
The above works only when ω100ZKS.Also,ifPSKv(ω10)=0 and is C1 then the above still works but then the integral does exist, so P!!0=0.

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