- #1

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- Homework Statement
- ##x(\omega) = x_1(\omega)+ix_2(\omega)##

##x_1(\omega ) = \frac{2}{\pi} P \int_{0}^{∞} \frac{\omega 'x_2(\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

Where ##P## are Cauchy principal value.

##P \int_a^b g(x)dx = lim_{\epsilon → 0} \int_a^{k - \epsilon} g(x) dx + \int_{k + \epsilon}^b g(x) dx ##

- Relevant Equations
- ##x(\omega) = x_1(\omega)+ix_2(\omega)##

##P \int_a^b g(x)dx = lim_{\epsilon → 0} \int_a^{k - \epsilon} g(x) dx + \int_{k + \epsilon}^b g(x) dx ##

In the Kramers-Kroning relation

Let ##x(\omega) = x_1(\omega)+ix_2(\omega)## be a complex function of the complex variable ##\omega## , Where ## x_1(\omega) ## and ## x_2(\omega) ## are real

We can find ##x_1(\omega) ## from this integral

##x_1(\omega ) = \frac{2}{\pi} P \int_{0}^{∞} \frac{\omega 'x_2(\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

Where ##P## are

If we know ##\alpha (\omega)## In relation to ## x_2(\omega) =\frac{k}{2 \omega} \alpha(\omega) ## this function defined interval ##a ≤ \omega ≤ b## and ##k## are real constant.

So I need find ##x_1(\omega)##

Let

##x_1(\omega ) = \frac{k}{\pi} P \int_{0}^{∞} \frac{\omega 'x_2(\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

##x_1(\omega ) = \frac{k}{\pi} P \int_{a}^{b} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

Let ##f(\omega') = \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2}##

if ##a ≤ \omega ≤ b## same ##a ≤ \omega '≤ b## Lead to ##f(\omega ')## go to infinity

But ##P \int_a^b g(x)dx = lim_{\epsilon → 0} \int_a^{c - \epsilon} g(x) dx + \int_{c + \epsilon}^b g(x) dx ##

How can i application the

I THINK

##x_1(\omega ) = \frac{k}{\pi} P \int_{a}^{b} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

##x_1(\omega ) = \frac{k}{\pi} lim_{\epsilon → 0}( \int_{a}^{c - \epsilon} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega ' + \int_{c + \epsilon}^{b} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega ' )##

if ##\alpha (\omega') ## maybe a ##c_1## constant we get

##x_1(\omega ) = \frac{k c_1}{\pi} lim_{\epsilon → 0}( \int_{a}^{c - \epsilon} \frac{1}{ \omega '^2 - \omega ^2} d \omega ' + \int_{c + \epsilon}^{b} \frac{ 1 }{ \omega '^2 - \omega ^2} d \omega ' )##

I don't know how can i set ##c## value

Let ##x(\omega) = x_1(\omega)+ix_2(\omega)## be a complex function of the complex variable ##\omega## , Where ## x_1(\omega) ## and ## x_2(\omega) ## are real

We can find ##x_1(\omega) ## from this integral

##x_1(\omega ) = \frac{2}{\pi} P \int_{0}^{∞} \frac{\omega 'x_2(\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

Where ##P## are

__Cauchy principal value__.If we know ##\alpha (\omega)## In relation to ## x_2(\omega) =\frac{k}{2 \omega} \alpha(\omega) ## this function defined interval ##a ≤ \omega ≤ b## and ##k## are real constant.

So I need find ##x_1(\omega)##

Let

##x_1(\omega ) = \frac{k}{\pi} P \int_{0}^{∞} \frac{\omega 'x_2(\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

##x_1(\omega ) = \frac{k}{\pi} P \int_{a}^{b} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

Let ##f(\omega') = \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2}##

if ##a ≤ \omega ≤ b## same ##a ≤ \omega '≤ b## Lead to ##f(\omega ')## go to infinity

But ##P \int_a^b g(x)dx = lim_{\epsilon → 0} \int_a^{c - \epsilon} g(x) dx + \int_{c + \epsilon}^b g(x) dx ##

How can i application the

__Cauchy principal value__. to this problem.I THINK

##x_1(\omega ) = \frac{k}{\pi} P \int_{a}^{b} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega '##

##x_1(\omega ) = \frac{k}{\pi} lim_{\epsilon → 0}( \int_{a}^{c - \epsilon} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega ' + \int_{c + \epsilon}^{b} \frac{\alpha (\omega ') }{ \omega '^2 - \omega ^2} d \omega ' )##

if ##\alpha (\omega') ## maybe a ##c_1## constant we get

##x_1(\omega ) = \frac{k c_1}{\pi} lim_{\epsilon → 0}( \int_{a}^{c - \epsilon} \frac{1}{ \omega '^2 - \omega ^2} d \omega ' + \int_{c + \epsilon}^{b} \frac{ 1 }{ \omega '^2 - \omega ^2} d \omega ' )##

I don't know how can i set ##c## value