So we can solve for ##c_1## and find ##x_1(\omega)##.

[Another]

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 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

 

[Asim Riaz]

to solve this problem$x_1(\omega ) = \frac{k c_1}{\pi} lim_{\epsilon → 0}( \int_{a}^{c - \epsilon} \frac{1}{ \omega '^2 - \omega ^2} d \omega ' + \int_{c + \epsilon}^{b} \frac{ 1 }{ \omega '^2 - \omega ^2} d \omega ' )$I think we can use the residue theorem to solve this problem. Let $f(\omega) = \frac{1}{ \omega '^2 - \omega ^2}$ we can find the residue of this function at $\omega$ .Res(f,ω) = lim (ω-→ω) (ω-ω)f(ω) =1/2ωThen$x_1(\omega ) = \frac{k c_1}{2\pi} \left(Res(f,a) + Res(f,b)\right) = \frac{k c_1}{\pi} \left(\frac{1}{2a} + \frac{1}{2b}\right)$