Non-local connection Between $D$ and $E$ (Kramers-Kronig relations) I'm trying to solve next exercise from Jackson's Electrodynamics, Chapter 7, page 348, 3th edition:
Consider the non-local (in time) connection between $D$ and $E$,
$$\vec{D}(\vec{x},t)=\epsilon_0 \left\{ \vec{E}(\vec{x},t) + \int d\tau G(\tau) \vec{E}(\vec{x},t-\tau) \right\}$$
with the $G(\tau)$ appropriate for the single-resonance model.
$$ \frac{\epsilon(\omega)}{\epsilon_0} =1+ \frac{\omega_p^2}{\omega_0^2-\omega^2-i\gamma \omega}. $$
a) Convert the nonlocal connection between D and E into an instantaneous relation involving derivatives of E with respect to time by expanding the electric field in the integral in a Taylor series in $\tau$. Evaluate the integrals over $G(\tau)$ explicitly up to at least a $ \frac{\partial^2\vec{E}}{\partial t^2} $
b) Show that the series obtained in part a can be obtained formally by converting the frequency-representation relation, $ \vec{D}(\vec{x},t) = \epsilon (\omega) \vec{E}(\vec{x},t) $ into a space-time relation
$$ \vec{D}(\vec{x},t) = \epsilon \left( i \frac{\partial}{\partial t} \right) \vec{E}(\vec{x},t) $$
where the variable $ \omega $ in $ \epsilon (\omega) $ is replaced by $ \omega -> i \frac{\partial}{\partial t} $.
On a) I tried replacing the electric field in the deduction from page 330, but I can't found anything. On b) A similar idea, same result.
 A: Problem (a)
The displacement field strength is
\begin{equation} 
  \vec{D}(x,t) = \epsilon_0( \vec{E}(x,t) + \int_0^\infty G(\tau)E(x,t)\,d\tau 
  \label{Eq:DandE} \tag{1}
 \end{equation}
We consider a Taylor series of electric field, namely :
\begin{equation}
    \vec{E}(x,t) = \vec{E}(x,0) + (t- \tau)\cdot \frac{\partial} {\partial t} E(x,0) + \frac{(t-\tau)^2}{2} \frac{\partial^2} {\partial t^2} E(x,0)
    \label{Eq:Tailor} \tag{2}
 \end{equation}
And $G(\tau)$
$$G(\tau) = \int_{-\infty}^{-\infty}f(\omega)e^{2\pi i f \tau}\,df$$
$$ f(\omega) = \frac{\omega_p^2} {\omega_0^2 - \omega^2 - i\gamma \omega}$$
$$ G(\tau) =  \int_{-\infty}^{-\infty} \frac{\omega_p^2} {\omega_0^2 - \omega^2 - i\gamma \omega}  e^{2\pi i f \tau}\,df $$
There are two residues as shown in the following picture, $\omega_\pm = 1/2 (-i\gamma \pm \sqrt{-\gamma^2 + 4 \omega_0^2)}$. It is simple to show that the part corresponding to the integration over half of circle diminishes when the radius is sent to $\infty$, so we can write the following formula.

$$ G(\tau)= \frac{1} {2\pi} \oint f(\omega) e^{i \omega \tau} \,d\omega = \frac{2\pi i \omega_p^2} {2\pi} \sum{ Res \frac{e^{i\omega \tau}} {\omega_0^2 - \omega^2 - i\gamma \omega}}$$
From the residual equation we have.
$$ Res(g(z))= \lim_{z \to z_0} g(z)(z-z_0) $$
And for G($\tau$)
\begin{equation}
 G(\tau) = -i \omega_p^2 
[
\frac{e^{-i \omega_+ \tau } }{ {(\omega-\omega_+)} (\omega-\omega_-) }\cdot {(\omega-\omega_+)} +
\frac{e^{-i \omega_- \tau } }{ {(\omega-\omega_-)} (\omega-\omega_+) }\cdot {(\omega-\omega_-)}
]=
\label{Eq:Gtau}
\end{equation}
$$ G(\tau)= 2\frac{\omega_p^2}{\breve{\omega}}e^{-\gamma \tau /2}\sin{\breve{\omega} \tau} \tag{3}$$
where $ \breve{\omega} = \sqrt{4\omega^2_0-\gamma^2} $
Using equations (1), (2) and (3) we get
$$
\vec{D}(x,t) =
\epsilon_0( \vec{E}(x,t) + 
\frac{2 \omega_p^2}{\breve{ \omega}}^2 \int_0^\infty \,d\tau
[
    e^{-\gamma \tau /2} \sin{ \breve{\omega}\tau}+
    \tau e^{-\gamma \tau /2} \sin{ \breve{\omega}\tau}+ 
    \tau^2 e^{-\gamma \tau /2} \sin{ \breve{\omega}\tau}
]
$$
With the help of Mathematica (I do not know how to take this integrals, if you know how, please explain me)
$$ 
    \int_0^\infty d\tau\, e^{-\gamma \tau /2} \sin{ \breve{\omega}\tau} = 
    \frac{\breve{\omega}} {2 \omega_0^2}
$$
$$ 
    \int_0^\infty \,d\tau\, \tau e^{-\gamma \tau /2} \sin{ \breve{\omega}\tau} = 
    \frac{\gamma \breve{\omega}} {2 \omega_0^4}
$$
$$ 
    \int_0^\infty \,d\tau\, \tau^2 e^{-\gamma \tau /2} \sin{ \breve{\omega}\tau} = 
    \frac{ ( \gamma^2 - \omega^2 ) \breve{\omega}} {2 \omega_0^4}
$$
So, in the end we have:
$$
    \vec{D}(x,t) = \epsilon_0[\vec{E}(x,t)(1+ \frac{\omega_p^2} {\omega_0^2})
    + \frac{\gamma^2 \omega_p}{\omega^4_0}\frac{\partial} {\partial t} \vec{E}(x,t)
    + \frac{\gamma^2 - \omega^2_0}{\omega^6_0}\frac{\partial^2} {\partial t^2} \vec{E}(x,t)]
$$
Problem (b)
You consider the following operator
$$
 \epsilon (\omega) = \epsilon_0 (1+ \frac{\omega_p^2} {\omega_0^2-\omega^2-i \gamma \omega} )  
$$
if we change $\omega$ to $i\partial_t$
$$
    \frac{\epsilon (i\partial_t)} {\epsilon_0} =1 + \frac{\omega_p^2} {\omega_0^2}\frac{1} {1+\frac{\partial^2_t}{\omega_0^2} + \frac{\gamma \partial_t} {\omega_0^2} }
$$
as a Taylor series of
$$
    (1+\hat{A})^n \approx 1-\hat{A} + \hat{A}^2
$$
where $\hat{A}$ is an operator, so we get
$$
     \frac{\epsilon (i\partial_t)}{\epsilon_0} \approx 1-\frac{\partial_t^2}{\omega_0^2} - \frac{\gamma \partial_t} {\omega_0^2} + \frac{\gamma^2 \partial_t^2} {\omega_0^4}
$$
After applying this operator to get displacement field strength
$$ \vec{D}(x,t) = \hat{\epsilon}(i \partial_t)\vec{E}(x,t)$$
$$
    \vec{D}(x,t) = \epsilon_0[\vec{E}(x,t)(1+ \frac{\omega_p^2} {\omega_0^2})
    + \frac{\gamma^2 \omega_p}{\omega^4_0}\frac{\partial} {\partial t} \vec{E}(x,t)
    + \frac{\gamma^2 - \omega^2_0}{\omega^6_0}\frac{\partial^2} {\partial t^2} \vec{E}(x,t)]
$$
The question is, however how are these equation are used?
