I'm unable to understand the steps between eq. 16 and 17 in Gaarde, M. B. et al. Phys. Rev. A - At. Mol. Opt. Phys. 83, (2011):
$\tilde{f}(\omega)$ is the Fourier transform of $f(t)$ and their convention is: $f(t) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}\tilde{f}(\omega)~e^{-i w t} dt$ and $\tilde{f}(\omega) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}f(t)~e^{i w t} d\omega$
What I have so far: $$ \Delta E= \int^{\infty}_{-\infty}z(t) \frac{\partial E}{\partial t} \text{dt} \text{ | insert FT}\\ = \frac{1}{2\pi}\iint^{\infty}_{-\infty} \tilde{z}(\omega)~\tilde{E}(\omega) \frac{\partial }{\partial t} \left(~e^{-i 2 \omega t} \right)\text{d}\omega~\text{dt} \text{ | do partial deriv.} \\ = - \frac{i}{\pi}\iint^{\infty}_{-\infty} \omega~\tilde{z}(\omega)~\tilde{E}(\omega)~e^{-i 2 \omega t} \text{d}\omega~\text{dt} $$ Now, I guess due the symmetry of z and E, I can do this, however I am unsure: $$ = - \frac{i}{\pi}\iint^{\infty}_{0}\omega~2~\tilde{z}(\omega)~\tilde{E}(\omega)~e^{-i 2 \omega t} \text{d}\omega~\text{dt} $$ Then, I could do the integration w.r.t. t, but then I end up with: $$ = \frac{1}{\pi}\int^{\infty}_{0} \tilde{z}(\omega)~\tilde{E}(\omega) \text{d}\omega $$
Where is the imag coming from? And why is the $2\omega$ still in eq. 17?
I did my best, but - obviously - I'm not an expert.. sry. any help is much appreciated.