Need help proving an equation for power loss in a dielectric

[BaghDal]

Homework Statement

2. Homework Equations
[/B]
I am trying to prove the equation
$$
\frac{\bar{P}}{V}=\frac{1}{2}E_0^2\sigma_{AC}
$$
which can be rewritten as
$$
\begin{align}
\frac{\bar{P}}{V} &= \frac{1}{2}E_0^2\sigma_{AC}\\
&=\frac{1}{2}E_0^2\ \omega\ \epsilon_0\ \epsilon^{''}_r\\
&=\frac{1}{2}E_0^2\ \omega\ \epsilon_0\ \epsilon_r^{'}\ \tan(\delta)
\end{align}
$$
Here $$\bar{P}$$ stands for the time-averaged power loss which satisfies the equation
$$
\bar{P}=\frac{1}{T}\int_0^T U\ I\ dt,
$$
where $$T=\frac{2\pi}{\omega}$$ is the time period, $$U=U_0 e^{j\omega t}$$ is the complex sinusoidal voltage, and $$I=j\omega\epsilon^{'}_rC_0U + \omega\epsilon^{''}_rC_0U$$.
The instructions say to use
$$
\begin{align}
U_0 &= E_0h\\
C &= \epsilon_r\epsilon_0\frac{A}{h}\\
V &= A\ h \\
\sigma_{AC}&=\omega\epsilon_0\epsilon^{''}_r=\epsilon_0\epsilon^{'}_r\tan(\delta)\\
\tan(\delta) &= \frac{\epsilon^{''}_r}{\epsilon^{'}_r}
\end{align}
$$

The Attempt at a Solution

The problem I face is after solving the main integral part, which is like:
$$
\epsilon^{''}_r*(F(T) - F(0)) + j*\epsilon^{'}_r((F(T) - F(0))
$$
where $$F(t) = e^{2j\omega t}$$ and I neglected all the constants for simplicity.
$$F(T)$$ is equal to $$\exp(j*4*\pi)$$ which is 1, making $$(F(T)-F(0))$$ zero and thus the whole equation zero.

I thought of root mean squaring both U and I to begin with, but this gives out $$\sqrt{\epsilon^{''2}_r+\epsilon^{'2}_r/2}$$ term which doesn't seem to lead to the proof result.

I have a hunch that I maybe missing some elementary calculation basics. I would really appreciate any help that you can offer.

 

[rude man]

Basic problem is with using complex expressions for volts and current to compute power. Can't be done.
Consider:
V = V0exp(jwt). Real part is V0 cos(wt).
I = I0exp(jwt). Real part is I0 cos(wt).
Multiply: P = VI = V0 I0 exp(2wt). Real part is V0 I0 cos(2wt) which has zero average over T = 2pi/w (no dc term).
Which we know is wrong since P = V0 sin(wt) x I0 sin(wt) = V0 I0 sin²(wt) = V0 I0 (1/2)[1 - cos(2w)]
with average P = (V0 I0)/2.