How does energy depend on frequency in an alternating current circuit? In what relation is the energy input in an alternating current circuit to its frequency?
I'd guess I have to compute something like
$$E=\int P(\omega,t) dt=\int U(\omega,t) I(\omega,t) dt, $$
but if say 
$$U(\omega,t)\propto\sin{(\omega t)},$$
then it seems part of the integral is $\propto\frac{1}{\omega}$, while I would expect the energy to grow with $\omega$.
 A: Consider just the power delivered:
$$
P(t) =  U(\omega t') I(\omega t') 
$$
Consider the simple case $ U(t')=U_0\sin(\omega t')$ and $ I(t')=I_0\sin(\omega t')$.  Then the power delivered is $$P(t) =  U_0 I_0\sin^2(\omega t)$$ 
$$={ U_0 I_0\over 2}\{1-cos(2\omega t)\}$$ 
We can break this into two terms:
$$P_{DC}={U_0 I_0\over 2}$$
and $$P_{2\omega} = { U_0 I_0\over 2}cos(2\omega t)$$
$P_{DC}$ represents the average power flowing out of our power supply.  It is $\propto$ voltage and current.  Since it is constant in time, the energy delivered is rising linearly in time.  That is, the longer we keep our device plugged in to our power supply, the more energy the power supply has delivered to it.
$P_{AC}$ delivers no net power.  For half of the $2\omega$ cycle, it is dumping some extra power in, for the other half, it extracts the extra power dumped in.  It is a reflection of the fact that current and voltage are not constant, but in fact are varying sinusoidally and only delivering $P_{DC}$ on average.  
And yes, if you integrate $P(t)$ to get $E(t)$ you will see a $1/\omega$ term in the answer.  I'll leave it to you to think about why.  Hint: higher frequency peak powers don't last as long as lower frequency ones.
A: If you have a circuit with static elements (e.g. combination of resistors, capacitors and inductors), then for
$$U(t) = U_0 \sin(\omega t)$$
you have
$$I(t) = I_0 \sin(\omega t + \phi)$$.
You get $I_0$ and $\phi$ from complex impedance $Z(\omega)$:
$$I_0 = \frac{U_0}{|Z|}, \tan \phi = \left(\frac{\text{Im}(Z)}{\text{Re}(Z)}\right).$$
You can also observe voltage and current as complex numbers:
$$U(t) = U_0 e^{i\omega t}, I(t) = I_0 e^{i\omega t}, U_0 = Z I_0$$
EDIT: If you are only concerned by term $\omega^{-1}$, use $T_0$ instead of $\omega$:
$$U(t) = U_0 \sin(\frac{2 \pi t}{T_0})$$
you have
$$I(t) = I_0 \sin(\frac{2 \pi t}{T_0} + \phi)$$.
You will end up with the expression that energy is proportional to $T_0$.  This only tells you that energy is smaller, because time in which it was transferred is smaller.
A: Note that your formula  
$$E=\int P(\omega,t) dt=\int U(\omega,t) I(\omega,t) dt$$ can be rewritten as  
$$E=\int U(\omega,t) I(\omega,t) dt=\int\frac{U^2(w,t)}{Z}dt$$ Now, let $U=U_0\sin(wt)$ and $Z=const$ which is reasonable during a short period of time $t$.Thus:  
$$E=\frac{1}{Z}\int_0^{t} U^2(w,t)dt=\frac{U_0^2}{Z}\int_0^{t} \sin^2(w,t)dt$$ Next:  
$$\int_0^{t} \sin^2(w,t)dt=\frac{t}{2}-\frac{\sin(2wt)}{4w}$$ So, if $w\rightarrow\infty$, then $E$ does not depend of $w$.  
Edit:
Additions: 
To be more specific, the circuit's impedance $Z$ depends on the frequency as  
$$Z=\sqrt{r(w)^2+\left ( wL-\frac{1}{wC}\right)^2}$$ where $L$ is the circuit's impedance and $C$ is the circuit's capacitance and $r$ is the resistance which depends also on $w$ due to the skin effect.  
That means $Z \rightarrow wL$ as $w\rightarrow\infty$  if $L\neq 0$ 
So an answer, closer to reality is that   
$$E=\frac{U_0^2}{wL}\frac{t}{2};w\rightarrow\infty$$  
