Energy of a signal "The energy" of a signal $x(t)$ is defined as :
$$E_s = \int_{-\infty}^{\infty}|x(t)|^2$$
Why is it called energy if it's not homogeneous to energy ?
What does it actually represent ?
Parseval's theorem states that :
$$\int_{-\infty}^{\infty}|x(t)|^2 = \int_{-\infty}^{\infty}|X(f)|^2$$
So it says that the "energy" of a signal is equal to the "energy" of it's spectrum, what does that mean ? It's a bit abstract...
 A: The intensity of your signal at every moment is given indeed by |x(t)|^2, s.t. the total intensity of the signal during the time is the integration of |x(t)|^2 over time.
About the connection with the frequency, see below.
You can describe the signal by its behavior in time x(t). But you can decompose it in sine and cosine functions, i.e.
(1) $x(t)$ = $\int [A(f) sin(2πft) + B(f) sin(2πft)] df$ .
A(f) and B(f) indicate with which amplitude each sine and cosine participates in your signal x(t). Notice that in the formula (1) the integration is from 0 to +∞, the frequencies are positive.
But the equality (1) can be written in another form, more compact, after introducing
(2) $sin(2πft) = [exp(2iπft) - exp(-2iπft)]/2i$
and
(3) $cos(2πft) = [exp(2iπft) + exp(-2iπft)]/2 $
You get
(4) $x(t)$ = $\int [C(f)exp(2iπft) - D(f)exp(-2iπft)] df$ ,
where
(5) $C(f) = [B(f) - iA(f)]/2$ ,
(6) $D(f) = [B(f) + iA(f)]/2 $.
For making the form (4) more compact, one makes a change of variable, f' = -f. Then one gets
(7) $x(t)$ = $\int C(f)exp(2iπft)df + ∫D(f')exp(2iπf't)df'$ ,
where the first integral is from 0 to +∞ and the second from -∞ to 0. So, introducing a function F(f) that behaves as C(f) for positive frequencies and as D(f) for negative frequencies
(8) $x(t)$ = $\int F(f)exp(2iπft)df$ ,
the integral being from -∞ to +∞ .
In all, you obtained $x(t)$ expressed by means of $F(f)$, its Fourier transform.
You see that the time description, and the description with the Fourier transform are equivalent - given F(f) you can restore x(t) as shown in (8).
What Parseval proved, and it's not difficult, is that instead of integrating the intensity along time, you can integrate the absolute square of the Fourier transform. 
(8) $\int|x(t)|^2 dt = \int|F(f)|^2 df$ .
Remember that F(t) is connected with the representation of your signal as a sum of sines and cosines, more exactly with their amplitudes.
Good luck !
A: If $x(t)$ is the current through a resistor $R$, then the voltage is $Rx(t)$ and then the instantaneous power dissipated is $Rx(t)\bar{x}(t)$ and the dissipated energy over all times is $E_s=\int R\vert {x(t)}\vert ^2 dt$. Because of Parseval's theorem you can write mean square in time as the mean square of the frequency spectrum, ie., Fourier transform $X(f)$ of $x(t)$.
A: An aspect of this question worth keeping in mind is that the "energy" in the signal may not be the energy in the system generating the signal. As an example, consider a signal that is the $x$-position of a mass, $m$, on a spring oscillating in time with an amplitude $A$ along the $x$-axis only. Just as in all of the discussion here suggests, the energy in the signal would be
\begin{equation}
E_{\textrm{signal}} = \int_{\infty}^{\infty}dt |x(t)|^2=A^2.
\end{equation}
But, we also know that the energy of a mass oscillating on a spring is
\begin{equation}
E_{\textrm{system}}=\frac{1}{2}mA^2\omega^2\ne E_{\textrm{signal}},
\end{equation}
where $\omega$ is $2\pi$ times the oscillation frequency.
