How to prove $\int_{-\infty}^{\infty}x^2(t)dt=\int_{-\infty}^{\infty}\hat{x}^2(t)dt$ where $\hat{x}(t)$ is the Hilbert Transformation of $x(t)$
The Hilbert transform of $x(t)$ denoted by $\hat{x}(t)$ has the following Fourier transform
\begin{equation}
 \mathcal{F}(\hat{x}) = H(w)X(w) = -j \operatorname{sgn}(w)X(w)
\end{equation}
because 
\begin{equation}
 \hat{x}(t) = \int\limits_{-\infty}^{\infty} \frac{x(u)}{\pi(t-u)} \ du
\end{equation}
Notice that 
\begin{equation}
 \vert \mathcal{F}(\hat{x}) \vert^2 = 
 \vert -j \operatorname{sgn}(w)X(w) \vert^2 
 =
 \vert X(w) \vert^2
\end{equation}
i.e.
\begin{equation}
\int\limits_{-\infty}^{\infty} \vert \mathcal{F}(\hat{x}) \vert^2 
\ dw
=
\int\limits_{-\infty}^{\infty}
\vert X(w) \vert^2
\ dw 
\end{equation}
Use Parsevals equation which gives you
\begin{equation}
\int\limits_{-\infty}^{\infty} \vert \hat{x}(t) \vert^2 
\ dt
=
\int\limits_{-\infty}^{\infty}
\vert x(t) \vert^2
\ dt
\end{equation}
P.S: For real signals, you could indeed remove the magnitudes (absolute values).