Demonstrate that $x(t)$ and $\hat{x}(t)$ are orthogonal I have to demonstrate that a function $x(t)$ and his Hilbert's transform are orthogonal, it is said:

$$\int^{\infty}_{-\infty} x(t) \cdot \hat{x}(t) dt = 0$$

I have tried the exercise using Parseval's theorem:

$$\int^{\infty}_{-\infty} x^2(t) \cdot \hat{x}^2(t) dt = \int^{\infty}_{-\infty} |X(f) \ast \hat{X}(f)|^2 df = \int^{\infty}_{-\infty} |X(f) \ast (-jsgn(f)X(f))|^2 df =$$

$$= \int^{0}_{-\infty} |X(f) \ast jX(f)|^2 df + \int^{\infty}_{0} |X(f) \ast (-jX(f))|^2 df$$

Using properties of convoution ($ax(t) \ast y(t)) = a(x(t) \ast y(t))$:
$$\int^{\infty}_{-\infty} x^2(t) \cdot \hat{x}^2(t) dt = \int^{0}_{-\infty} | j(X(f) \ast X(f))|^2 df + \int^{\infty}_{0} |-j(X(f) \ast X(f))|^2 df =$$ $$=\int^{0}_{-\infty} |X(f) \ast X(f)|^2 df + \int^{\infty}_{0} |X(f) \ast X(f)|^2 df = $$
$$= -\int^{\infty}_{0} |X(f) \ast X(f)|^2 df + \int^{\infty}_{0} |X(f) \ast X(f)|^2 df = 0$$

But I don't know how to relationate both integrals. Maybe that isn't the way to demonstrate the exercise.

Thanks for your help.
You have to use the following form of Parseval's identity:

$$\int_{-\infty}^{\infty}x(t)y^*(t)dt=\int_{-\infty}^{\infty}X(f)Y^*(f)df\tag{1}$$

where $^*$ denotes complex conjugation, and where $X(f)$ and $Y(f)$ are the Fourier transforms of $x(t)$ and $y(t)$, respectively. With $y(t)=\hat{x}(t)$ and $Y(f)=-j\,\textrm{sgn}(f)X(f)$, we get from $(1)$

$$\int_{-\infty}^{\infty}x(t)\hat{x}^*(t)dt=j\int_{-\infty}^{\infty}|X(f)|^2\textrm{sgn}(f)df\tag{2}$$

For real-valued $x(t)$, $|X(f)|^2$ is an even function, and, consequently, the integrand on the right-hand side of $(2)$ is odd. This implies that the integral on the right-hand side of $(2)$ equals zero if it is interpreted as a Cauchy principal value.

Note that we've used the fact that $x(t)$ is real-valued. For complex-valued $x(t)$, the integrals in $(2)$ are generally not zero.