Demonstrate that $x(t)$ and $\hat{x}(t)$ are orthogonal

Question

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

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

where:

$$\hat{x}(t) = x(t) \ast \frac{1}{\pi t} =\frac{1}{\pi} \int^{\infty}_{-\infty} \frac{x(s)}{t-s} ds$$

or, in the frequency domain:

$$\hat{x}(t) = \mathcal{F}^{-1} \lbrace -jX(f)sgn(f)\rbrace$$

with $\mathcal{F}^{-1}$ is the inverse Fourier transform and $sgn(f)$ is the sign of $f$

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.

Answer 1

$$<x, \hat{x}> = \int^{\infty}_{-\infty} x(t) \cdot \hat{x}(t) dt = \int_{-\infty}^{\infty} x(t) \int_{-\infty}^{\infty} \frac{x(s)}{t-s} ds dt = \\ \int_{-\infty}^{\infty} x(s) \int_{-\infty}^{\infty} -\frac{x(t)}{s-t} dt ds = - \int_{-\infty}^{\infty} x(s) \hat{x}(s) ds = - <x,\hat{x}> $$

So what must $<x,\hat{x}>$ be?