To prove that $\int_{-\infty} ^\infty \mathcal{H}(g(t))(t)\text{d}t\neq\mathcal{H}(\int_{-\infty} ^\infty g(t) \text{d}t)$, where $\mathcal{H}(\cdot)$ is the Hilbert transform operator
My approach to solution:
$\int_{-\infty} ^\infty \mathcal{H}(g(t))(t)\text{d}t$
The inside integral can be written as: $-jG(f)$ when $f>0$ and $jG(f)$ when $f<0$ where $G(f)$ is the Fourier transform of $g(t)$
Similarly, integral can be represented as:
$\frac{-1}{j2\pi|f|}$ when $f<0$ and $\frac{1}{j2\pi|f|}$ when $f>0$
Multiplying the positing and negative sides of integral with the corresponding Hilbert transform sides, I got:
$\frac{-G(f)}{j2\pi|f|}$ when $f<0$ and $\frac{-G(f)}{j2\pi|f|}$ when $f>0$
Similarly, for:
$\mathcal{H}(\int_{-\infty} ^\infty g(t) \text{d}t)$
$\frac{-G(f)}{j2\pi|f|}$ when $f<0$ and $\frac{G(f)}{j2\pi|f|}$ when $f>0$
Furthermore, For Hilbert Transform we have:
$-j$ when $f>0$ and $j$ when $f<0$
Multiplying the positing and negative sides of integral with the corresponding Hilbert transform sides, I got:
$\frac{-G(f)}{j2\pi|f|}$ when $f<0$ and $\frac{-G(f)}{j2\pi|f|}$ when $f>0$
which is the same as the upper equation and hence no difference.
Ironically the same property holds for differentiation, which is provable by my approach. However, my solution does not work for integral so I am not sure if my approach is correct or not. Can somebody please help me out with this question?
