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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?

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    $\begingroup$ It's only because of the constant of integration. The Hilbert transform of a constant is zero. If the DC term coming out of the integral is zero, then the order of processing does not matter. $\endgroup$ Commented Oct 2, 2021 at 19:42
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    $\begingroup$ And the same reasoning applies with diiferentiation. $\endgroup$ Commented Oct 2, 2021 at 19:44
  • $\begingroup$ I was under the impression that the Hilbert transform, being linear as is the Integral, should be permutative with other linear operations. Is this a fallacy? $\endgroup$
    – Keegs
    Commented Oct 4, 2021 at 21:26

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