# 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 $x(t)$ [closed]

$\hat{x}(t)$ is the Hilbert Transform of $x(t)$. Can anybody help me proving the above.

## closed as off-topic by Tendero, MBaz, lennon310, A_A, Stanley PawlukiewiczSep 15 '18 at 19:25

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• Yes, with pleasure, if you provide us with the point at where you are stuck or have doubts – Laurent Duval Sep 13 '18 at 12:47
• I'm voting to close this question as off-topic because the OP hasn't showed any effort to solve the problem by himself. – Tendero Sep 13 '18 at 13:13

The Hilbert transform of $x(t)$ denoted by $\hat{x}(t)$ has the following Fourier transform $$\mathcal{F}(\hat{x}) = H(w)X(w) = -j \operatorname{sgn}(w)X(w)$$ because $$\hat{x}(t) = \int\limits_{-\infty}^{\infty} \frac{x(u)}{\pi(t-u)} \ du$$ Notice that $$\vert \mathcal{F}(\hat{x}) \vert^2 = \vert -j \operatorname{sgn}(w)X(w) \vert^2 = \vert X(w) \vert^2$$ i.e. $$\int\limits_{-\infty}^{\infty} \vert \mathcal{F}(\hat{x}) \vert^2 \ dw = \int\limits_{-\infty}^{\infty} \vert X(w) \vert^2 \ dw$$ Use Parsevals equation which gives you $$\int\limits_{-\infty}^{\infty} \vert \hat{x}(t) \vert^2 \ dt = \int\limits_{-\infty}^{\infty} \vert x(t) \vert^2 \ dt$$ P.S: For real signals, you could indeed remove the magnitudes (absolute values).
• Thanks, I am not sure whether $x(t)$ is real or not. – RAKESH GANDHI Sep 14 '18 at 5:51