# Show that $\operatorname{Im}(G)$ is the Hilbert Transform of $\operatorname{Re}(G)$

Given that $$\operatorname{G}: \mathbb{R} \rightarrow \mathbb{C}$$ is such that $$\widehat{\operatorname{G}}(w) = 0$$ for $$w < 0$$, show that $$\operatorname{Im}(\operatorname{G})$$ is the Hilbert Transform of $$\operatorname{Re}(\operatorname{G})$$.

So far I have showed that if $$\widehat{\operatorname{G}} = \operatorname{h}(w) + i\operatorname{f}(w)$$, then:

$$\operatorname{Re}(\operatorname{G}) = \frac{1}{2\pi} \int_{0}^{\infty}(\operatorname{h}(w)\cos(wt) - \operatorname{f}(w)\sin(wt)) dw$$

$$\operatorname{Im}(\operatorname{G}) = \frac{1}{2\pi} \int_{0}^{\infty}(\operatorname{f}(w)\cos(wt) + \operatorname{h}(w)\sin(wt)) dw$$

Then I tried calculating the Hilbert Transform of $$\operatorname{Re}(\operatorname{G})$$, but did not get to $$\operatorname{Im}(\operatorname{G})$$. Anybody can give me advice on how to proceed with this problem?

Update:

I tried again and got the expected result.

$$\mathcal{F}[\operatorname{Re}(\operatorname{G})] = A + B$$

$$A = \frac{1}{2\pi} \int_{0}^{\infty} \operatorname{h}(x) \int_{-\infty}^{\infty} cos(xt)e^{-iwt}dt ~ dx$$

$$B = - \frac{1}{2\pi} \int_{0}^{\infty} \operatorname{f}(x) \int_{-\infty}^{\infty} sin(xt)e^{-iwt}dt ~ dx$$

Now I will show for A, but the same logic is applied B.

$$A = \frac{1}{2\pi} \int_{0}^{\infty} \operatorname{h}(x) \pi[\delta(w+x) + \delta(w-x)] dx$$

Now applying the Hilbert Transform

$$\mathcal{H}[A] = \mathcal{F}^{-1}[-isign(w)A]$$ $$\mathcal{H}[A] = \mathcal{F}^{-1} \left[\frac{1}{2\pi} \int_{0}^{\infty} \operatorname{h}(x) i\pi[\delta(w+x) - \delta(w-x)] dx \right]$$ $$\mathcal{H}[A] = \frac{1}{2\pi} \int_{0}^{\infty} \operatorname{h}(x) sin(xt) dx$$

Doing the same for $$B$$ we arrive at the desire result.