Hilbert transform of sinusoids

Example

Let $\omega > 0$, and define the Hilbert transform of a bounded continuous function $f$ by the principal value at $0$ and at $\infty$ $$Hf(x) = \frac{1}{\pi} \lim_{\varepsilon \to 0^+,\, R \to \infty} \int_{\varepsilon < \abs{u} < R} \frac{f(x - u)}{u}\, du.$$ Then $$H[\cos(\omega \cdot)](x) = \sin(\omega x), \qquad H[\sin(\omega \cdot)](x) = -\cos(\omega x).$$ Check for the cosine: $\cos(\omega(x - u)) = \cos(\omega x) \cos(\omega u) + \sin(\omega x) \sin(\omega u)$. On the symmetric set $\varepsilon < \abs{u} < R$, the odd function $\cos(\omega u)/u$ integrates to $0$. The even function $\sin(\omega u)/u$ integrates to $2 \int_{\varepsilon}^{R} \frac{\sin(\omega u)}{u}\, du$, which tends to $2 \cdot \frac{\pi}{2} = \pi$. So $H[\cos(\omega \cdot)](x) = \frac{1}{\pi} \cdot \pi \sin(\omega x) = \sin(\omega x)$. The same computation with $\sin(\omega(x - u)) = \sin(\omega x)\cos(\omega u) - \cos(\omega x)\sin(\omega u)$ gives $-\cos(\omega x)$. For $\omega < 0$ the signs change, since $\int_0^\infty \frac{\sin(\omega u)}{u}\, du = -\frac{\pi}{2}$.