Hilbert Transform of cos wt = sinwt. Hilbert Transform of cos wt = sin wt. Can anyone help me with the proof.
in Last Step 

how this become pi
 A: Consider the integral
$$\oint_C dz \frac{e^{i \omega z}}{z} $$
where $\omega \gt 0$ and  $C$ is a semicircle of radius $R$ in the upper half complex plane, with a semicircular detour of radius $\epsilon$ about the origin into the upper half plane.  The contour integral is then
$$\int_{-R}^{-\epsilon} dx \frac{e^{i \omega x}}{x} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{i \omega \epsilon e^{i \phi}}}{\epsilon e^{i \phi}}+\int_{\epsilon}^{R} dx \frac{e^{i \omega x}}{x}+i R \int_{0}^{\pi} d\theta \, e^{i \theta} \frac{e^{i \omega R e^{i \theta}}}{R e^{i \theta}}$$
As $R\to\infty$, the magnitude of the fourth integral is bounded by
$$\int_0^{\pi} d\theta \, e^{-\omega R \sin{\theta}} \le 2 \int_0^{\pi/2} d\theta\, e^{-2 \omega R \theta/\pi} \le \frac{\pi}{\omega R}$$
As $\epsilon \to 0$, the second integral approaches $-i \pi$.  Further, by Cauchy's thereom, the contour integral is $0$.  Therefore,
$$PV \int_{-\infty}^{\infty} dx \frac{e^{i \omega x}}{x} = i \pi$$
Taking real and imaginary parts, we get
$$PV \int_{-\infty}^{\infty} dx \frac{\cos{\omega x}}{x} = 0$$
$$\int_{-\infty}^{\infty} dx \frac{\sin{\omega x}}{x} = \pi$$
When $\omega \lt 0$, we need to use a contour in the lower half-plane.  The analysis is the same, but the result is that the sign of the second integral is flipped.  Therefore the general result is that
$$\int_{-\infty}^{\infty} dx \frac{\sin{\omega x}}{x} = \pi \operatorname*{sgn}{\omega}$$
A: There is a property of the Fourier transform you can use to solve the integral easily:
$$ F(x')=\int_\mathbb{R} f(x)e^{-i2\pi xx'}dx $$
Property: The integral of a function is equal to the Fourier transform of the function evaluated in zero
$$ F(0)=\int_\mathbb{R} f(x) dx $$
This way you can Fourier transform your $ \frac{sin(x)}{x} $ to see very easily that it correspond to a rectangle function with amplitude $A=\pi$. Then you evaluate
$$ rect(0)=\pi $$
Which is the solution to the integral for the sake of the property.
