Compute \[ I_1 \da \operatorname{PV}\int_\RR {\sin(x) \over x}\dx = \pi \\ I_2 \da \operatorname{PV}\int_\RR {\cos(x) \over x}\dx = 0 .\]
Solution
Take \(\Gamma\) an upper-half-plane semicircular contour indented at the origin; considering \(f(z) = e^{iz}\), by Jordan’s lemma \(\int_{C_R}f \to 0\) and the pieces along \(\RR\) converge to \(\PV \int f\). The singularity at \(z_0 = 0\) contributes a fractional residue: \[ \int_{C_\eps}f(z)\dz \to i\pi \Res_{z=z_0} f(z) = i\pi \lim_{z\to 0} {(z-0) e^{iz}\over z} = i\pi \cdot 1 .\] Thus \[ I \da \PV \int_\RR f(z)\dz = 0 + i\pi \implies I_1 = \Im(I) = \pi, \quad I_2 = \Re(I) = 0 .\]