I need to calculate this $\int_{-\infty}^\infty \frac{sin(z)}{z}dz$ using the residue of $\frac{sin(z)}{z}$.
Then I write $\int_{-\infty}^\infty \frac{sin(z)}{z}dz$ = $\lim_{R\to\infty}Im(\int_{-R}^R \frac{e^{iz}}{z}dz+\int_{CR}\frac{e^{iz}}{z}dz)=2\pi i Res(\frac{e^{iz}}{z},0)$ where CR is the the half circunference of radius R>0 over the plane.
Then I need to show that $\lim_{R\to\infty}|\int_{CR}\frac{e^{iz}}{z}dz|=0$
Can someone help me?