Integrating $\frac {e^{iz}}{z}$ over a semicircle around $0$ of radius $\epsilon$ I am trying to find the value of $\int_{-\infty}^{\infty} \frac{\sin (x)}{x}$ using residue theorem and a contour with a kink around $0$. For this, I need to find $\int_{C_\epsilon} \frac {e^{iz}} {z}$ where $C_\epsilon$ is the semicircle centred at $0$ with radius $\epsilon$ from $-\epsilon$ to $\epsilon$. I guess it is equal to half the residue of $\frac {e^{iz}} {z}$ at $0$. Is this true? Any help is appreciated.

Another approach you can use is Jordan's lemma. They provided you the hint to use $\int_{C_\epsilon} \frac {e^{iz}} {z}$ because the integral $\int_{-\infty}^{\infty} \frac{\sin (x)}{x}$ is an improper integral and does not satisfy the conditions of p.v. We can use Jordan's lemma here. Thus we have $$\int_{C_\epsilon} \frac {e^{iz}} {z}$$ take the limit of the pole, which is at $z=0$. Apply Jordan lemma, and since we used the identity $e^{iz}$ to replace $\sin(z)$, you will take the imaginary parts and then you are through. 
