Proving:$\frac{1}{2i}\int\limits_{-\infty}^{\infty}\frac{e^{ix}-1}{x}dx = \frac{\pi}{2}$ via Complex Methods?
Complex analysis: Calculating $\int_{-\infty}^{\infty} \frac{\sin x}{x} dx$ by using $f(z) = \frac{e^{iz} - 1}{iz}$