\int_{-\infty}^{\infty} e^{ix}\frac{1-e^{2ix}}{x^2}\,dx = . . . =
 \int_{-\infty}^{\infty}4e^{ix}\frac{\sin^2(x)}{x^2} - 2 i \int_{-\infty}^{\infty} \frac{\sin(x)}{x^2}= 2\pi