\int_{-\infty}^{\infty} e^{iax}\frac{1-e^{2ix}}{x^2}\,dx =\int_{-\infty}^{\infty} e^{iax}\frac{1-\cos 2x-i\sin 2x}{x^2}\,dx=2\int_{-\infty}^{\infty}e^{iax}\frac{\sin^2x}{x^2}-i\int_{-\infty}^{\infty}e^{iax}\frac{\sin 2x}{x^2}