integration - $int^infty_0frac{sin x}{x} , dx = frac{1}{2i}int^infty_{-infty} frac{e^{ix}-1}{x} , dx$, why?

How comes this true?

$$\int^\infty_0\frac{\sin x} x \, dx = \frac{1}{2i}\int^\infty_{-\infty} \frac{e^{ix}-1} x \, dx$$