Evaluating the integral $\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$?

A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: $$\int\limits_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$ Well, can anyone prove this without using Residue theory? I actually thought of using the series representation of $\sin x$: $$\int\limits_0^\infty \frac{\sin x} x \, dx = \lim\limits_{n \to \infty} \int\limits_0^n \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm dt$$ but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.