calculus - 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:

$$\displaystyle\int_0^\infty \frac{\sin x} x \,\mathrm dx = \frac \pi 2$$

Well, can anyone prove this without using Residue theory. I actually thought of doing this:
$$\int_0^\infty \frac{\sin x} x \, dx = \lim_{t \to \infty} \int_0^t \frac{1}{t} \left( t - \frac{t^3}{3!} + \frac{t^5}{5!} + \cdots \right) \,\mathrm dt$$
but I don't see how $\pi$ comes here, since we need the answer to be equal to $\dfrac{\pi}{2}$.

Answer

Here's another way of finishing off Derek's argument. He proves
$$\int_0^{\pi/2}\frac{\sin(2n+1)x}{\sin x}dx=\frac\pi2.$$
Let
$$I_n=\int_0^{\pi/2}\frac{\sin(2n+1)x}{x}dx=

\int_0^{(2n+1)\pi/2}\frac{\sin x}{x}dx.$$
Let
$$D_n=\frac\pi2-I_n=\int_0^{\pi/2}f(x)\sin(2n+1)x\ dx$$
where
$$f(x)=\frac1{\sin x}-\frac1x.$$
We need the fact that if we define $f(0)=0$ then $f$ has a continuous
derivative on the interval $[0,\pi/2]$. Integration by parts yields
$$D_n=\frac1{2n+1}\int_0^{\pi/2}f'(x)\cos(2n+1)x\ dx=O(1/n).$$
Hence $I_n\to\pi/2$ and we conclude that
$$\int_0^\infty\frac{\sin x}{x}dx=\lim_{n\to\infty}I_n=\frac\pi2.$$