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 $\dfrac{\pi}{2}$.

Let's consider the integrals
$$I_1(t)=\int_t^{\infty}\frac{\sin(x-t)}{x}dx\qquad\mbox{ and }\qquad
I_2(t)=\int_0^{\infty}\frac{e^{-tx}}{1+x^2}dx,\qquad t\geq 0.$$
A direct calculation shows that $I_1(t)$ and $I_2(t)$ satisfy the ordinary differential equation
$$y''+y=\frac{1}{t},\qquad t>0.$$
Therefore, the difference $I(t)=I_1(t)-I_2(t)$ satisfy the homogeneous differential equation
$$y''+y=0,\qquad t>0,$$
hence it should be of the form 
$$I(t)=A\sin (t+B) $$
with some constants $A$, $B$. But  $I_1(t)$ and $I_2(t)$ both converge to $0$ as $t\to\infty$. This implies that $A=0$ and $I_1(t)=I_2(t)$ for all $t\geq 0$. Finally, we have that
$$\int_0^{\infty}\frac{\sin x}{x}dx=\int_{0}^{\infty}\frac{1}{1+x^2}dx=\lim_{n\to\infty}\left(\arctan(n)\right)-\arctan(0)=\frac{\pi}{2}.$$
