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

By definition (Laplace Transform):
\begin{equation*}
            F(s)=L\left[\frac{\sin(t)}{t}\right]=\int_{0}^{\infty}\frac{\sin(t)}{t}e^{-st}dt=\arctan^{-1}\left(\frac{1}{s}\right)
\end{equation*}
Then, for $s=0$,
\begin{equation*}
            F(0)=\int_{0}^{\infty}\frac{\sin(t)}{t}dt = \lim_{s\to0}\arctan^{-1}\left(\frac{1}{s}\right)=\lim_{u\to\infty}\arctan^{-1}(u)=\frac{\pi}{2}
\end{equation*}
