Here's a way using minimum machinery, but it requires the simple lemma that if $\displaystyle f\in\mathcal{C}[a,b]$ then $\displaystyle \displaystyle \lim_{n\to\infty}\int_a^b f(x)\sin((2n+1)x)\text{ }dx=0$ and the fact that $\displaystyle \displaystyle \int_0^{\infty}\frac{\sin(x)}{x}$ exists, although this follows immediately from Dirichlet's test for integrals.

Ok, so note that since $\displaystyle \displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx$ exists we have that

$\displaystyle \displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx=\lim_{\lambda\to\infty}\int_0^\lambda \frac{\sin(x)}{x}\text{ }dx=\lim_{\substack{n\to\infty\\ n\in\mathbb{N}}}\int_0^{\frac{\pi n}{2}}\frac{\sin(x)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\ sin(nx)}{x}\text{ }dx$

and thus clearly $\displaystyle \displaystyle \int_0^\infty \frac{\sin(x)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\ sin\left((2n+1)x\right)}{x}\text{ }dx$. We next claim that $\displaystyle \displaystyle \int_0^{\frac{\pi}{2}}\frac{\sin\left((2n+1)x\righ t)}{\sin(x)}\text{ }dx=\frac{\pi}{2}$ for every $\displaystyle n\in\mathbb{N}$. But, this follows immediately from the common trigonometric identity associated with the Dirichlet kernel. We then note that

$\displaystyle \displaystyle \begin{aligned}\lim_{n\to\infty}\left(\int_0^{\fra c{\pi}{2}}\frac{\sin\left((2n+1)x\right)}{x}\text{ }dx-\int_0^{\frac{\pi}{2}}\frac{\sin\left((2n+1)x\righ t)}{\sin(x)}\text{ }dx\right) &=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\si n(x)-x}{\sin(x)}\sin\left((2n+1)x\right)\text{ }dx\\ &=0\end{aligned}$

where we've made use of the lemma mentioned at the beginning of the post. Thus, it follows that

$\displaystyle \displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\ sin\left((2n+1)x\right)}{x}\text{ }dx=\lim_{n\to\infty}\int_0^{\frac{\pi}{2}}\frac{\ sin\left((2n+1)x\right)}{\sin(x)}\text{ }dx=\frac{\pi}{2}$