Improper integral $\sin(x)/x $ converges absolutely, conditionaly or diverges?

Let $N \in \Bbb N, N > 1$, we have:

\begin{align} \int_0^{2\pi N} \left|\frac{\sin x}{x}\right|\,dx &= \sum_{n=0}^{N-1} \int_{2\pi n}^{2\pi(n+1)} \left|\frac{\sin x}{x}\right|\,dx \\ &\ge \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{2\pi n}^{2\pi(n+1)} \left|\sin x\right|\,dx \\ &= \sum_{n=0}^{N-1} \frac{1}{2\pi (n+1)} \int_{0}^{2\pi} \left|\sin x\right|\,dx \\ &= \sum_{n=0}^{N-1} \frac{2}{\pi (n+1)} \end{align}

The last sum diverges as $N \to \infty$, and so does the original integral.

Your integral is on $[1, \infty]$, but it also diverges because $\left|\frac{\sin{x}}{x}\right|$ is continuous on $[0, 1]$. My proof is on $[0, \infty]$ because it makes managing the summation slightly easier.