The integral of sin x / x

Near $0$. Since $\frac{\sin x}x\to1$, the integrand extends to a continuous function on $[0,1]$: no problem there.

1. At infinity. For $R>1$, integrate by parts with $u=\frac1x$, $v'=\sin x$:
$$\int_1^R\frac{\sin x}x\,dx=\left[-\frac{\cos x}x\right]_1^R-\int_1^R\frac{\cos x}{x^2}\,dx=\cos1-\frac{\cos R}R-\int_1^R\frac{\cos x}{x^2}\,dx.$$
As $R\to+\infty$, $\frac{\cos R}R\to0$, and $\int_1^\infty\frac{|\cos x|}{x^2}dx\le\int_1^\infty\frac{dx}{x^2}<\infty$, so $\int_1^R\frac{\cos x}{x^2}dx$ converges. Hence $\int_1^R\frac{\sin x}xdx$ has a limit: the integral converges.

2. Not absolutely. On $[k\pi,(k+1)\pi]$, $x\le(k+1)\pi$ and $\int_{k\pi}^{(k+1)\pi}|\sin x|\,dx=2$, so
$$\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}x\,dx\ge\frac2{(k+1)\pi}.$$
Summing over $k=0,\dots,N-1$ gives at least $\frac2\pi\sum_{k=1}^N\frac1k=\frac2\pi H_N\to+\infty$ (harmonic series). So the absolute integral diverges.

Remark. The integral converges only thanks to cancellations between positive and negative lobes. Its exact value is $\frac\pi2$ (the Dirichlet integral).