Contradiction to a Theorem for Lebesgue integrability proof

Question

We know that $f$ is Lebesgue Integrable iff $|f|$ is Lebesgue Integrable. We have shown by contour integration that $\int^{\infty}_0 \frac{\sin(x)}{x}$ is $\frac{\pi}{2}$ - yet we can also show using a summation trick on the integral and considering limits $n\pi$ and $0$ and letting n tend to $\infty $ -which I'm sure many are aware of- that $\int^{\infty}_0 \frac{\left|\sin(x)\right|}{x}$ tends to $\infty$ we means by our iff that $\frac{\sin(x)}{x}$ is not Lebesgue Integrable on $(0, \infty)$. This confuses me greatly- we've assigned a value to it but it's not Lebesgue Integrable? How?