Showing $\frac{\sin x}{x}$ is NOT Lebesgue Integrable on $\mathbb{R}_{\ge 0}$ Let $$f(x) = \frac{\sin(x)}{x}$$ on $\mathbb{R}_{\ge 0}$.
$$\int |f| < + \infty\quad\text{iff}\quad \int f^+ < \infty \color{red}{\wedge \int f^- < \infty}$$
EDIT: So what I'm trying to do is to show that in fact $\int f^+ > \infty$ so that therefore $\int |f| = \infty$
Now consider the assertion that:
$$\int f+ = \sum_{k=1}^\infty \int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{x} \right) dx \le \sum_{k=1}^\infty\int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{2 \pi k + \pi}\right) dx$$
Two Questions:
(1) Is the first step of asserting that 
$$
\int f^+ = \sum_{k=1}^\infty \int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{x} \right) dx
$$
correct, in the sense that you can partition a Lebesgue integral into an INFINITE series of integrals being added together (whose individual term domains cover all of the overall domain of the original integral s.t. they are also pairwise disjoint)?
(2) Is there a well known lower bound of $\sum_{k=1}^\infty\int_{[2\pi k , 2\pi k + \pi]} \left( \frac{\sin(x)}{2 \pi k + \pi}\right) dx$ that diverges whose existence establishes that $f^+$ is in fact not integrable?

Let $f(x)=\frac{\sin x}{x}$. Surely the function is measurable, so we need to prove that $$\int|f(x)|dx=+\infty$$ 
In order to do that we need to look closer to the function. 
To this end, note that it is almost like $x\mapsto\frac{1}{x}$, however our function has zeros so we cannot compare them directly. 
A better approach is to use that the local maximums are attained at $x=\frac{\pi}{2}+\pi \cdot n$, in fact we have 
$$|f(x)| \geq \frac{1}{\sqrt{2}x}, \quad\text{for $ \left|\frac{(2n+1)\pi}{2}-x\right|\leq \frac{\pi}{4}$}$$
Now, do you see how to estimate the function from below?
