Showing $\frac{\sin x}{x}$ is NOT Lebesgue Integrable on $\mathbb{R}_{\ge 0}$ Let $$f(x) = \frac{\sin(x)}{x}$$ on $\mathbb{R}_{\ge 0}$.

$$\int |f| &lt; + \infty\quad\text{iff}\quad \int f^+ &lt; \infty \color{red}{\wedge \int f^- &lt; \infty}$$

EDIT: So what I'm trying to do is to show that in fact $\int f^+ &gt; \infty$ 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?
You might want to start by showing that $f$ is conditionally convergent. This would show that $f$ is not Lebesgue integrable. Use the fact that
$$\int_{(n - 1)\pi}^{n\pi} \frac{\left| \sin x \right|}{x} dx \geq \frac{1}{n\pi}\int_{(n - 1)\pi}^{n\pi} \left| \sin x \right| dx = \frac{2}{n\pi}$$
to get
$$\int_0^{N\pi} \frac{\left| \sin x \right|}{x} dx \geq \frac{2}{\pi}\sum_{n = 1}^N \frac{1}{n}$$
which shows that
$$\int_0^\infty \frac{\left| \sin x \right|}{x} dx$$
is divergent.
It is well known that the integral $ \int_{0}^{\infty}\left|\frac{\sin(x)}{x}\right|dx $ does not converge. Now use the theorem "a function $f$ is Lebesgue integrable iff |f| is" to show, that $f(x)=\frac{\sin(x)}{x}$ is not Lebesgue integrable..