TITLE: Showing $\frac{\sin x}{x}$ is NOT Lebesgue Integrable on $\mathbb{R}_{\ge 0}$
QUESTION [31 upvotes]: 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?

REPLY [12 votes]: By definition, a measurable function $f=f(x)$ is Lebesgue integrable over a measurable set $E$ if and only if the Lebesgue integrals$\int_E f^+(x)\,dx$ and $\int_Ef^-(x)\,dx$ are both finite. Let us look at $f^+(x)$, which in this case is $\sin^+(x)/x$. Since $f^+(x)\geqslant 0$, $f^+(x)$ is integrable if and only if $\int f^+(x)\,dx<\infty$. Again, since $f^+(x)\geqslant 0$, we can compute its Lebesgue integral with the Monotone Convergence Theorem, and then estimate it from below:
\begin{align*}
\int_0^\infty f^+(x)\,dx &= \int_0^\infty \frac{\sin^+(x)}{x}\,dx \\
&\color{red}{=}\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}{\color{green}{x}}\,dx & \color{red}{\text{Monotone Convergence Theorem}} \\
&\color{green}{\geqslant}\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}{\color{green}{(2k+1)\pi}}\,dx \\
&= \sum_{k=0}^\infty\frac{1}{(2k+1)\pi}\color{blue}{\int_{2k\pi}^{(2k+1)\pi}\sin(x)\,dx} \\
&\color{blue}{=} \sum_{k=0}^\infty\frac{\color{blue}{2}}{(2k+1)\pi}  \\
&= \sum_{k=0}^\infty\frac{1}{(k+1/2)\pi}
\end{align*}
By the Limit Comparison Test applied to the series obtained above and the harmonic series, we conclude that the series diverges, hence $\int_0^\infty f^+(x)\,dx = \infty$, so $f(x)$ is not integrable.