Spivak, Ch. 23, "Infinite Series", Problem 17: Prove that $\int_0^\infty \left |\frac{\sin{x}}{x}\right |dx$ diverges. The following problem is from Ch. 23 "Infinite Series" of Spivak's Calculus



*Problem 19-43 shows that the improper integral $\int_0^\infty \frac{\sin{x}}{x} dx$ converges.

Prove that $\int_0^\infty \left |\frac{\sin{x}}{x}\right |dx$
diverges.

In a separate question, I ask about the solution manual solution.
The current question regards if the following solution is correct, and how to make it more rigorous if it is correct.
Let $f(x)=\left |\frac{\sin{x}}{x}\right |$. I first plotted the function and considered triangular areas as shown below.

The areas of the triangles form a sequence
$$\frac{\pi}{2},\frac{1}{5},\frac{1}{7},\frac{1}{9},\frac{1}{11},\frac{1}{13},...=\{b_n\}$$
It can be shown that the sequence $\left \{ \frac{1}{3+2n} \right\}=\frac{1}{5},\frac{1}{7},\frac{1}{9},...$ diverges using the ratio test with $\{\frac{1}{n}\}$
$$\lim\limits_{n\to\infty} \frac{\frac{1}{n}}{\frac{1}{3+2n}}=2$$
Thus, $\{b_n\}$ diverges.
Claim: $\int_{k\pi}^{(k+1)\pi} f>b_k, k=0,1,...$
If this claim is true, then
$$\int\limits_0^\infty f = \sum\limits_{k=0}^\infty \int\limits_{k\pi}^{(k+1)\pi} f>\sum\limits_{n=0}^\infty b_n=\infty$$
Is this proof correct?
How can we prove the claim made above?
 A: I'm not looking at the book so I don't know what properties are available to you, but one of the most basic properties of integrals is that if $f\geq g$ then
$$
\int_a^b f\geq\int_a^b g,
$$
which justifies your claim.
The inequality can be justified by noticing that the equation of the triangle on the interval $[k\pi, (k+1)\pi]$ is
$$
g(x)=\frac2{\pi(2k+1)}-\frac4{\pi^2(2k+1)}\Big|x-\big(k\pi+\frac\pi2\big)\Big|
$$
Let us assume $k$ even to simplify notation, but the computatios don't change much. On $[k\pi,k\pi+\frac\pi2]$,
\begin{align}
\frac{|\sin x|}x
&=\frac{\sin x}x
\geq\frac{\sin x}{k\pi+\frac\pi2}
=\frac2{\pi(2k+1)}\,\sin x.
\end{align}
So, writing $x=k\pi+t$, with $0\leq t\leq\frac\pi2$,
\begin{align}
\frac{\pi(2k+1)}{2}\,(f(x)-g(x))
&\geq\frac{\pi(2k+1)}{2}\,\frac{\sin x}x-1+\frac2\pi\,\Big(k\pi+\frac\pi2-x\Big)\\[0.3cm]
&\geq\sin x-1+\frac2\pi\,\Big(k\pi+\frac\pi2-x\Big)\\[0.3cm]
&=\sin t-1+\frac2\pi\,\Big(\frac\pi2-t\Big)\\[0.3cm]
&=\sin t-\frac2\pi\,t.
\end{align}
We want to check that $h(t)=\sin t-\frac2\pi\,t\geq0$ on $[0,\frac\pi2]$. This is a function with $h(0)=0$, and $h''(t)=-\sin t\leq 0$ on the whole interval $[0,\frac\pi2]$. This means that $h$ is concave, so for all $r\in[0,1]$
$$
0=r\,h(\frac\pi2)=h(0)+r\,\big[h(\frac\pi2)-h(0)\big]
\leq h(0+r (\frac\pi2-0))=h(r\,\frac\pi2).
$$
Thus $h(t)\geq0$ for all $t\in[0,\frac\pi2]$.

A slightly easier proof, where you don't have to worry about calculating the $b_k$, is to notice that $|\sin x|\geq\frac{\sqrt2}2$ if $k\pi+\frac\pi4\leq x\leq (k+1)\pi-\frac\pi4$. Then
\begin{align}
\int_0^\infty f&=\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}f
\geq \sum_{k=0}^\infty\int_{k\pi+\frac\pi4}^{(k+1)\pi-\frac\pi4}f\\[0.3cm]
&\geq\sum_{k=0}^\infty\int_{k\pi+\frac\pi4}^{(k+1)\pi-\frac\pi4}\frac{\sqrt2}{2(k+1)\pi}\\[0.3cm]
&=\frac{\sqrt2}4\sum_{k=0}^\infty\frac1{k+1}=\infty.
\end{align}
A: The simplest in my opinion is the following
$$\int\limits_{\pi}^{N\pi}{|\sin x|\over x}\,dx =
\sum_{n=2}^N\int\limits_{(n-1)\pi}^{n\pi}{|\sin x|\over x}\,dx\ge \sum_{n=2}^N{1\over n\pi}\int\limits_{(n-1)\pi}^{n\pi}|\sin x|\,dx={2\over \pi}\sum_{n=2}^N{1\over n}$$
Another proof can be made by applying the convergence of
$$\int\limits_{2\pi}^\infty {\cos x\over x}\,dx=\int\limits_{\pi}^\infty {\cos (2x)\over x}\,dx\quad (*)$$
We have
$$|\sin x|\ge \sin^2x={1\over 2}[1-\cos(2x)]$$
Then $$\int\limits_{2\pi}^N {|\sin x|\over x}\,dx \ge {1\over 2}\int\limits_{2\pi}^N \left [{1\over x}-{\cos(2x)\over x}\right ]\,dx
\\ = {1\over 2}\int\limits_{2\pi}^N {1\over x}\,dx-{1\over 2}\int\limits_{2\pi}^N {\cos(2x)\over x}\,dx$$ When $N\to +\infty$ the first term tends to $+\infty,$ while the second is convergent.
Remark We can avoid applying the convergence of $(*).$ If $(*)$ diverges then the similar integral of $\sin x/x$ diverges. Hence it cannot be absolutely convergent. That's a pretty artificial argument, as $(*)$ is convergent.
