Certain step in proof of conditional but not absolute convergence of $\int_1^{\infty}\frac{\sin(x)}{x}dx$.

Question

I'm struggling to understand why from

$|\int_{n\pi}^{(n+1)\pi}\frac{\sin(x)}{x}dx|\geq \frac{1}{(1+n)\pi}|\int_{n\pi}^{(n+1)\pi}\sin(x)dx|=\frac{2}{(1+n)\pi}$ (which I still understand perfectly)

we can say:

$\int_{\pi}^{N\pi}|\frac{\sin(x)}{x}|dx\geq \frac{2}{\pi}\sum_{n=1}^{N-1}\frac{1}{n+1}$

I also understand that from the last line we can conclude that that the absolute integral of the Integralsin (that's what it is called in my language) diverges, I'm just struggling with understanding why the absolute value bars just move into the integral in the next step. Any help is highly appreciated.

Answer 1

This is just the triangle inequality for integrals and summing the result of the previous line.

$$\text{Triangle Inequality:}\\|a|+|b|\ge|a+b|\implies\int_a^b|f(x)|~\mathrm dx\ge\left|\int_a^bf(x)~\mathrm dx\right|$$

\begin{align}\int_\pi^{N\pi}\left|\frac{\sin(x)}x\right|~\mathrm dx&=\sum_{n=1}^{N-1}\int_{n\pi}^{(n+1)\pi}\left|\frac{\sin(x)}x\right|~\mathrm dx\\&\ge\sum_{n=1}^{N-1}\left|\int_{n\pi}^{(n+1)\pi}\frac{\sin(x)}x~\mathrm dx\right|\tag{$\triangle$ ineq.}\\&\ge\sum_{n=1}^{N-1}\frac2{(n+1)\pi}\\&=\frac2\pi\sum_{n=1}^{N-1}\frac1{n+1}\end{align}

Answer 2

$$\left|\int_{n\pi}^{(n+1)\pi}\frac{\sin x}{x}\,dx\right|=\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}{x}\,dx \stackrel{\text{since }x\leq(n+1)\pi}{\geq} \frac{1}{(n+1)\pi}\int_{n\pi}^{(n+1)\pi}|\sin(x)|\,dx = \frac{2}{(n+1)\pi}$$ so $$ \int_{0}^{N\pi}\frac{|\sin(x)|}{x}\,dx \geq \frac{2}{\pi}\sum_{k=1}^{N}\frac{1}{k}=\frac{2}{\pi}H_N\geq \frac{2}{\pi}\log(N+1).$$ Notice that $\frac{2}{\pi}$ is exactly the mean value of $|\sin(x)|$. In general, Kronecker's lemma/integration by parts gives that if $\int_{0}^{+\infty}\frac{f(x)}{x}\,dx$ (as an improper Riemann integral) is convergent, then $f(x)$ has mean zero.