How to prove using the Abel test that $\int_ {1}^\infty \frac{|\sin x|}{x}$ is divergent? 
How to prove using the Abel test that $\int_ {1}^\infty    \frac{|\sin x|}{x}$ is divergent?

 A: For $n>0$, we have
$$\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}{x}dx=$$
$$\int_0^{\pi}\frac{|\sin(t)|}{t+n\pi}\geq \frac{2}{(n+1)\pi}.$$
since $\int_0^{\pi}\sin(t)dt=2$.
Thus,
$$\int_{\pi}^{(n+1)\pi}\frac{|\sin(x)|}{x}dx$$
$$\geq \frac{2}{\pi}\sum_{k=1}^n\frac{1}{(k+1)}$$ 
which goes to $+\infty$ when $n$ tends to $+\infty$ as an harmonic series. The integral is therefore divergent.
A: \begin{align}
&\int_1^M \frac{|\sin x|}xdx>\int_1^M \frac{\sin^2 x}xdx=I(M)\\
\\
I(M)&=\int_{1+\pi/2}^{M+\pi/2} \frac{\sin^2 (x-\pi/2)}{x-\pi/2}dx=\int_{1+\pi/2}^{M+\pi/2} \frac{\cos^2 x}{x-\pi/2}dx\\
&>\int_{1+\pi/2}^{M+\pi/2} \frac{\cos^2 x}{x}dx\\
&=\int_{1}^M \frac{\cos^2 x}{x}dx-\int_{1}^{1+\pi/2} \frac{\cos^2 x}{x}dx+\int_M^{M+\pi/2} \frac{\cos^2 x}{x}dx\\
&>\int_{1}^M \frac{\cos^2 x}{x}dx-\int_{1}^{1+\pi/2} \frac{\cos^2x}{x}dx\\
&=\int_{1}^M \frac{\cos^2 x}{x}dx+K\\
\\
2I(M)&>\int_1^M \frac{\sin^2 x}xdx+\int_1^M \frac{\cos^2 x}xdx+K\\
&=\int_1^M \frac1xdx+K=\ln M+K\\
\end{align}
which means $\lim_{M\to\infty} I(M)$ diverges and so does the original integral.
