Does $\int_{1}^{\infty}|\sin(x)/x|dx$ converge? $$\int_{1}^{\infty} \left| \frac{\sin(x)}{x}\right|dx$$ I'd like to know if this integral converges or not. I tried Wolframalpha but it didn't give me answer.
 A: Note that
$$\left.\frac{|\sin x|}{|x|}\right|_{[n, n+1]} \ge\frac1{n+1}|\sin(x)|\tag 1$$
And that
$$\int_{\alpha}^{\alpha +1}|\sin(x)| dx \ge \int_{-\frac12}^{\frac12} |\sin x| dx = 2\int_0^{\frac12} \sin x dx = 2(1-\cos(\frac12)) =: C > 0\tag 2$$
So we have
$$\int_1^\infty \left|\frac{\sin x}x\right| dx =\sum_{n=1}^\infty \int_n^{n+1}\left|\frac{\sin x}x\right| dx \stackrel{(1)}\ge \sum_{n=1}^\infty \frac1{n+1} \int_n^{n+1} |\sin x| dx \stackrel{(2)}\ge C \sum_{n=2}^\infty \frac1n$$
The latter is the harmonic series (minus the first term) and is well-known to diverge, or can be shown by comparison to $\ln$.
A: First, for $0 < x < \frac{\pi}{2}$, $\sin x \ge \dfrac{2x}{\pi}$.  Likewise for any interval $[k\pi, (k + \frac12)\pi]$ (with integer $k$), $|\sin{x}| \ge \dfrac{2(x-k\pi)}{\pi}$.
For each interval $[k\pi, (k + \frac12)\pi]$, $\dfrac{|\sin x|}{x} \ge \dfrac{2(x - k\pi)}{\pi x} \gt \dfrac{2(x-k\pi)}{(k+1)\pi} = f_k(x)$.
$f_k$ over the $k$th interval forms a triangle with base $\dfrac{\pi}2$ and height $\dfrac{2((k+\dfrac12)\pi - k\pi)}{(k+1)\pi} = \dfrac{1}{k+1}$, and area $\dfrac{\pi}{4(k+1)} = g_k$
The $f_k$ cover the left half of each "bump" in $\dfrac{|\sin x|}{x}$.  This isn't a problem because the integrand is non-negative for all of $[0,\infty]$.  $\lim_{n \rightarrow \infty} \sum_{k=0}^{n} g_k$ diverges as $O(\log n)$.  Since $\dfrac{|\sin x|}{x} \ge f_k(x)$ over every respective interval, the integral must diverge also.
