TITLE: Prove $\int_{0}^{\infty}\frac{|\sin x|\sin x}{x}dx=1$
QUESTION [8 upvotes]: Prove 
$$\int_{0}^{\infty}\frac{|\sin x|\sin x}{x}dx=1.$$
I know how to calculate $\int_{0}^{\infty}\frac{\sin x}{x}dx=\frac{\pi}{2}$, but the method cannot be applied here. So I am thinking 
$$\sum_{k=0}^n(-1)^k\int_{k\pi}^{(k+1)\pi}\frac{\sin^2 x}{x}dx$$
but I don't know how to proceed.

REPLY [6 votes]: We have
$$ \left|\sin x\right|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1} \tag{1} $$
$$\forall n\in\mathbb{N}^+,\quad \int_{0}^{+\infty}\frac{\sin(x)\cos(2nx)}{x}\,dx=0\tag{2} $$
hence
$$ \int_{0}^{+\infty}\frac{\left|\sin x\right|\sin x}{x}\,dx = \frac{2}{\pi}\int_{0}^{+\infty}\frac{\sin x}{x}\,dx = 1.\tag{3}$$