Prove $\int_{0}^{\infty}\frac{|\sin x|\sin x}{x}dx=1$ 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.
 A: 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
$$ \int_{0}^{+\infty}\frac{\left|\sin x\right|\sin x}{x}\,dx = \frac{2}{\pi}\int_{0}^{+\infty}\frac{\sin x}{x}\,dx = 1.\tag{3}$$
A: Let
$$I:=\int_0^\infty\,\frac{\big|\sin(x)\big|\,\sin(x)}{x}\,\text{d}x\,.$$
Therefore,
$$2I=\int_{-\infty}^{+\infty}\,\frac{\big|\sin(x)\big|\,\sin(x)}{x}\,\text{d}x=\int_0^\pi\,\sin^2(x)\,\left(\sum_{k=-\infty}^{+\infty}\,\frac{(-1)^k}{x+k\pi}\right)\,\text{d}x\,.$$
It can be proven by residue calculus that
$$\text{csc}(z)=\sum_{k=-\infty}^{+\infty}\,\frac{(-1)^k}{z+k\pi}\text{ for all }z\in\mathbb{C}\setminus\pi\mathbb{Z}\,.$$
Thus,
$$2I=\int_0^\pi\,\sin^2(x)\,\text{csc}(x)\,\text{d}x=\int_0^\pi\,\sin(x)\,\text{d}x=2\,,$$
whence $I=1$.
A: By Lobachevsky integral formula: https://en.wikipedia.org/wiki/Lobachevsky_integral_formula
$$\int_{0}^{\infty}\frac{\sin x}{x}|\sin x|\,\mathrm{d}x=\int_0^{\pi/2}|\sin x|\,\mathrm{d}x=1.$$
