Prove $\int_{0}^{\infty}\frac{|\sin x|\sin x}{x}dx=1$

By Lobachevsky integral formula: https://en.wikipedia.org/wiki/Lobachevsky_integral_formula $$\int_{0}^{\infty}\frac{\sin x}{x}|\sin x|\,\mathrm{d}x=\int_0^{\pi/2}|\sin x|\,\mathrm{d}x=1.$$

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 $$ \int_{0}^{+\infty}\frac{\left|\sin x\right|\sin x}{x}\,dx = \frac{2}{\pi}\int_{0}^{+\infty}\frac{\sin x}{x}\,dx = 1.\tag{3}$$

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 $I=1$.