On another post (here), I saw the following manipulation. Let $n$ be a positive integer,
$$ \int_0^{\pi/2}\frac{|\sin(2nx)|}{x} = \sum_{k=1}^n\int_0^\pi\frac{\sin(x)}{x+(k-1)\pi} $$
Is this correct? If so how does one arrive at this?
Edit: Numerical tests support it so I believe it. Just don't see how you get to it.
$$\begin{align} \int_0^{\pi/2}\frac{|\sin(2nx)|}{x}dx &= \int_0^{n\pi}\frac{|\sin(x)|}{x}dx \tag{i} \\ &= \sum_{k=1}^n \int_{(k-1)\pi}^{k\pi} \frac{|\sin (x)|}{x}dx \tag{ii} \\ &= \sum_{k=1}^n\int_{0}^{\pi} \frac{|\sin (x + (k-1)\pi)|}{x + (k-1)\pi}dx \tag{iii} \\ &= \sum_{k=1}^n\int_{0}^{\pi} \frac{\sin (x)}{x + (k-1)\pi}dx \tag{iv} \end{align}$$
(i): Substitution $x\to x/2n$
(ii): Using the rule $\int_a^b + \int_b^c = \int_a^c$
(iii): Substitution $x\to x+(k-1)\pi$ in the kth integral
(iv): $|\sin(x+(k-1)\pi)|=\sin(x)$ for $x\in (0,\pi)$ and $k\in\mathbb Z$
