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Show that $$\int_{0}^{\pi}\left|\frac{\sin nx}{x}\right|~dx \geqslant \frac{2}{\pi}\left(1+\frac{1}{2}+ \dots +\frac{1}{n}\right)$$

I know $0 \leqslant \left|\sin nx\right| \leqslant 1 $. But with this I can't solve. Please help.

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This is a cute question (+1) – Chris's sis the artist Jan 2 '13 at 14:43

1 Answer 1

up vote 8 down vote accepted


1) Perform a variable change to $y= n x$ then split the integral up into parts $y \in S_j$ with $S_j=[(j-1)\pi,j\pi]$, $j=1,...,n$.

2) $1/x$ is monotonously decaying.

Spoiler below:

$$\int_0^\pi \left| \frac{\sin nx}{x} \right| dx=\int_0^{n\pi} \left| \frac{\sin y}{y} \right| dy = \sum_{j=1}^n \int_{(j-1)\pi}^{j\pi} \overbrace{\frac{|\sin y|}{y}}^{|\sin y|/j\pi} \ge \sum_{j=1}^n \frac{1}{j\pi} \overbrace{\int_{(j-1)\pi}^{j\pi} |\sin y| dy}^{2} =\frac{2}\pi \sum_{j=1}^n \frac1j$$

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it looks nice! (+1) – Chris's sis the artist Jan 2 '13 at 17:11

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