Jenny Branch

2022-01-29

Evaluating $\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}{\int }_{0}^{x}|\mathrm{sin}\left(t\right)|dt$

Darrell Boone

Note that $|\mathrm{sin}\left(t\right)|$ is non-negative, periodic with period $\pi$, and that
${\int }_{0}^{\pi }|\mathrm{sin}\left(t\right)|dt=2$
Let f(x) be the largest integer smaller than or equal to $\frac{x}{\pi }$. Then it holds that
${\int }_{0}^{f\left(x\right)\pi }|\mathrm{sin}\left(t\right)|dt\le {\int }_{0}^{x}|\mathrm{sin}\left(t\right)|dt\le {\int }_{0}^{\left[f\left(x\right)+1\right]\pi }|\mathrm{sin}\left(t\right)|dt$
This can be written as
$2f\left(x\right)\le {\int }_{0}^{x}|\mathrm{sin}\left(t\right)|dt\le 2\left[f\left(x\right)+1\right]$
Dividing by x and noting that $\underset{x\to \mathrm{\infty }}{lim}\frac{f\left(x\right)}{x}=\frac{1}{\pi }$ it follows that
$\frac{2}{\pi }\le \underset{x\to +\mathrm{\infty }}{lim}\frac{1}{x}{\int }_{0}^{x}|\mathrm{sin}\left(t\right)|dt\le \frac{2}{\pi }$

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