Evaluating $\lim_{x \to \infty}\frac{1}{x}\int_0^x|\sin(t)|dt$

Question

I would appreciate some help with this problem:

Evaluate:

$$\lim_{x \to \infty}\frac{1}{x}\displaystyle\int_0^x|\sin(t)|dt$$

Answer 1

Note that $\vert\sin(t)\vert$ is non-negative, periodic with period $\pi$, and that $$\int_0^\pi\vert \sin(t)\vert dt=2.$$ Let $f(x)$ be the largest integer smaller than or equal to $x/\pi$. Then it holds that $$\int_0^{f(x)\pi}\vert\sin(t)\vert dt\leq\int_0^x\vert\sin(t)\vert dt\leq\int_0^{[f(x)+1]\pi}\vert\sin(t)\vert dt.$$ This can be written as $$2f(x)\leq\int_0^x\vert\sin(t)\vert dt\leq2[f(x)+1].$$ Dividing by $x$ and noting that $\lim_{x\to\infty}f(x)/x=1/\pi$ it follows that $$\frac2\pi\leq\lim_{x\to+\infty}\frac1x\int_0^x\vert\sin(t)\vert dt\leq\frac2\pi.$$

Answer 2

Ok, thanks for your comments I think I get it now:

$\displaystyle\int_0^{2\pi}|\sin (t)|dt$ = 4

And due to the periodicity of $|\sin (t|)|$,

$\displaystyle\int_0^{2k\pi}|\sin (t)|dt = k\displaystyle\int_0^{2\pi}|\sin (t)|dt = 4k$

So the limit is equivalent to:

$$\lim_{k\to\infty}\left(\frac{1}{2k\pi}\cdot k\displaystyle\int_0^{2\pi}|\sin (t)|dt\right) = \lim_{k\to\infty}\frac{4k}{2k\pi} = \frac{2}{\pi}$$

... so in reality this question probably wasn't as hard as I first thought it was :/