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

I would appreciate some help with this problem:

Evaluate:

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

• What have you tried towards solving this integral? – Parcly Taxel Nov 3 '18 at 16:31
• HINT: Start by considering what happens when $x$ is an integer. – Ted Shifrin Nov 3 '18 at 16:33
• Maybe not exactly when $x$ is an integer, but when it's of the form $2k\pi$ for some integer $k$ – Jakobian Nov 3 '18 at 16:43
• Thanks for your hints, it's helped my understanding and I've realised it isn't as hard as I first thought, so I've been able to answer the question! – Gentleman_Narwhal Nov 3 '18 at 16:54
• Oh, yikes, of course I meant $x=n\pi$ for $n$ an integer. – Ted Shifrin Nov 3 '18 at 16:58

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 $$\frac2\pi\leq\lim_{x\to+\infty}\frac1x\int_0^x\vert\sin(t)\vert dt\leq\frac2\pi.$$

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 :/

• Almost. You want to use the 3 functions theorem. This doesn't actually prove that the limit is $\frac{2}{\pi}$ – Jakobian Nov 3 '18 at 16:53
• @Jakobian Is that equivalent to the pinching/squeezing thm? – Gentleman_Narwhal Nov 3 '18 at 16:56
• I meant squeeze theorem. Maybe I said it wrong, I'm not from an english-speaking country – Jakobian Nov 3 '18 at 16:59