1
$\begingroup$

I would appreciate some help with this problem:

Evaluate:

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

$\endgroup$
  • 2
    $\begingroup$ What have you tried towards solving this integral? $\endgroup$ – Parcly Taxel Nov 3 '18 at 16:31
  • 3
    $\begingroup$ HINT: Start by considering what happens when $x$ is an integer. $\endgroup$ – Ted Shifrin Nov 3 '18 at 16:33
  • 1
    $\begingroup$ Maybe not exactly when $x$ is an integer, but when it's of the form $2k\pi$ for some integer $k$ $\endgroup$ – Jakobian Nov 3 '18 at 16:43
  • $\begingroup$ 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! $\endgroup$ – Gentleman_Narwhal Nov 3 '18 at 16:54
  • 1
    $\begingroup$ Oh, yikes, of course I meant $x=n\pi$ for $n$ an integer. $\endgroup$ – Ted Shifrin Nov 3 '18 at 16:58
3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 :/

$\endgroup$
  • 1
    $\begingroup$ Almost. You want to use the 3 functions theorem. This doesn't actually prove that the limit is $\frac{2}{\pi}$ $\endgroup$ – Jakobian Nov 3 '18 at 16:53
  • $\begingroup$ @Jakobian Is that equivalent to the pinching/squeezing thm? $\endgroup$ – Gentleman_Narwhal Nov 3 '18 at 16:56
  • $\begingroup$ I meant squeeze theorem. Maybe I said it wrong, I'm not from an english-speaking country $\endgroup$ – Jakobian Nov 3 '18 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.