I want to compute $\lim_{s \to \infty} \int_0^1 \left|\cos(sx)\right| \, dx$ where $s \in \mathbb{R}$, given that it exists.

I tried the obvious substitution $u = sx$ to get $\lim_{s \to \infty} \frac{1}{s} \int_0^s \left|\cos(u)\right| \, du$. But I haven't been able to proceed. In particular, I can't find a good way to break up the integral in order to use that $\cos(x)$ is $2\pi$-periodic and/or that it's bounded by $1$.

Any help is appreciated.

EDIT: I forgot to add the absolute value in my original post.

  • 7
    $\begingroup$ $$ \lim_{s\,\to\,\infty} \frac 1 s \int_0^s \cos u\,du = \lim_{s\,\to\,\infty} \frac 1 s \Big[ \sin u\Big]_{u\,=\,0}^{u\,=\,s} = \lim_{s\,\to\,\infty} \frac{\sin s} s. $$ $${}$$ $$ \frac{-1} s \le \frac{\sin s} s \le \frac 1 s. $$ $${}$$ $$ \text{And } \lim\limits_{s\,\to\,\infty} \dfrac 1 s = 0. $$ $\endgroup$ Aug 18 '18 at 19:17
  • $\begingroup$ @MichaelHardy Thank you, but I mistyped my question. $\endgroup$ Aug 18 '18 at 19:33
  • $\begingroup$ I would expect this limit to be $\sqrt{2}/2$. My motivation comes from the approximation $\int_0^t \sin^2(x) \,\mathrm{d} x \approx t/2$. Thus $\sin^2$ behaves approximately as the constant $1/2$ and thus $|\sin(x)|$ behaves approximately as $\sqrt{1/2}$ which gives the result I gave. This is of course not a complete proof and I am not sure if the approximation implications are valid. But it could provide as a start for you to work out! Btw, I expect that it doesn't matter if you use $\sin$ or $\cos$. $\endgroup$ Aug 18 '18 at 19:44

$\int_0^1 \left|\cos(2\pi x)\right| \, dx=\frac{2}{\pi}$. This is the integral over 1 cycle. For large $s$, you are integrating over $n$ cycles, each of width $1/n$, so the integral (ignoring a fraction of a cycle) remains (approximately) the same, i.e. the limit $=\frac{2}{\pi}$.

  • $\begingroup$ Shouldn't we divide by $2\pi$ because of the $1/s$ factor in the OP's argument from substitution? This gives $2/\pi$ which, unlike $4$, could actually be right; the integral is clear $\le(1-0)\max|\cos x|=1$. $\endgroup$
    – J.G.
    Aug 18 '18 at 20:05
  • $\begingroup$ @J G You are right (I simply forgot) - I corrected it. $\endgroup$ Aug 18 '18 at 20:25
  • $\begingroup$ I like these cualitative arguments. $\endgroup$ Aug 19 '18 at 1:35

You've practically solved it yourself: $$\int_0^1 \left| \cos \pi s x \right| dx = \frac 1 s \int_0^s \left| \cos \pi u \right| du = \frac 1 s \left( \int_0^{\lfloor s \rfloor} + \int_{\lfloor s \rfloor}^s \right) \left| \cos \pi u \right| du = \\ \frac {2 \lfloor s \rfloor} {\pi s} + O \left( \frac 1 s \right) = \frac 2 \pi + O \left( \frac 1 s \right).$$


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