Computing $\lim_{s \to \infty} \int_0^1 \left|\cos(sx)\right| \, dx$

Question

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.

Answer 1

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

Answer 2

$\int_0^1 \left|\cos(2\pi x)\right| \, dx=\frac{2}{\pi}$. 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 $=\frac{2}{\pi}$.