$$ \lim_{k\rightarrow\infty} \int_o^1 |\cos{(kx)}|\,dx $$
$$ \int_o^1 |cos(kx)|\,dx = \frac{1}{k}\int_o^k |\cos{y}|\,dy =\frac{f(k)}{k} $$ where I used $y=kx$ and $f(x)=\int_0^x |\cos{y}|\,dy$.
How can I proceed or what would be the easier approach?
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Your answer is not true. Since $$\int_0^1 |\cos{(kx)}|dx = \frac{1}{k} \int_0^k |\cos{y}|dy.$$ Note that $$\dfrac2k\lfloor \frac{k}{\pi} \rfloor=\frac{1}{k} \int_0^{\lfloor k/\pi\rfloor\pi} |\cos{y}|dy\le\frac{1}{k} \int_0^k |\cos{y}|dy\le \frac{1}{k} \int_0^{(\lfloor k/\pi\rfloor+1)\pi} |\cos{y}|dy=\frac{2}{k} \left(\lfloor \frac{k}{\pi} \rfloor+1\right)$$ This inequality follows from that we integrate via the first $(\lfloor k/\pi\rfloor+1)$ intervals ( i.e. $[0,(\lfloor k/\pi\rfloor+1)\pi]$ to get the right inequality since $k\le (\lfloor k/\pi\rfloor+1)\pi$. The left inequality follows the same way).Then by inequality we know that $$\lim_{k\to\infty} \int_0^1 |\cos{(kx)}|dx=\frac{2}{\pi}$$
Golbez
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2Fine, since $\lfloor k/\pi\rfloor$ is the integer $n$ such that $k/\pi-n\in[0,1)$. Then $k/\pi-n<1$, we know then $k-n\pi<\pi$, and $k<(n+1)\pi$ – Golbez Aug 23 '14 at 13:12
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Let's split the unit interval range into the sets where $\cos $ is positive or negative:
$$[0,1] = \{ x: cos(kx) > 0 \} \cup \{ x: cos(kx) < 0 \}$$
As $k \to \infty$, the positive and negative sets should be throughly mixed, and $\cos (kx)$ should converge to a function violently oscillating between 0 and 1, with average value 1/2
$$ \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} \cos x \; dx = \frac{2}{\pi} $$
This can be proven rigorously with the Birkhoff Ergodic Theorem.
To summarize we replace the violent function $\cos kx$ with its average value
$$ \int_0^1 |\cos{(kx)}|\,dx \approx \int_0^1 \frac{2}{\pi} \; dx= \frac{2}{\pi}$$
I thought the integral is $\frac{1}{2}$, then I found the correct average.
cactus314
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