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$$ \lim_{k\rightarrow\infty} \int_o^1 |\cos{(kx)}|\,dx $$

$$ \int_o^1 |cos(kx)|\,dx = \frac{1}{k}\int_o^k |\cos{y}|\,dy =\frac{f(k)}{k} $$ 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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Remember that $n=k/\pi$ is not necessarily an integer. –  JiK Aug 23 at 12:38
    
First note that if $y = kx$ then $dy =kdx$ so the second should have a $\frac{1}{k}$ in front. –  JHance Aug 23 at 12:45

2 Answers 2

up vote 3 down vote accepted

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 $$\lim_{k\to\infty} \int_0^1 |\cos{(kx)}|dx=\frac{2}{\pi}$$

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Can you explain the inequality $k\leq(⌊k/π⌋+1)\pi$ –  user150391 Aug 23 at 13:09
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Fine, 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 at 13:12
    
Very nice proof Golbez, congratulations. –  Juan Ospina Aug 23 at 13:17

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.

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