A more general version of this question.

Consider the series

$$\lim_{N\to \infty}\frac{1}{N}\sum_1^N |\sin(f(k))|$$

We could apply Weyl equidistribution theorem in the case $f(k)=k$ and compute as $\int_0^\pi \frac{1}{\pi}\cdot \sin(x)dx$

I have been experimenting with $f(k) = k\pi/4$ which gives me $0.60355\cdots$.

The sum $\frac{1}{4}(\sin(0) + \sin(\pi/2) + \sin(\pi) + \sin(3\pi/4)) = \frac{1}{4}(\sqrt{2}+1) \approx 0.60355\cdots$ which is close to above.

This would correspond to the integral

$$\int_{-\infty}^{\infty}\frac{1}{4}(\delta(x-0)+\delta(x-\pi/4)+\delta(x-\pi/2)+\delta(x-3\pi/4))\cdot \sin(x)dx$$

Where $\delta$ is the Dirac delta distribution which when working on a function ( multiplied and integrated together with ) picks out value at $x=0$:

$$\delta : \int_{-\infty}^{\infty}f(x)\delta(x)dx = f(0)$$

Is this a coincidence for this particular choice or can we (if we are able to calculate density) estimate the series as a integral weighted with the density in this sense? What do we need to demand of our function $f$ for this to be valid?

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    $\begingroup$ With $f(k)=k\pi/4$, I think the limit is obvious since $\sin$ is periodic. $\endgroup$ – Simply Beautiful Art Mar 23 '17 at 13:28
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    $\begingroup$ Your (legitimate)question could/should be transformed into a question about limit of (discrete) measures. See for example (math.stackexchange.com/q/489658). $\endgroup$ – Jean Marie Mar 23 '17 at 13:28
  • $\begingroup$ @SimplyBeautifulArt Yes that is why I chose that particular function as I hoped it would be easy to realize what the answer should be in that particular case. $\endgroup$ – mathreadler Mar 23 '17 at 13:39

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