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?