Consider the limit
$$\lim_{L\rightarrow\infty }f_L(t)=\lim_{L\rightarrow\infty }\frac {1}{L^2}\left|\sum_{k=0}^{L-1}e^{it\cos\left(\frac{2\pi}{L} k\right)}\right|^2 $$
I think I can get to this by computing the simpler limit
$$ g(t)=\lim_{L\to\infty}\frac{1}{L}\sum_{k=0}^{L-1}e^{it\cos\left(\frac{2\pi}{L} k\right)}$$
I suspect that $g(t)=0$ for all $t\neq 0$, but I have no idea how to prove it. For $L$ even it is easy to see
$$ g_L(t)=\frac{1}{L}\sum_{k=0}^{L-1}e^{it\cos\left(\frac{2\pi}{L} k\right)}=\frac 1L \left[\sum_{k=0}^{L/2-1}e^{it\cos\left(\frac{2\pi}{L} k\right)}+\sum_{k=L/2}^{L-1}e^{it\cos\left(\frac{2\pi}{L} k\right)}\right]=\frac 1L \left[\sum_{k=0}^{L/2-1}e^{it\cos\left(\frac{2\pi}{L} k\right)}+\sum_{k=0}^{L/2-1}e^{it\cos\left(\frac{2\pi}{L} k+L/2\right)}\right]=\\=\frac 1L \left[\sum_{k=0}^{L/2-1}e^{it\cos\left(\frac{2\pi}{L} k\right)}+\sum_{k=0}^{L/2-1}e^{-it\cos\left(\frac{2\pi}{L} k\right)}\right]=\frac{2}{L}\sum_{k=0}^{L/2-1}\cos\left({t\cos\left(\frac{2\pi}{L} k\right)}\right)$$
I think that by playing with floor functions I can get the same for $L$ odd, but I don't thnik that'll help me. How can I compute this?
Edit: I think the limit can be expressed as an integral by thinking of it as a Riemann sum
$$ g(t)=\int_0^1 \mathrm dk\,e^{it \cos(2\pi k)}$$
