What is $\lim_{x\to\infty}\frac{\int_{0}^{x}\cos\{t-\cos t\}dt}{x}$?

Question

I want to find a closed form for the average value of $\cos\{t-\cos t\}$ where $\{n\}$ denotes the fractional part of $n$. I do not have experience finding an average value over an infinite domain but I assume it would be something like this: $$\lim_{x\to\infty}\frac{\int_{0}^{x}\cos\{t-\cos t\}dt}{x}$$ I am not sure if the limit converges. Here is what I have from Desmos:

$$\begin{array}{|c|c|} \hline x & \frac{\int_{0}^{x}\cos\{t-\cos t\}dt}{x} \\\hline 10 & 0.829164368874\\\hline 10*10^3 & 0.834299716027\\\hline 10*10^6 & 0.859423358961\\\hline 10*10^9 & 0.840428861971\\\hline 10*10^{14} & 0.826897135363\\\hline 10*10^{15} & 0.958730802913\\\hline 10*10^{16} & 0.995401790748\\\hline 10*10^{26} & 0.999999999999\\\hline 10*10^{27} & 1\\\hline \end{array}$$

As you can see, the limit appears to oscillate which makes sense after looking at the graph. When it gets to 10$^{\text{15}}$, it suddenly jumps up and begins to approach 1. My intuition tells me this is probably a bug because I am using such high numbers. I tried to solve it by hand but failed to figure out how to convert it to a summation.

I have a few ideas to what it converges to (assuming it does and is not 1) but they are basically wild guesses. $$\cos(\cos(1))\approx 0.857553215846$$ $$\frac{\pi^2}{12}\approx 0.822467033424$$ All help is appreciated :)

Answer 1

The limit is $$\sin(1) \approx 0.8414709848.$$ As others have said, invalid approximations using floating point arithmetic are likely to blame for the values in the table with $x > 10^{16}$.

Let $f(t) = \cos(\{t - \cos t\})$ and let $A(x) = x^{-1}\int_0^x f(t)\, dt$. Note that $A(x)$ is the average value of $f$ on $[0,x]$. The derivative of $A(x)$ is $A'(x) = -x^{-2} \int_0^x f(t)\, dt + x^{-1} f(x)$, so $|A'(x)| \le 2/x$. By the mean value theorem, $|A(x) - A(2 \pi n)| \le 2/n$ for $n = \lfloor x / (2 \pi)\rfloor$, so it is enough to find $\lim_{n\to \infty} A(2 \pi n)$ (for $n$ integer). Now we use the fact that $f(t)$ is not $2\pi$-periodic, but still rather easy to study along steps by $2\pi$. We have $$\begin{aligned}A(2 \pi n) &= \frac1{2\pi n} \int_0^{2 \pi n} f(t)\, dt\\ &= \frac1{2\pi n} \sum_{k=0}^{n-1} \int_0^{2 \pi} f(t + 2\pi k)\, dt\\ &= \frac1{2\pi}\int_0^{2 \pi} \frac1n \left( \sum_{k=0}^{n-1} \cos(\{t - \cos t + 2\pi k\}) \right)\,dt.\end{aligned}$$ Since $2\pi$ is irrational, its mutliples modulo $1$ are asymptotically equidistributed, so for any fixed $t \in [0,2 \pi]$ the integrand $$\frac1n \sum_{k=0}^{n-1} \cos(\{t - \cos t + 2 \pi k\}) \longrightarrow \int_0^1 \cos(s)\,ds = \sin(1)$$ as $n \to \infty$. By the dominated convergence theorem it follows that $A(2 \pi n) \to \sin(1)$.

Answer 2

The integrand $2\pi$-periodic and odd with respect to reflection $ t\to \pi-t$, $$\cos \left( (2 n+1)\ \pi - t - \cos \left( (2 n+1) \pi -t\right )\right)=-\cos (t-\cos (t))$$ so the integrals over all subintervals $t\in (\mathbb Z \ \pi, (\mathbb Z+1) \pi$ are zero.

Don't know is that helps, but approximations by numeric integration should be done per period.

Its an expectation of a random variable with values in (-1,1), continuous with random jumps. Since the argument is in $(0,\pi/2) + (\pi/2, \cos 1 )$ the value positive.

A histogram of the numeric integral over consecutive intervals
$\left(2 \ n\ \pi ,(2n+2) \ \pi \right)$ is showing a zig-zag in a small interval around 5.5

    tb = Array[(NIntegrate[Cos[FractionalPart[t - Cos[t]]],
               {t, # 2 \[Pi], (# + 1) 2 \[Pi]}] &), 128] 

The sorted

Answer 3

I do not know what is the answer but, working with a required high precision, there are severe numerical problems as soon as $x>97$ because the integrand is highly oscillatory and discontinuous.

What does happen for larger values of $x$ ? that is the question.

Answer 4

$\lim_{x\to\infty}\frac{\int_0^x\cos{\{t-\cos{t}}\}dt}{x}$

Use L'Hospital's rule

$\lim_{x\to\infty}\cos{\{x-\cos{x}}\}$

Hmmm, idk man, seems like it doesn't converge