Interesting conjecture about integrals.

Question

My question is whether following conjecture is true:

If $f: \>{\mathbb R}_{\geq0}\to{\mathbb R}$ is continuous, and $\lim_{x\to\infty}\bigl|f(x)\bigr|=\infty$, then $$\lim_{x\to\infty}{1\over x}\int_0^x\bigl|\sin(f(t))\bigr|\>dt={2\over\pi}\ .$$

In other words it says that average value of function $|\sin(f(t))|$ from $0$ to infinity is equal to average value of $|\sin(t)|$ from $0$ to infinity (which is $\frac{2}{\pi}$), when $f(t)$ satisfies conditions shown above.

I don't know how to prove it. I've checked it numerically for few cases using desmos. There is problem with proving it for single cases, because the integral $\int_0^x|\sin(f(t))|dt$ is almost always very hard to calculate and requires using special functions like Fresnel S integral.

If you have any ideas, how to prove it, please let me know.

Thanks for all the help.

Answer 1

The conjecture is not true. Let $f$ be a continuous approximation to the step function $\pi\lfloor x\rfloor$, where the portions of the curve that connect consecutive horizontal steps get steeper and steeper as $x\to\infty$ in a way that the total length of the intervals where $f(x)$ is not an integer multiple of $\pi$ is finite. Then $\lim_{x\to\infty}{1\over x}\int_0^x|\sin(f(t))|\,dt=0$, not $2/\pi$.

Answer 2

No, this conjecture is not true. Consider a continuous function such that $f(x) = 2\pi n$ for $x\in [n+1/n^2, n+1]$. Then we get for $n\leq x \leq n+1$ $$ \frac{1}{x} \int_0^x \vert \sin(f(t)) \vert dt \leq \frac{1}{n} \sum_{k=1}^\infty \int_{k}^{k+1/k^2} \vert \sin(f(t)) \vert dt \leq \frac{1}{n} \sum_{k\geq 1} \frac{1}{k^2} \rightarrow 0 $$ for $n\rightarrow \infty$.