Evaluate the integral : $\int_0^\infty (-1)^{ \lfloor x \sin x \rfloor } dx$

How to evaluate the following integral? $$\int_0^\infty (-1)^{ \lfloor x \sin x \rfloor } dx$$

I have no idea how to calculate this improper integral. Maybe I have to use some property of floor functions to reduce $$(-1)^{ \lfloor x \sin x \rfloor }$$ to something simpler I can work with, but I don't know what property.
Thanks in advance for help!

Edit: I reached the following,
For every $$k \in \mathbb{Z_+}$$ we have that $$\int_0^{\pi \cdot (k+1)} (-1)^{ \lfloor x \sin x \rfloor } dx = \sum_{m=0}^k \int_{\pi \cdot m}^{ \pi \cdot (m+1)} (-1)^{ \lfloor x \sin x \rfloor } = \{ x = t + \pi \cdot m , dx = dt \} = \sum_{m=0}^k \int_{0}^{ \pi} (-1)^{ \lfloor ( t+\pi \cdot m)\cdot( \sin(t+\pi \cdot m) ) \rfloor } dt = \sum_{m=0}^k \int_{0}^{ \pi} (-1)^{ \lfloor ( t+\pi \cdot m)\cdot( \sin(\pi \cdot m )\cos(t) + \sin(t)\cos(\pi \cdot m) ) \rfloor } dt = \sum_{m=0}^k \int_{0}^{ \pi} (-1)^{ \lfloor ( t+\pi \cdot m)\cdot (-1)^m \cdot \sin(t) \rfloor } dt$$ .
If I can get a formula for the last integral from $$0$$ to $$\pi$$ I'll have an answer whether the integral from $$0$$ to $$\infty$$ converges or not.

• Decompose $(0,\infty)$ into sets $E_k = \{x > 0 : k \le x \sin x < k+1\}$ with $k \in \mathbb Z$. Nov 5, 2021 at 15:46
• Will that imply that ${ \int_{ x \in E_k,k\in \mathbb{Z_{odd}} } (-1)^k + \int_{ x \in E_k,k\in \mathbb{Z_{even}} } (-1)^k } \xrightarrow[k \to \infty]{} 0?$ Nov 5, 2021 at 16:04
• It is not obvious, so (if true) it would require work. Nov 5, 2021 at 16:49
• The integral fails the alternating series test if you break the interval up into $\pi$ long intervals and integrate to get coefficients. It diverges. Nov 5, 2021 at 16:58
• You can also split up $$(-1)^{\lfloor x\sin(x)\rfloor}=\cos(\pi\lfloor x\sin)x)\rfloor)+ i\sin(\pi\lfloor x\sin)x)\rfloor)$$ which kind of resembles this function, but there are better methods. Nov 7, 2021 at 19:12

(Too long for a comment)

1. Using the Fourier series

$$(-1)^{\lfloor x \rfloor} = \frac{4}{\pi} \sum_{k=1,3,5,\ldots} \frac{\sin(k\pi x)}{k} \quad\text{for } x \in \mathbb{R}\setminus\mathbb{Z}$$

that converges uniformly over any compact subset of $$\mathbb{R}\setminus\mathbb{Z}$$,

it follows that

\begin{align*} \int_{0}^{R} (-1)^{\lfloor x\sin x\rfloor} \, \mathrm{d}x &= \frac{4}{\pi} \sum_{k=1,3,5,\ldots} \frac{1}{k} \int_{0}^{R} \sin(k \pi x\sin x) \, \mathrm{d}x \\ &= \frac{4}{\pi} \sum_{k=1,3,5,\ldots} \frac{1}{k^2} \int_{0}^{kR} \sin(\pi x\sin (x/k)) \, \mathrm{d}x \end{align*}

I suspect that the following holds:

$$\sum_{k=1,3,5,\ldots} \frac{1}{k^2} \sup_{R \geq 0} \left| \int_{0}^{kR} \sin(\pi x\sin (x/k)) \, \mathrm{d}x \right| < \infty$$

If this is the case, then the above formula shows that $$\int_{0}^{\infty} (-1)^{\lfloor x \sin x \rfloor} \, \mathrm{d}x$$ converges as an improper Riemann integral and admits the series representation

$$\int_{0}^{R} (-1)^{\lfloor x\sin x\rfloor} \, \mathrm{d}x = \frac{4}{\pi} \sum_{k=1,3,5,\ldots} \frac{1}{k^2} \int_{0}^{\infty} \sin(\pi x\sin (x/k)) \, \mathrm{d}x.$$

But of course, proving (or disproving) the above claim would be quite hard.

2. Here is a numerical simulation of the map $$r \mapsto \int_{0}^{r} (-1)^{\lfloor x \sin x \rfloor} \, \mathrm{d}x$$ for $$0 \leq r \leq 100\pi$$.

The graph is generated by finding all the points of discontinuity of $$x \mapsto \lfloor x \sin x \rfloor$$ in the range $$0 \leq x \leq 100\pi$$. So I believe it is much more precise than simply throwing the function $$(-1)^{\lfloor x \sin x \rfloor}$$ to a numerical integrator.

• Using a stationary phase argument it is probably true that $$\int_0^{ + \infty } {\sin (\pi x\sin (x/k))dx} = \Im \int_0^{ + \infty } {e^{\pi ix\sin (x/k)} dx} = k\Im \int_0^{ + \infty } {e^{\pi ikt\sin t} dt} = \mathcal{O}(\sqrt k ).$$
– Gary
Nov 6, 2021 at 2:42

Comment
Some pictures for the integral from $$0$$ to $$2\pi$$ .

$$x\sin x$$

$$(-1)^{\lfloor x\sin x\rfloor}$$

$$\int_0^{2\pi}(-1)^{\lfloor x\sin x\rfloor}\;dx \approx -1.210856$$.