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

Question

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.

Answer 1

(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.

Answer 2

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

$x\sin x$

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

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