Can we prove that the solutions of $\int_0^y \sin(\sin(x)) dx =1$ are irrational?

Question

Can we prove that the solutions of $$\int_0^y \sin(\sin(x)) dx =1$$ are irrational? Wolfram Alpha gives two approximate sets of solutions as $\{4.58+2\pi k|k\in\mathbb{Z}\}$ and $\{1.69+2\pi k|k\in\mathbb{Z}\}$. Can we prove they are irrational?

Answer 1

Here is some partial progress towards the desired integral.

The Jacobi-Anger expansion provides a direct line of attack. The Fourier series of $\sin( z\sin (x))$ is $$\sin(z \sin(x)) = 2\sum_{k\text{ odd}}^\infty J_{k}(z)\sin kx$$ where $J_k(x)$ is a Bessel function of order $k$. Hence we may immediately integrate, obtaining

\begin{align} I_z(y)\equiv \int_0^y \sin(z \sin(x))\,dx &=2\sum_{k\text{ odd}}^\infty J_k(z) \int_0^y \sin kx \,dx\\ &=\sum_{k\text{ odd}}^\infty \frac{4}{k}J_k(z)\sin^2\left(\frac{1}{2}k y\right) \end{align} (As an aside, for small $y$ this gives $I_z(y)\approx Cy^2$ with $C=\sum_{k\text{ odd}}^\infty k\,J_k(z).$)

Hence the question has been converted to a summation problem; more specifically, this is a Neumann series expansion for $I_z(y)$. (The Jacobi-Anger expansion was this kind of series as well.) The question is then how one proceeds further, with the goal of finding roots in the special case $I(1;y)=1$. Does anyone see a path forward?