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?
$\begingroup$
$\endgroup$
11
-
$\begingroup$ Doesn't this follow immediately from the fact that $\pi$ is irrational? $\endgroup$ – user155385 Jul 2 '14 at 15:12
-
$\begingroup$ @dleggas Oh, how? $\endgroup$ – Sawarnik Jul 2 '14 at 15:13
-
$\begingroup$ A rational number times an irrational is always irrational. Then an irrational number plus a rational number is also irrational $\endgroup$ – user155385 Jul 2 '14 at 15:15
-
2$\begingroup$ @dleggas $4.58$ and $1.69$ are just approximations, I doubt that the actual values would be rational at all! You can try it on WA. $\endgroup$ – Sawarnik Jul 2 '14 at 15:17
-
1$\begingroup$ One could also ask the following question, which seems to me just as interesting: Suppose we consider a variety of 'target values' on the RHS. Are there cases with rational $y$? $\endgroup$ – Semiclassical Jul 22 '14 at 14:58
|
Show 6 more comments
$\begingroup$
$\endgroup$
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?
answered Jul 25 '14 at 3:34
