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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?

Antonio Vargas
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Sawarnik
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  • Doesn't this follow immediately from the fact that $\pi$ is irrational? –  Jul 02 '14 at 15:12
  • @dleggas Oh, how? – Sawarnik Jul 02 '14 at 15:13
  • A rational number times an irrational is always irrational. Then an irrational number plus a rational number is also irrational –  Jul 02 '14 at 15:15
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    @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. – Sawarnik Jul 02 '14 at 15:17
  • I edited your post slightly because your upper limit of integration was the same as your integration variable. I hope this does not change what you meant in your post. – Cameron Williams Jul 02 '14 at 15:20
  • @CameronWilliams Yes, no problem, thanks :) – Sawarnik Jul 02 '14 at 15:22
  • A line of attack ? Consider the inverse function of the square root of the integral, and find its McLaurin development around $0$. (The square root is to avoid an infinite slope at the origin.) My guess is that all derivatives yield an integer value. By evaluating the series at $1$, you'll get (hopefully convergent) rational approxiamtions. –  Jul 02 '14 at 16:22
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    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 $y$? – Semiclassical Jul 22 '14 at 14:58
  • The two solutions in $[0,2\pi]$ sum to $2\pi$ due to symmetry, $f(\pi+t)=f(\pi-t)$. Moreover, the Taylor series of the inverse function of $f$ has the form $$f^{-1}(x)=\frac{1}{\sqrt{2}}\sum_{n=0}^{+\infty}q_n x^{(2n+1)/2}$$ where all the $q_n$s are rational numbers. This brings the approximation $\zeta_0\approx\frac{6043}{2520\sqrt{2}}$, for instance, but it looks really hard to find useful arithmetic informations for such coefficients. – Jack D'Aurizio Jul 25 '14 at 04:55
  • I would try to understand if the inverse function can be expressed as a generalized continued fraction with rational terms by checking if $f^{-1}(x)$ satisfies a good-looking second-order differential equation, just to mimic the continued fraction proof of the irrationality of $e$. I do not really believe this will happen, however. – Jack D'Aurizio Jul 25 '14 at 04:59
  • The first continued fraction approximation for $f^{-1}(x)$ is $\frac{\sqrt{2x}}{1-x/6}$ and the second one is $\frac{\sqrt{2x}}{1-\frac{x}{6-\frac{x}{10}}}$ that leads to the pretty good approximation $\zeta_0\approx\frac{59}{49}\sqrt{2}$. – Jack D'Aurizio Jul 25 '14 at 05:14

1 Answers1

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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?

Semiclassical
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