A basic problem in transcendental number theory is to determine the arithmetic properties of values of special functions. Many special functions, such as Bessel functions and certain hypergeometric functions, are E-functions which are a natural generalization of the exponential function and satisfy certain linear differential equations. In this case, there exists an algorithm which determines if $f(\alpha )$ is transcendental or algebraic if $f(z)$ is an E-function and $\alpha \in {\bar{\mathbb{Q}}}^{⁎}$ is a non-zero algebraic number. In this paper, we consider the analogous question when $f(z)$ satisfies an integral equation, in particular, a Fredholm integral equation of the first or second kind where the kernel and forcing term satisfy strong arithmetic properties. We show that in both periodic and non-periodic cases, there exists no algorithm to determine if $f(0)\in \mathbb{Q}$ is rational. Our results are an application of the undecidability of the Generalized Collatz Problem due to Conway [6].

超越数论的一个基本问题是确定特殊函数值的算术性质。许多特殊函数，例如贝塞尔函数和某些超几何函数，都是E函数，它们是指数函数的自然推广，并且满足某些线性微分方程。在这种情况下，存在一种算法来确定是否$F(\alpha )$是超越的或代数的，如果$F(z)$是一个E函数并且$\alpha \in {\overline{\mathbb{问}}}^{⁎}$是一个非零代数数。在本文中，我们考虑了类似的问题，当$F(z)$满足积分方程，特别是第一或第二类 Fredholm 积分方程，其中核和强制项满足强算术性质. 我们表明，在周期性和非周期性情况下，不存在确定是否$F(0)\in \mathbb{问}$是理性的。我们的结果是对 Conway [6] 的广义 Collat​​z 问题的不可判定性的应用。