Consider the integral \begin{equation} \int_{0}^{t}J_1(f(t)-f(s))\mathop{ds}=\int_{0}^{t}\sum_{m=0}^{\infty} c_m (f(t)-f(s))^{2m+1}\mathop{ds}, \end{equation} such that $J_1$ is the Bessel function of first kind and order, $f(t)\in\mathbb{R}$, and \begin{equation} c_m=\frac{(-1)^m}{2^{2m+1}\Gamma(m+1)\Gamma(m+2)}. \end{equation} If $\sum_{m=0}^{\infty} c_m (f(t)-f(s))^{2m+1}$ exists for all $s$, and there is an integrable function \begin{equation} \sum_{m=0}^{\infty} \left|c_m (f(t)-f(s))^{2m+1}\right|\geq\left|\sum_{m=0}^{\infty} c_m (f(t)-f(s))^{2m+1}\right|, \end{equation} then the integration and summation symbols may be swapped in the first equation. So, is there a way to show that this sum converges to an integrable function? Any help would be much appreciated.

We also may assume that \begin{equation} f(t)^k = \sum_{n=0}^{\infty} \alpha_{k,n} t^n, \end{equation} and is differentiable/integrable for all $k\in\mathbb{N}$ (including $0$) such that $\alpha_{k,n}\in\mathbb{R}$.

yousay about your $f$? $\endgroup$