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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 $\alpha_{k,n}\in\mathbb{R}$.

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  • $\begingroup$ The answer is no without further assumptions on $f$. $\endgroup$
    – Iosif Pinelis
    Commented Jan 28, 2022 at 16:35
  • $\begingroup$ @IosifPinelis What further assumptions on $f$ would be required to make the answer yes? $\endgroup$
    – UNOwen
    Commented Jan 28, 2022 at 16:37
  • $\begingroup$ The boundedness of $f$ will certainly be enough. What can you say about your $f$? $\endgroup$
    – Iosif Pinelis
    Commented Jan 28, 2022 at 16:42
  • $\begingroup$ I can say that $f$ is ergodic, if that helps. I think boundedness would be fine in certain situations (i.e. specific values of $\kappa$ and $\beta$). The integro-differential equation $\kappa \ddot{f}+\dot{f}=\beta\int_{0}^t J_1(f_t-f_s)e^{s-t}\mathop{ds}$ describes $f$ for real constants (but I'm interested in analysing the integral for a more general $f$). $\endgroup$
    – UNOwen
    Commented Jan 28, 2022 at 16:50

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