# Does $\sum_{m=0}^{\infty} \left|c_m g(x)^{2m+1}\right|$ converge absolutely to an integrable function?

Consider the integral $$$$\int_{0}^{t}J_1(f(t)-f(s))\mathop{ds}=\int_{0}^{t}\sum_{m=0}^{\infty} c_m (f(t)-f(s))^{2m+1}\mathop{ds},$$$$ such that $$J_1$$ is the Bessel function of first kind and order, $$f(t)\in\mathbb{R}$$, and $$$$c_m=\frac{(-1)^m}{2^{2m+1}\Gamma(m+1)\Gamma(m+2)}.$$$$ If $$\sum_{m=0}^{\infty} c_m (f(t)-f(s))^{2m+1}$$ exists for all $$s$$, and there is an integrable function $$$$\sum_{m=0}^{\infty} \left|c_m (f(t)-f(s))^{2m+1}\right|\geq\left|\sum_{m=0}^{\infty} c_m (f(t)-f(s))^{2m+1}\right|,$$$$ 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 $$$$f(t)^k = \sum_{n=0}^{\infty} \alpha_{k,n} t^n,$$$$ and is differentiable/integrable for all $$k\in\mathbb{N}$$ (including $$0$$) such that $$\alpha_{k,n}\in\mathbb{R}$$.

• The answer is no without further assumptions on $f$. Jan 28, 2022 at 16:35
• @IosifPinelis What further assumptions on $f$ would be required to make the answer yes? Jan 28, 2022 at 16:37
• The boundedness of $f$ will certainly be enough. What can you say about your $f$? Jan 28, 2022 at 16:42
• 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 $f$). Jan 28, 2022 at 16:50