I want to show that the following function is continuously differentiable:
$$g(x)=\sum_{n=1}^{\infty}\frac{1}{n^{2}}e^{\int_{0}^{x}t\sin(n/t)\,dt}.$$
I tried using the idea that series where the first n terms of the infinite series converge uniformly if it converges pointwise at a time and derivative series converges uniformly. So I tried to show derivative series converges by the Cauchy criterion but I can't seem to bound the tail.
Best Answer
Let $f_n(x)=\int_0^xt\sin(n/t)\,dt$. Since $t\sin(n/t)$ is even and continuous on $\mathbb{R}$ (assuming it takes the value $0$ at $t=0$), $f_n$ is odd and differentiable and $f_n'(x)=x\sin(n/x)$. Let $R>0$. Then $$ |f_n(x)|\le\int_0^{|x|}t\,dt\le\frac{R^2}{2}\quad\forall x\in[-R,R]. $$ By the Weierstrass convergence theorem (also known as the $M$-test) the series converges uniformly on $[-R,R]$ to a continuous function. This shows that $g$ is continuous on $\mathbb{R}$.
Moreover $$ |\Bigl(e^{f_n(x)}\Bigr)'|=|f_n'(x)|e^{f_n(x)}|\le Re^{R^2/2}\quad\forall x\in[-R,R]. $$ It follows that the series obtained by term by term differentiation is uniformly convergent on $[-R,R]$ to a continuous function, that $g$ is continuously differentiable and that $$ g'(x)=\sum_{n=1}^\infty\frac{x\sin(n/x)}{n^2}e^{f_n(x)}. $$