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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.

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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)}. $

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    I'll try to explain it with a simpler example. The sequence $x^n$ converges uniformly to $0$ on $[0,a]$ for 0, but converges point wise on $[0,1]$ to the discontinuous function that equals $0$ on $[0,1)$ and to $1$ for $x=1$. As for the derivatives, I have edited the answer.2012-11-22