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Prove that the formula $$g(x)=\sum_{n\ge 1} \frac{1}{n^2}e^{\int_0^x t\sin(t/n) dt}$$ defines a function $\mathbb R\to \mathbb R$. Show that it is continuously differentiable.

I guess for the first part I need to show that the series converges (not necessarily uniformly), and for the second part do I need to show that the $\sum_{n\ge 1} (\frac{1}{n^2}e^{\int_0^x t\sin(t/n) dt})_x'$ converge uniformly? If so, how do I differentiate that? For the convergence of the original series, the standard way is to use the M-test, which is not clear how to apply at all.

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  • $\begingroup$ You cannot expect that $\sum_{n=1}^{\infty}\frac{d}{dx}\left(\frac{1}{n^2}\exp\left\{\int_{0}^{x}t\sin(t/n)\,dt\right\}\right)$ converges uniformly on all of $\mathbb{R}$, but you can expect this to happen on every compact interval of $\mathbb{R}$. $\endgroup$ Jul 5, 2018 at 1:27

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If $\sum f_n$ converges uniformly to $f$ and $\sum f_n'$ converges uniformly to $g$ say on $[0,N]$then $f$ is differentiable and $f'=g$. Te see this denote the partial sums of $\sum f_n$ by $s_1,s_2,...$ so that $ s_n$ converges uniformly to $f$ and $s_n'$ converges uniformly to $g$ on $[0,N]$. Since $s_n (x)=s_n(0)+\int_0^{x} s_n'(t) \, dt$ we can pass to the limit to get $f(x)=f(0)+\int_0^{x} g(t) \, dt$. Since $g$ is continuous we get $f' =g$. Back to the original question: If you use the fact that $\sin x \leq x $ on $(0,\infty )$ we get $e^{t \sin (t/n)} \, dt \leq e^{t^{2}/n} \, dt$. From this it is fairly easy to prove that the original series and the differentiated series converge uniformly on $[0,N]$ for each positive integer $N$. (If you need more details I would be glad to provide them).

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