Is there any conditions for a function to be derivative function?

Question

For $0<x<2\pi$, prove the infinite sum $$S = \sum_{n=1}^\infty \frac{\cos(nx)}{n}$$ converges.

I want to prove it by the same idea given by the text(Arfken mathematical physics).

$$S = \int_{1}^{\infty} f(n)\mathrm{d}x + \int_{1}^{\infty}(n-[n])f'(n)\mathrm{d}x$$ Now, substitute $f(n)$ by the given series then, it is easy to show the first integral converges through some simple calculations. I'm wondering how to check the second integral. $$ S' = \int_{1}^{\infty}(n-[n])(-\frac{x}{n}\sin (nx) - \frac{cos(nx)}{n^2} )\mathrm{d}n$$ The text says "we note that its term $\cos(nx) / n^2$ also leads to a convergent integral". I don't understand why it works. It seems to ignore the $(n-[n])$ term which is in the integral.

Moreover, it says that "setting $(n-[n])\sin(nx)= g'(n)$, we write $$\int_{1}^{\infty}(n-[n]) \frac{\sin(nx)}{n} \mathrm{d}n = [\frac{g(n)}{n}]_{n=1}^{\infty}+ \int_{1}^{\infty}\frac{g(n)}{n^2}\mathrm{d}n$$ We do not have an explicit expression for $g(n)$, but we do know that it is bounded because $\sin x$ oscillates with a period incommensurate with that of the sawtooth periodicity of $(n-[n])$. " Is there any condition for a function to be derivative function? I guess that there are some more conditions for being derivative function such as Darboux's thm. It is odd to me to set $g'(n)$ without considering these derivative conditions. Even if the setting is logically safe, how we can verify the integral converges? I don't know how to deal with $\lim_{n \rightarrow \infty}\frac{g(n)}{n}$ and $\int_{1}^{\infty}\frac{g(n)}{n^2}\mathrm{d}n$.

Answer 1

$\sum_{n=1}^{N}\frac{\cos(nx)}{n}$ is the real part of $$ \sum_{n=1}^{N}\frac{e^{inx}}{n} = \sum_{n=0}^{N-1}\int_{0}^{e^{ix}}z^n\,dz = \int_{0}^{e^{ix}}\frac{1-z^N}{1-z}\,dz$$ and since $z^N$ uniformly converges to zero over any compact subset of $\{z:|z|<1\}$, it is enough to prove that the integral $$ \int_{0}^{e^{ix}}\frac{dz}{1-z} = \int_{0}^{-1}\frac{dz}{1-z}+\int_{\gamma}\frac{dz}{1-z} $$ is convergent, where $\gamma$ is the arc of unit circle joining $-1$ and $e^{ix}$, by avoiding the point $1$ where $\frac{1}{1-z}$ has a pole. This decomposition proves $$ \sum_{n\geq 1}\frac{\cos(nx)}{n} = -\text{Re}\log(1-e^{ix})=-\log\left\|1-e^{ix}\right\|=-\frac{1}{2}\log(2-2\cos x) $$ or

$$ \forall x\in(0,2\pi),\qquad \sum_{n\geq 1}\frac{\cos(nx)}{n} = \color{red}{-\log\left|2\sin\frac{x}{2}\right|}. $$

The convergence of the original series can be deduced from Dirichlet's test too, since for any $x\in(0,2\pi)$ the sequence $\{\cos(nx)\}_{n\geq 1}$ has bounded partial sums.