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Find this sum $$S(x)=\sum_{k=1}^{\infty}\dfrac{\cos{(2kx\pi)}}{k},x\in R$$

my idea: since

$$S'(x)=2x\pi\cdot\sum_{k=1}^{\infty}\sin{(2kx\pi)}$$ then I can't.

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$S(x)$ is real part of $\displaystyle\sum_{k=1}^{\infty}\frac{e^{2x\pi i}}k$

But,$\displaystyle\sum_{k=1}^{\infty}\frac{e^{2kx\pi i}}k=-\ln(1-e^{2x\pi i})$

Now $\displaystyle1-e^{2x\pi i}=-e^{x\pi i}(e^{x\pi i}-e^{-x\pi i})=-e^{x\pi i}(2i\sin x\pi)$

For $\displaystyle \sin x\pi>0, \ln(1-e^{2x\pi i})=\ln(2\sin x\pi)+x\pi i+\ln(-i)$

Again, $\displaystyle -i=e^{-\dfrac{i\pi}2}\implies \ln(-i)=\left(2n\pi-\dfrac{i\pi}2\right)i$ for some integer $n$

But we are interested in the real part only

Similarly, if $\displaystyle \sin x\pi<0$

If $\displaystyle \sin x\pi=0,\cos2kx\pi=\cdots=1,$ then $S(x)=?$

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  • $\begingroup$ maybe with $\ln{(\cos{x})}?$, $\endgroup$ – china math May 31 '14 at 5:55
  • $\begingroup$ @chinamath, Please find the edited version $\endgroup$ – lab bhattacharjee May 31 '14 at 10:43
  • $\begingroup$ @labbhattacharjee when you used the Taylor expansion for $-\ln(1-x)$, why is the $k$ still there? That is, shouldn't it be: $$\sum_{k=1}^{\infty}\frac{e^{2kx\pi i}}{k} =\sum_{k=1}^{\infty}\frac{(e^{2x\pi i})^k}{k} = -\ln(1-e^{2x\pi i})?$$ $\endgroup$ – chs21259 May 31 '14 at 21:50
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    $\begingroup$ @chs21259, Thanks for your observation $\endgroup$ – lab bhattacharjee Jun 1 '14 at 5:24
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm S}\pars{x}\equiv\sum_{k = 1}^{\infty}{\cos\pars{2kx\pi} \over k}:\ {\large ?}.\qquad\qquad x\in {\mathbb R}}$

\begin{align} &\color{#66f}{\large{\rm S}\pars{x}} =\Re\sum_{k = 1}^{\infty}\expo{2kx\pi\,\ic} \int_{0}^{1}t^{k - 1}\,\dd t =\Re\int_{0}^{1}\sum_{k = 1}^{\infty}\pars{\expo{2x\pi\,\ic}t}^{k}\,{\dd t \over t} =\Re\int_{0}^{1}{\expo{2x\pi\,\ic}t \over 1 - \expo{2x\pi\,\ic}t}\,{\dd t \over t} \\[3mm]&=-\left.\Re\ln\pars{1 - \expo{2x\pi\,\ic}t} \right\vert_{\,t\ =\ 0}^{\,t\ =\ 1} =-\Re\ln\pars{1 - \expo{2x\pi\,\ic}} =-\Re\ln\pars{\expo{x\pi\,\ic}\bracks{\expo{-x\pi\,\ic} - \expo{x\pi\,\ic}}} \\[3mm]&=-\Re\ln\pars{-2\ic\expo{x\pi\,\ic}\sin\pars{\pi x}} =-\Re\ln\pars{2\bracks{\sin\pars{\pi x} - \ic\cos\pars{\pi x}}\sin\pars{\pi x}} \\[3mm]&=\color{#66f}{\large-\ln\pars{\root{2}\verts{\sin\pars{\pi x}}}} \end{align}

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