# How find this sum $S(x)=\sum_{k=1}^{\infty}\frac{\cos{(2kx\pi)}}{k}$

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.

$S(x)$ 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 $S(x)=?$

• maybe with $\ln{(\cos{x})}?$, – china math May 31 '14 at 5:55
• @chinamath, Please find the edited version – lab bhattacharjee May 31 '14 at 10:43
• @labbhattacharjee when you used the Taylor expansion for $-\ln(1-x)$, why is the $k$ still there? That is, shouldn't it be: $$\sum_{k=1}^{\infty}\frac{e^{2kx\pi i}}{k} =\sum_{k=1}^{\infty}\frac{(e^{2x\pi i})^k}{k} = -\ln(1-e^{2x\pi i})?$$ – chs21259 May 31 '14 at 21:50
• @chs21259, Thanks for your observation – lab bhattacharjee Jun 1 '14 at 5:24


\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}