Infinite series of sin(nx)/n and cos(nx)/n

In today´s post I want to evaluate two basic infinite series that will be useful in the derivation of the Fourier series of the Log Gamma function

$\sum_{k=1}^{\infty}\frac{\sin(2 \pi k x)}{k} \qquad \tex{and}\qquad \sum_{k=1}^{\infty}\frac{\cos(2 \pi k x)}{k}$

Lets start showing the series expansion for $\log(1-x)$

Note

$\int_{0}^{x}\frac{du}{1-u}=-\log(1-x)$ (1)

on the other hand

$\int_{0}^{x}\frac{du}{1-u}=\int_{0}^{x}\sum_{k=0}^{\infty}u^kdu=\sum_{k=0}^{\infty}\int_{0}^{x}u^kdu=\sum_{k=0}^{\infty}\frac{x^{k+1}}{k+1}=\sum_{k=1}^{\infty}\frac{x^k}{k}$ (2)

Equating (1) and (2) we get

$\boxed{-\log(1-x)=\sum_{k=1}^{\infty}\frac{x^k}{k}}$ (3)

Now let´s try to evaluate the first sum:

$S=\sum_{k=1}^{\infty}\frac{\sin(2 \pi k x)}{k}$

By Euler´s formula

$\boxed{\sin(2 \pi k x)= \frac{e^{2 \pi i k x}-e^{-2 \pi i k x}}{2i}}$

$\sum_{k=1}^{\infty}\frac{\sin(2 \pi k x)}{k}= \sum_{k=1}^{\infty}\frac{e^{2 \pi i k x}-e^{-2 \pi i k x}}{2ik}$

$=\frac{1}{2i}\Big\{\sum_{k=1}^{\infty}\frac{e^{2 \pi i k x}}{k}- \sum_{k=1}^{\infty}\frac{e^{-2 \pi i k x}}{k}\Big\}$

From (3) above we can rewrite the last equation as

$=\frac{1}{2i}\Big\{-\log(1-e^{2 \pi i x})+\log(1-e^{-2 \pi i x})\Big\}$

$=\frac{1}{2i}\Big\{-\log(1-e^{2 \pi i x})+\log(1-\frac{1}{e^{2 \pi i x}})\Big\}$

$=\frac{1}{2i}\Big\{-\log(1-e^{2 \pi i x})+\log(\frac{e^{2 \pi i x}-1}{e^{2 \pi i x}})\Big\}$

$=\frac{1}{2i}\Big\{-\log(1-e^{2 \pi i x})+\log({e^{2 \pi i x}-1})-\log(e^{2 \pi i x})\Big\}$

$=\frac{1}{2i}\Big\{-\log(1-e^{2 \pi i x})+\log(-1(1-e^{2 \pi i x}))-2 \pi i x\Big\}$

$=\frac{1}{2i}\Big\{-\log(1-e^{2 \pi i x})+\log(1-e^{2 \pi i x})+\log(e^{i \pi})-2 \pi i x\Big\}$

$=\frac{1}{2i}\Big\{i \pi-2 \pi i x\Big\}$

And finally

$\boxed{\sum_{k=1}^{\infty}\frac{\sin(2 \pi k x)}{k} =\pi\Big\{\frac{1}{2}- x\Big\}}\qquad \qquad(4)$

Similarly, we can try to evaluate the second sum in the same fashion

$S=\sum_{k=1}^{\infty}\frac{\cos(2 \pi k x)}{k}$

By Euler´s formula

$\boxed{\cos(2 \pi k x)= \frac{e^{2 \pi i k x}+e^{-2 \pi i k x}}{2}}$

$\sum_{k=1}^{\infty}\frac{\cos(2 \pi k x)}{k} =\frac{1}{2}\Big\{\sum_{k=1}^{\infty}\frac{e^{2 \pi i k x}}{k}+ \sum_{k=1}^{\infty}\frac{e^{-2 \pi i k x}}{k}\Big\}$

$=\frac{1}{2}\Big\{-\log(1-e^{2 \pi i x})-\log(1-e^{-2 \pi i x})\Big\}$

$=\frac{1}{2}\Big\{-\log\Big[(1-e^{2 \pi i x})(1-e^{-2 \pi i x})\Big]\Big\}$

$= - \frac{1}{2}\log(1-e^{-2 \pi i x}-e^{2 \pi i x}+1)$

$= - \frac{1}{2}\log \Big(2-2(\frac{e^{-2 \pi i x}+e^{2 \pi i x}}{2})\Big)$

$= - \frac{1}{2}\log(2-2 \cos(2 \pi x))$

Now recall the double angle formula for $\cos(2A)$

$\boxed{\cos(2A)=1-2\sin^2(A)}$

$A=\pi x$ for us, therefore

$\sum_{k=1}^{\infty}\frac{\cos(2 \pi k x)}{k} = - \frac{1}{2}\log(2-2(1- 2\sin^2( \pi x)))$

$= - \frac{1}{2}\log(2^2\sin^2( \pi x)))$

And finally

$\boxed{\sum_{k=1}^{\infty}\frac{\cos(2 \pi k x)}{k} = -\log(2)-\log( \sin( \pi x))}\qquad \qquad(5)$

Evaluating $\zeta(2)$

As a corollary from (4) we can derive an amazing result, $\sum_{k=1}^{\infty}\frac{1}{k^2}=\zeta(2)$ . First, lets make the change of variable $2 \pi x \longmapsto x$ in (4)