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Clausen sum

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Evaluate the following


\(\displaystyle \sum_{k\geq 1}\frac{\cos \left(\frac{\pi k}{2} \right)}{k^2}\)​
 

M R

Active member
Jun 22, 2013
51
It looks like part of the Fourier series for \(\displaystyle x(\pi-x)\) at \(\displaystyle x=\pi/4\)

So I'm getting \(\displaystyle \pi/4(\pi-\pi/4)=\pi^2/6-S\).

Leading to S=\(\displaystyle -\pi^2/48\)...?

Bit of a cheat really.
 

chisigma

Well-known member
Feb 13, 2012
1,704
Evaluate the following


\(\displaystyle \sum_{k\geq 1}\frac{\cos \left(\frac{\pi k}{2} \right)}{k^2}\)​
It is relatively easy...

$\displaystyle \sum_{k = 1}^{\infty} \frac{\cos (\frac{\pi k}{2})}{k^{2}} = - \frac {1}{4}\ (1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + ...) = - \frac{\pi^{2}}{48}$

Kind regards

$\chi$ $\sigma$
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Evaluate the following


\(\displaystyle \sum_{k\geq 1}\frac{\cos \left(\frac{\pi k}{2} \right)}{k^2}\)​

Hello Z! (Sun) Nice little problem there...

I hope no one minds the bump, but there's a general closed form for this series... Define the second order Clausen functions as follows:


\(\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}\log\Biggr|2\sin\frac{x}{2}\Biggr| \,dx=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^2}\)

\(\displaystyle \text{Sl}_2(\theta)=\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^2}\)


[N.B. that integral definition holds without use of the absolute value sign within the range \(\displaystyle 0 < \theta < 2\pi\,\)]

Now we convert the CL-integral into complex exponential form

\(\displaystyle \text{Cl}_2(\theta)=-\int_0^{\theta}\log\left(2\sin\frac{x}{2}\right)\,dx=-\int_0^{\theta}\log\left[2\left(\frac{e^{ix/2}-e^{-ix/2}}{2i}\right)\right]\,dx=\)

\(\displaystyle -\int_0^{\theta}\log\left(\frac{1-e^{-ix}}{ie^{-ix/2}}\right)\,dx=\)

\(\displaystyle \int_0^{\theta}\left(\log i-\frac{ix}{2}\right)\,dx-\int_0^{\theta}\log(1-e^{-x})\,dx=\)

\(\displaystyle \frac{\pi i\theta}{2}-\frac{i\theta^2}{4}-\int_0^{\theta}\log(1-e^{-ix})\,dx\)


Now we expand the complex logarithm within the integrand as a power series:

\(\displaystyle \log(1-z)=-\sum_{k=0}^{\infty}\frac{z^k}{k}\)

\(\displaystyle \Rightarrow\)

\(\displaystyle \int_0^{\theta}\log(1-e^{-ix})\,dx=-\sum_{k=0}^{\infty}\frac{1}{k}\int_0^{\theta}e^{-ikx}\,dx=\)

\(\displaystyle -\sum_{k=0}^{\infty}\frac{1}{k}\int_0^{\theta}(\cos x-i \sin x)^{k}\,dx=-\sum_{k=0}^{\infty}\frac{1}{k}\int_0^{ \theta}(\cos kx-i \sin kx)\,dx=\)

\(\displaystyle -\sum_{k=0}^{\infty}\frac{1}{k}\Biggr[\frac{1}{k}\left(\sin kx+i \cos kx\right)\Biggr]_0^{\theta}=\)

\(\displaystyle -\sum_{k=0}^{\infty}\frac{1}{k}\Biggr[\frac{1}{k}\left(\sin k\theta+i \cos k\theta-i\right)\Biggr]=\)

\(\displaystyle -\sum_{k=0}^{\infty}\frac{\sin k\theta}{k^2}-i\sum_{k=0}^{\infty}\frac{\cos k\theta}{k^2}+i\sum_{k=0}^{\infty}\frac{1}{k^2}=\)

\(\displaystyle -\text{Cl}_2(\theta)-i\text{Sl}_2(\theta)+i\zeta(2)=-\text{Cl}_2(\theta)-i\text{Sl}_2(\theta)+\frac{i\pi^2}{6}\)


Combining this with our earlier results, and equating the imaginary parts to zero, we deduce that

\(\displaystyle \text{Sl}_2(\theta)=\frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}\)



Or, equivalently...


\(\displaystyle \sum_{k\ge 1}\frac{\cos k\theta}{k^2}= \frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}\, .\, \Box\)


-------------------------------------------

For \(\displaystyle m\in \mathbb{N}^+\,\), all series of the form


\(\displaystyle \text{Sl}_{2m}(\theta)=\sum_{k=0}^{\infty}\frac{ \cos k\theta}{k^{2m}}\)

\(\displaystyle \text{Sl}_{2m+1}(\theta)=\sum_{k=0}^{\infty}\frac{\sin k\theta}{k^{2m}}\)


Are expressible in closed form, as a polynomial in \(\displaystyle \theta\,\) and \(\displaystyle \pi\,\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Nice , generalization !
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Thanks Z! :D


It can make tricky-looking results quite easy. Such as


\(\displaystyle \sum_{k=1}^{\infty}\frac{\cos (\pi k/10)}{k^2}=\frac{\pi^2}{6}-\frac{\pi}{4}\sqrt{\frac{5+\sqrt{5}}{2}}+\frac{5+ \sqrt{5}}{16}\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Thanks Z! :D


It can make tricky-looking results quite easy. Such as


\(\displaystyle \sum_{k=1}^{\infty}\frac{\cos (\pi k/10)}{k^2}=\frac{\pi^2}{6}-\frac{\pi}{4}\sqrt{\frac{5+\sqrt{5}}{2}}+\frac{5+ \sqrt{5}}{16}\)

Oops!!! :eek::eek::eek:

Sorry folks... Wasn't really concentrating yesterday... I put in value of \(\displaystyle \cos \theta\) rather than the argument, \(\displaystyle \theta\)...


[hangs head in shame]