# Clausen sum

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Evaluate the following

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

#### M R

##### Active member
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
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
Evaluate the following

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

Hello Z! 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...

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

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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
Nice , generalization !

#### DreamWeaver

##### Well-known member
Thanks Z! 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
Thanks Z! 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!!!   Sorry folks... Wasn't really concentrating yesterday... I put in value of $$\displaystyle \cos \theta$$ rather than the argument, $$\displaystyle \theta$$...