ZaidAlyafey said:
Evaluate the following
$$\sum_{k\geq 1}\frac{\cos \left(\frac{\pi k}{2} \right)}{k^2}$$
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:$$\text{Cl}_2(\theta)=-\int_0^{\theta}\log\Biggr|2\sin\frac{x}{2}\Biggr| \,dx=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^2}$$
$$\text{Sl}_2(\theta)=\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^2}$$[N.B. that integral definition holds without use of the absolute value sign within the range $$0 < \theta < 2\pi\,$$]
Now we convert the CL-integral into complex exponential form
$$\text{Cl}_2(\theta)=-\int_0^{\theta}\log\left(2\sin\frac{x}{2}\right)\,dx=-\int_0^{\theta}\log\left[2\left(\frac{e^{ix/2}-e^{-ix/2}}{2i}\right)\right]\,dx=$$
$$-\int_0^{\theta}\log\left(\frac{1-e^{-ix}}{ie^{-ix/2}}\right)\,dx=$$
$$\int_0^{\theta}\left(\log i-\frac{ix}{2}\right)\,dx-\int_0^{\theta}\log(1-e^{-x})\,dx=$$
$$\frac{\pi i\theta}{2}-\frac{i\theta^2}{4}-\int_0^{\theta}\log(1-e^{-ix})\,dx$$Now we expand the complex logarithm within the integrand as a power series:
$$\log(1-z)=-\sum_{k=0}^{\infty}\frac{z^k}{k}$$
$$\Rightarrow$$
$$\int_0^{\theta}\log(1-e^{-ix})\,dx=-\sum_{k=0}^{\infty}\frac{1}{k}\int_0^{\theta}e^{-ikx}\,dx=$$
$$-\sum_{k=0}^{\infty}\frac{1}{k}\int_0^{\theta}(\cos x-i \sin x)^{k}\,dx=-\sum_{k=0}^{\infty}\frac{1}{k}\int_0^{ \theta}(\cos kx-i \sin kx)\,dx=$$
$$-\sum_{k=0}^{\infty}\frac{1}{k}\Biggr[\frac{1}{k}\left(\sin kx+i \cos kx\right)\Biggr]_0^{\theta}=$$
$$-\sum_{k=0}^{\infty}\frac{1}{k}\Biggr[\frac{1}{k}\left(\sin k\theta+i \cos k\theta-i\right)\Biggr]=$$
$$-\sum_{k=0}^{\infty}\frac{\sin k\theta}{k^2}-i\sum_{k=0}^{\infty}\frac{\cos k\theta}{k^2}+i\sum_{k=0}^{\infty}\frac{1}{k^2}=$$
$$-\text{Cl}_2(\theta)-i\text{Sl}_2(\theta)+i\zeta(2)=-\text{Cl}_2(\theta)-i\text{Sl}_2(\theta)+\frac{i\pi^2}{6}$$Combining this with our earlier results, and equating the imaginary parts to zero, we deduce that
$$\text{Sl}_2(\theta)=\frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}$$
Or, equivalently...$$\sum_{k\ge 1}\frac{\cos k\theta}{k^2}= \frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}\, .\, \Box$$-------------------------------------------
For $$m\in \mathbb{N}^+\,$$, all series of the form$$\text{Sl}_{2m}(\theta)=\sum_{k=0}^{\infty}\frac{ \cos k\theta}{k^{2m}}$$
$$\text{Sl}_{2m+1}(\theta)=\sum_{k=0}^{\infty}\frac{\sin k\theta}{k^{2m}}$$Are expressible in closed form, as a polynomial in $$\theta\,$$ and $$\pi\,$$
