A Clausen function triplication formula

[DreamWeaver]

Define the second order Clausen function by:\(\displaystyle \text{Cl}_2(\varphi) = -\int_0^{\varphi} \log\Bigg| 2\sin \frac{x}{2} \Bigg|\, dx = \sum_{k=1}^{\infty}\frac{\sin k\varphi}{k^2}\)Prove the triplication formula:\(\displaystyle \text{Cl}_2(3\varphi) = 3\text{Cl}_2(\varphi) + 3\text{Cl}_2\left(\varphi+ \frac{2\pi}{3} \right) + 3\text{Cl}_2\left(\varphi+ \frac{4\pi}{3} \right)\)Hint:

Spoiler

Consider the triplication formula for the Sine:

\(\displaystyle \sin 3x = 3\sin x-4\sin^3 x\)

 

[Almsco]

Use the Fourier series.The triplication formula can be proven by starting with the Fourier series representation of the second order Clausen function:

\text{Cl}_2(\varphi) = \sum_{k=1}^{\infty}\frac{\sin k\varphi}{k^2}

We can then use the identity $\sin(3x) = 3\sin x - 4\sin^3 x$ to expand $\sin 3\varphi$ in terms of sines of multiples of $\varphi$. This leads to the following expression for $\text{Cl}_2(3\varphi)$:

\text{Cl}_2(3\varphi) = \sum_{k=1}^{\infty}\frac{3\sin k\varphi - 4\sin^3 k\varphi}{k^2}

By expanding the right-hand side of this equation into separate sums for each term, we can rearrange it to obtain the triplication formula:

\text{Cl}_2(3\varphi) = 3\text{Cl}_2(\varphi) + 3\text{Cl}_2\left(\varphi+ \frac{2\pi}{3} \right) + 3\text{Cl}_2\left(\varphi+ \frac{4\