MHB A Clausen function triplication formula

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The second order Clausen function is defined as Cl₂(φ) = -∫₀^φ log|2sin(x/2)| dx, which can also be expressed as a series involving sine functions. The triplication formula Cl₂(3φ) = 3Cl₂(φ) + 3Cl₂(φ + 2π/3) + 3Cl₂(φ + 4π/3) can be proven using the Fourier series representation of Cl₂(φ). By applying the identity sin(3x) = 3sin(x) - 4sin³(x), the expression for Cl₂(3φ) is derived. This leads to a rearrangement that confirms the triplication formula holds true. The discussion emphasizes the mathematical derivation and properties of the Clausen function.
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Define the second order Clausen function by:$$\text{Cl}_2(\varphi) = -\int_0^{\varphi} \log\Bigg| 2\sin \frac{x}{2} \Bigg|\, dx = \sum_{k=1}^{\infty}\frac{\sin k\varphi}{k^2}$$Prove the triplication formula:$$\text{Cl}_2(3\varphi) = 3\text{Cl}_2(\varphi) + 3\text{Cl}_2\left(\varphi+ \frac{2\pi}{3} \right) + 3\text{Cl}_2\left(\varphi+ \frac{4\pi}{3} \right)$$Hint:
Consider the triplication formula for the Sine:

$$\sin 3x = 3\sin x-4\sin^3 x$$
 
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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\
 
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