Since the Clausen functions are intimately related to a number of other important special functions, such as Inverse Tangent Integrals, Polylogarithms, Polygamma Functions, Zeta Functions, and more besides - many of which are at the forefront of modern mathematical research.
Define the Clausen Functions \(\displaystyle \text{Cl}_m(\theta)\) and \(\displaystyle \text{Sl}_m(\theta)\) as follows:
\[\text{Cl}_m(\theta) = \begin{cases} \displaystyle\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^m} & \text{if }m\text{ is even} \\ \displaystyle\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^m} & \text{if }m\text{ is odd} \end{cases}~,~\text{Sl}_m(\theta) = \begin{cases} \displaystyle\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^m} & \text{if }m\text{ is even} \\ \displaystyle\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^m} & \text{if }m\text{ is odd} \end{cases}
\]
As one would therefore expect, basic trigonometry can be used to develop all sorts of Clausen function properties. For example, consider the difference
\[\begin{align*}
&\text{Cl}_{2m}(\theta)-\text{Cl}_{2m}(\pi-\theta)=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^{2m}}-\sum_{k=1}^{\infty}\frac{\sin k(\pi-\theta)}{k^{2m}}\\
&=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^{2m}}-\sum_{k=1}^{\infty}\frac{(\sin\pi k\cos k\theta-\cos\pi k\sin k\theta)}{k^{2m}}\\
&=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^{2m}}+\sum_{k=1}^{\infty}\frac{(\cos\pi k\sin k\theta)}{k^{2m}}=\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^{2m}}+\sum_{k=1}^{\infty}(-1)^k\frac{\sin k\theta}{k^{2m}}\\
&=2\,\left(\frac{\sin 2\theta}{2^{2m}}+\frac{\sin 4\theta}{4^{2m}}+\frac{\sin 6\theta}{6^{2m}}+\cdots\,\right)\\
&=\displaystyle \frac{2}{2^{2m}}\,\left(\sin 2\theta+\frac{\sin 4\theta}{2^{2m}}+\frac{\sin 6\theta}{3^{2m}}+\cdots\,\right)=\frac{1}{2^{2m-1}}\text{Cl}_{2m}(2\theta)
\end{align*}\]
We now have the duplication formula for the CL-type Clausen function of even order:
\(\Large\mathbf{\color{Purple}{Result ~1:}}\)
\[\text{Cl}_{2m}(2\theta)=2^{2m-1}\Big[\text{Cl}_{2m}(\theta)-\text{Cl}_{2m}(\pi-\theta)\Big]
\]
Next, let's take the above duplication formula, replace \(\theta\) with the variable \(x\), and integrate both sides:
\[\int_0^{\varphi}\text{Cl}_{2m}(2x)\,\mathrm{d}x=2^{2m-1}\left[\int_0^{\varphi}\text{Cl}_{2m}(x)\,\mathrm{d}x-\int_0^{\varphi}\text{Cl}_{2m}(\pi-x)\,\mathrm{d}x\right]
\]
The L.H.S. is equal to
\[\begin{align*}
&\sum_{k=1}^{\infty}\frac{1}{k^{2m}}\,\int_0^{\varphi}\sin 2kx\,\mathrm{d}x=-\frac{1}{2}\,\sum_{k=1}^{\infty}\frac{1}{k^{2m+1}}\Big[\cos 2kx\Big]_0^{\varphi}\\
&=-\frac{1}{2}\,\sum_{k=1}^{\infty}\frac{\cos 2k\varphi}{k^{2m+1}}+\frac{1}{2}\,\sum_{k=1}^{\infty}\frac{1}{k^{2m+1}}=\frac{1}{2}\,\Big[\zeta(2m+1)-\text{Cl}_{2m+1}(2\varphi)\Big]
\end{align*}\]
Whereas the difference of the two integrals on the R.H.S. is
\[\begin{align*}
&2^{2m-1}\,\sum_{k=1}^{\infty}\frac{1}{k^{2m}}\int_0^{\varphi}\left[\sin kx-\sin k(\pi-x)\right]\,\mathrm{d}x\\
=&2^{2m-1}\,\sum_{k=1}^{\infty}\frac{1}{k^{2m+1}}\Big[-\cos kx+\cos k(\pi-x)\Big]_0^{\varphi}\\
=&2^{2m-1}\,\sum_{k=1}^{\infty}\frac{[\cos k(\pi-\varphi)-\cos k\varphi]}{k^{2m+1}}=2^{2m-1}[\text{Cl}_{2m+1}(\pi-\varphi)-\text{Cl}_{2m+1}(\varphi)]
\end{align*}\]
We now have the duplication formula for a CL-type Clausen function of odd order:
\(\Large\mathbf{\color{Purple}{Result ~2:}}\)
\[\text{Cl}_{2m+1}(2\theta)=\zeta(2m+1)+2^{2m}\Big[\text{Cl}_{2m+1}(\theta)-\text{Cl}_{2m+1}(\pi-\theta)\Big]
\]
Let's return to the first of the two duplication formulae, and derive a few particular values and correlations:
\[\text{Cl}_{2m}(2\theta)=2^{2m-1}\Big[\text{Cl}_{2m}(\theta)-\text{Cl}_{2m}(\pi-\theta)\Big]
\]
Let \(\theta =0\), to obtain the obvious results (\(n=\) integer):
\[\text{Cl}_{2m}(0) = 0~,~\text{Cl}_{2m}(\pi n) = 0
\]
Let \(\theta =\pi/3\), to obtain
\[\text{Cl}_{2m}\left ( \frac{\pi }{3} \right )=\frac{(1+2^{2m-1})}{2^{2m-1}}\text{Cl}_{2m}\left ( \frac{2\pi }{3} \right )
\]
Similarly, let \(\theta =\pi/4\), to obtain
\[\text{Cl}_{2m}\left ( \frac{\pi }{2} \right )=2^{2m-1}\text{Cl}_{2m}\left ( \frac{\pi }{4} \right )-2^{2m-1}\text{Cl}_{2m}\left ( \frac{3\pi }{4} \right )
\]
The leftmost Clausen function is expressible in terms of the Dirichlet Beta function:
\[\beta(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^x}
\]
Since
\[\text{Cl}_{2m}\left ( \frac{\pi }{2} \right )=\sum_{k=1}^{\infty}\frac{\sin (\pi k/2)}{k^{2m}}=1-\frac{1}{3^{2m}}+\frac{1}{5^{2m}}-\frac{1}{7^{2m}}+\,\cdots\, =\beta(2m)
\]
So, in summation, thus far we have the following useful values/relations:
\(\Large\mathbf{\color{Purple}{Result ~3:}}\)
\[\text{Cl}_{2m}(0) = 0~,~\text{Cl}_{2m}(\pi n) = 0~,~\text{Cl}_{2m}\left ( \frac{\pi }{2} \right ) = \beta(2m)
\]
\[\text{Cl}_{2m}\left ( \frac{\pi }{3} \right )=\frac{(1+2^{2m-1})}{2^{2m-1}}\text{Cl}_{2m}\left ( \frac{2\pi }{3} \right )~,~\text{Cl}_{2m}\left ( \frac{\pi }{4} \right )=\frac{\beta(2m)}{2^{2m-1}}+\text{Cl}_{2m}\left ( \frac{3\pi }{4} \right )
\]
Continuing on with the first of the two duplication formulae, note that we can apply it to itself as follows:
\[\begin{align*}
&\text{Cl}_{2m}(2\theta)=2^{2m-1}\left[\text{Cl}_{2m}(\theta)-\text{Cl}_{2m}(\pi-\theta)\right]\\
\Rightarrow &\text{Cl}_{2m}(4\theta)=2^{2m-1}\left[\text{Cl}_{2m}(2\theta)-\text{Cl}_{2m}(\pi-2\theta)\right]\\
=&2^{2m-1}\left[2^{2m-1}\left(\text{Cl}_{2m}(\theta)-\text{Cl}_{2m}(\pi-\theta)\right)\right]-2^{2m-1}\text{Cl}_{2m}(2(\pi/2-\theta))\\
=&4^{2m-1}\left[\text{Cl}_{2m}(\theta)-\text{Cl}_{2m}(\pi-\theta)\right]-4^{2m-1}\left[\text{Cl}_{2m}(\pi/2-\theta)-\text{Cl}_{2m}(\pi/2+\theta)\right]
\end{align*}\]
\(\Large\mathbf{\color{Purple}{Result ~4:}}\)
\[\text{Cl}_{2m}(4\theta)=4^{2m-1}\Bigg[\text{Cl}_{2m}(\theta)-\text{Cl}_{2m}(\pi-\theta) -\text{Cl}_{2m}\left(\frac{\pi}{2}-\theta\right)+\text{Cl}_{2m}\left(\frac{\pi}{2}+\theta\right)\Bigg]
\]
By applying the exact same process to the second of the two duplication formulae, then for CL-type Clausen function of odd index we have:
\(\Large\mathbf{\color{Purple}{Result ~5:}}\)
\[\begin{align*}
&\text{Cl}_{2m+1}(4\theta)\\
&=\zeta(2m+1)+4^{2m}\Bigg[\text{Cl}_{2m+1}(\theta)-\text{Cl}_{2m+1}(\pi-\theta) -\text{Cl}_{2m+1}\left(\frac{\pi}{2}-\theta\right)+\text{Cl}_{2m+1}\left(\frac{\pi}{2}+\theta\right)\Bigg]
\end{align*}\]
In \(\mathbf{\color{Purple}{Result~4}}\), set \(\displaystyle \theta=\pi/8\,\) to obtain:
\[\text{Cl}_{2m}\left(\frac{\pi}{2}\right)=4^{2m-1}\left[\text{Cl}_{2m}\left(\frac{\pi}{8}\right)-\text{Cl}_{2m}\left(\frac{7\pi}{8}\right) -\text{Cl}_{2m}\left(\frac{3\pi}{8}\right)+\text{Cl}_{2m}\left(\frac{5\pi}{8}\right)\right]
\]
We've already seen that the leftmost Clausen function is expressible in terms of the Dirichlet Beta function, so we quickly conclude that:
\(\Large\mathbf{\color{Purple}{Result ~6:}}\)
\[\text{Cl}_{2m}\left(\frac{\pi}{8}\right) -\text{Cl}_{2m}\left(\frac{3\pi}{8}\right) +\text{Cl}_{2m}\left(\frac{5\pi}{8}\right) -\text{Cl}_{2m}\left(\frac{7\pi}{8}\right)=\frac{\beta(2m)}{4^{2m-1}}
\]
And in particular:
\[\text{Cl}_{2}\left(\frac{\pi}{8}\right) -\text{Cl}_{2}\left(\frac{3\pi}{8}\right) +\text{Cl}_{2}\left(\frac{5\pi}{8}\right) -\text{Cl}_{2}\left(\frac{7\pi}{8}\right)=\frac{\beta(2)}{4}=\frac{\mathbf{G}}{4}
\]
To express a CL-type Clausen function of rational argument and even order as a sum of polygamma functions, proceed as follows:
\[\text{Cl}_{2m}\left(\frac{p\pi}{q}\right)=\sum_{k=1}^{\infty}\frac{\sin(p\pi/q)}{k^{2m}}
\]
Split this into \(q\) parts
\[\begin{align*}
&\sum_{k=0}^{\infty}\frac{\sin\left[(kq+1)\dfrac{p\pi}{q}\right]}{(kq+1)^{2m}}+ \sum_{k=0}^{\infty}\frac{\sin\left[(kq+2)\dfrac{p\pi}{q}\right]}{(kq+2)^{2m}}+ \sum_{k=0}^{\infty}\frac{\sin\left[(kq+3)\dfrac{p\pi}{q}\right]}{(kq+3)^{2m}}+ \,\cdots\,+\\
&\sum_{k=0}^{\infty}\frac{\sin\left[(kq+(q-1))\dfrac{p\pi}{q}\right]}{(kq+(q-1))^{2m}}+ \sum_{k=0}^{\infty}\frac{\sin\left[(kq+q)\dfrac{p\pi}{q}\right]}{(kq+q)^{2m}}
\end{align*}\]
Convert into a finite sum:
\[\begin{align*}
&\sum_{j=1}^{j=q}\sum_{k=0}^{\infty}\frac{\sin\left(kp\pi+\dfrac{jp\pi}{q}\right)}{(kq+j)^{2m}}=\frac{1}{q^{2m}}\sum_{j=1}^{j=q}\sum_{k=0}^{\infty}\frac{\sin\left(kp\pi+\dfrac{jp\pi}{q}\right)}{(k+j/q)^{2m}}=\frac{1}{q^{2m}}\sum_{j=1}^{j=q}\sum_{k=0}^{\infty}\frac{\sin\left(kp\pi+\dfrac{jp\pi}{q}\right)}{(k+j/q)^{2m}}\\
&=\frac{1}{q^{2m}}\sum_{j=1}^{j=q}\sum_{k=0}^{\infty}\frac{\cos(kp\pi)\sin\left(\dfrac{jp\pi}{q}\right)}{(k+j/q)^{2m}}=\frac{1}{q^{2m}}\sum_{j=1}^{j=q}\,\sin\left(\frac{jp\pi}{q}\right)\,\sum_{k=0}^{\infty}\frac{(-1)^{kp}}{(k+j/q)^{2m}}
\end{align*}\]
We now have two distinct cases to consider, since
\[(-1)^{kp} = \begin{cases} 1 & \text{if }p\text{ is even} \\ (-1)^k & \text{if }p\text{ is odd} \end{cases}
\]
\(\mathbf{\color{Teal}{Case ~1. ~when ~\boldsymbol{p} ~is ~even:}}\)
This is, naturally, the simpler of the two, since the result above reduces to:
\[\frac{1}{q^{2m}}\sum_{j=1}^{j=q}\,\sin\left(\frac{jp\pi}{q}\right)\,\sum_{k=0}^{\infty}\frac{1}{(k+j/q)^{2m}}
\]
We then apply the series definition for the polygamma function
\[\psi_n(z)= (-1)^{n+1}\,n!\,\sum_{k=0}^{\infty}\frac{1}{(z+k)^{n+1}}
\]
to obtain
\[\frac{1}{q^{2m}}\sum_{j=1}^{j=q}\,\sin\left(\frac{jp\pi}{q}\right)\,\frac{\psi_{2m-1}\left ( \dfrac{j}{q}\right )}{(2m-1)!}
\]
This, in turn, can be expressed in the infinitely more elegant form:
\(\Large\mathbf{\color{Purple}{Result ~7A:}}\)
\[(2m-1)!\,q^{2m}\,\text{Cl}_{2m}\left(\frac{p\pi}{q}\right)=\sum_{j=1}^{j=q}\,\sin\left(\frac{jp\pi}{q}\right)\,\psi_{2m-1}\left(\frac{j}{q}\right)
\]
\(\mathbf{\color{Teal}{Case ~2. ~when ~\boldsymbol{p} ~is ~odd:}}\)
When \(p\) is odd, we are faced with an alternating series, which must be split in two:
\[\begin{align*}
&\frac{1}{q^{2m}}\sum_{j=1}^{j=q}\,\sin\left(\frac{jp\pi}{q}\right)\,\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(k+j/q)^{2m}}\\
=&\frac{1}{q^{2m}}\sum_{j=1}^{j=q}\,\sin\left(\frac{jp\pi}{q}\right)\,\left[ \sum_{k=0}^{\infty}\frac{1}{(2k+j/q)^{2m}}- \sum_{k=0}^{\infty}\frac{1}{(2k+1+j/q)^{2m}} \right]\\
=&\frac{1}{(2q)^{2m}}\sum_{j=1}^{j=q}\,\sin\left(\frac{jp\pi}{q}\right)\,\left[ \sum_{k=0}^{\infty}\frac{1}{[k+(j/2q)]^{2m}}- \sum_{k=0}^{\infty}\frac{1}{[k+((j+q)/2q)]^{2m}} \right]\\
=&\frac{1}{(2q)^{2m}}\sum_{j=1}^{j=q}\,\sin\left(\frac{jp\pi}{q}\right)\,\left[ \frac{\psi_{2m-1}\left(\dfrac{j}{2q}\right)-\psi_{2m-1}\left(\dfrac{j+q}{2q}\right) }{(2m-1)!} \right]
\end{align*}\]
From which we obtain the final result:
\(\Large\mathbf{\color{Purple}{Result ~7B:}}\)
\[\displaystyle (2m-1)!\,(2q)^{2m}\,\text{Cl}_{2m}\left(\frac{p\pi}{q}\right)=\sum_{j=1}^{j=q}\,\sin\left(\frac{jp\pi}{q}\right)\,\left[ \psi_{2m-1}\left(\frac{j}{2q}\right)-\psi_{2m-1}\left(\frac{j+q}{2q}\right) \right]
\]
Here's an extremely elementary application of trigonometry to express a couple of trig series in terms of Clausen functions. Start of with a CL-type Clausen function of odd order and double argument:
\[\text{Cl}_{2m+1}(2\theta)=\sum_{k=1}^{\infty}\frac{\cos 2k\theta}{k^{2m+1}}
\]
Next, split the sum using the double angle formula for the cosine:
\[\text{Cl}_{2m+1}(2\theta)=\sum_{k=1}^{\infty}\frac{(\cos^2 k\theta-\sin^2 k\theta)}{k^{2m+1}}= \sum_{k=1}^{\infty}\frac{\cos^2 k\theta}{k^{2m+1}}- \sum_{k=1}^{\infty}\frac{\sin^2 k\theta}{k^{2m+1}}
\]
Again, using nothing but the most basic level of trigonometry, ie \(\displaystyle \sin^2 x+\cos^2 x=1\,\), we see that the sum of the last two series must be
\[\sum_{k=1}^{\infty}\frac{\cos^2 k\theta}{k^{2m+1}}+ \sum_{k=1}^{\infty}\frac{\sin^2 k\theta}{k^{2m+1}}=\sum_{k=1}^{\infty}\frac{1}{k^{2m+1}}=\zeta(2m+1)
\]
Taking the sum/difference of these two results gives:
\(\Large\mathbf{\color{Purple}{Result ~8:}}\)
\[\sum_{k=1}^{\infty}\frac{\sin^2 k\theta}{k^{2m+1}}=\frac{1}{2}\Big[\zeta(2m+1)-\text{Cl}_{2m+1}(2\theta)\Big]
\]
\[\sum_{k=1}^{\infty}\frac{\cos^2 k\theta}{k^{2m+1}}=\frac{1}{2}\Big[\zeta(2m+1)+\text{Cl}_{2m+1}(2\theta)\Big]
\]
As mentioned right at the start, there are deep connections between the Clausen functions and - amongst others - polylogarithms. Here's a simple example.
For \(\displaystyle |z|\le 1\), the Polylogarithm of order \(m\) has the series expansion:
\[\text{Li}_m(z)=\sum_{k=1}^{\infty}\frac{z^k}{k^m}
\]
Setting \(\displaystyle z=e^{i\theta}\,\) in the series above gives:
\[\begin{align*}
\text{Li}_m(e^{i\theta})&=\sum_{k=1}^{\infty}\frac{(e^{i\theta})^k}{k^m}=\sum_{k=1}^{\infty}\frac{(\cos \theta+i\sin \theta)^k}{k^m}\\
&=\sum_{k=1}^{\infty}\frac{(\cos k\theta+i\sin k\theta)}{k^m}=\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^m}+i\,\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^m}
\end{align*}\]
So, for a polylogarithm of even order, we have:
\[\text{Li}_{2m}(e^{i\theta})=\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^{2m}}+i\,\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^{2m}}= \text{Sl}_{2m}(\theta)+i\,\text{Cl}_{2m}(\theta)
\]
And similarly, for a polylogarithm of odd order:
\[\text{Li}_{2m+1}(e^{i\theta})=\sum_{k=1}^{\infty}\frac{\cos k\theta}{k^{2m+1}}+i\,\sum_{k=1}^{\infty}\frac{\sin k\theta}{k^{2m+1}}= \text{Cl}_{2m+1}(\theta)+i\,\text{Sl}_{2m+1}(\theta)
\]
\(\Large\mathbf{\color{Purple}{Result ~9:}}\)
\[\begin{align*}
&\text{Li}_{2m}(e^{i\theta})= \text{Sl}_{2m}(\theta)+i\,\text{Cl}_{2m}(\theta)~,~\text{Li}_{2m+1}(e^{i\theta})= \text{Cl}_{2m+1}(\theta)+i\,\text{Sl}_{2m+1}(\theta)\\
&\text{Li}_{2m}(e^{-i\theta})= \text{Sl}_{2m}(\theta)-i\,\text{Cl}_{2m}(\theta)~,~ \text{Li}_{2m+1}(e^{-i\theta})= \text{Cl}_{2m+1}(\theta)-i\,\text{Sl}_{2m+1}(\theta)
\end{align*}\]
As later posts in this thread will demonstrate, the (definite integral) moments of Clausen functions have a number of important applications, so it would, therefore, be quite useful to find a closed form for these moments:
\[\mathbf{Cl}_{(m, n)}(\theta)=\int_0^{\theta}x^n\,\text{Cl}_{m}(x)\,\mathrm{d}x
\]
In order to find these, we start off with the integral definition of \(\displaystyle \text{Cl}_2(\theta)\):
\[\text{Cl}_2(\theta)=-\int_0^{\theta}\ln\Big|2\sin\frac{x}{2}\Big|\,\mathrm{d}x
\]
We could just as easily use the series definition, but it important to observe that, within the range \(\displaystyle 0 <\theta\le 2\pi\,\), the absolute value of the logsine term can be omitted. Hence while within this range, we can write
\[\text{Cl}_2(\theta)=-\int_0^{\theta}\ln\left(2\sin\frac{x}{2}\right)\,\mathrm{d}x~,~ \text{Cl}_1(\theta)=-\ln\left(2\sin\frac{\theta}{2}\right)
\]
with no loss of generality. Note that the omission of \(\theta =0\) was to avoid the divergence of \(\displaystyle \text{Cl}_1(\theta)\,\) at this point (although, in the integrals presented below, this singularity is of no consequence).
The generalized moments of Clausen functions will be dealt with shortly, but for the moment, let's consider the special - and perhaps most important - case of the moments of \(\displaystyle \text{Cl}_1(\theta)\,\).
\[\mathbf{Cl}_{(1, n)}(\theta)=\int_0^{\theta}x^n\,\text{Cl}_{1}(x)\,\mathrm{d}x= -\int_0^{\theta}x^n\ln\left(2\sin\frac{x}{2}\right)\,\mathrm{d}x
\]
Applying the series definition for \(\displaystyle \text{Cl}_1(\theta)\) this can be written as
\[\mathbf{Cl}_{(1, n)}(\theta)=-\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{\theta}x^n\cos kx\,\mathrm{d}x
\]
Although tedious to derive, the closed form of that last trigonometric integral is easily deduced. Define
\[\mathcal{I}_{(p)}=\int_0^{\theta}x^p\cos kx\,\mathrm{d}x
\]
Then we note that, for \(p > 2\):
\[\mathcal{I}_{(p)}=\frac{x^p}{k}\sin kx\,\Biggr|_0^{\theta}+\frac{px^{p-1}}{k^2}\cos kx\,\Biggr|_0^{\theta}-\frac{p(p-1)}{k^2}\mathcal{I}_{(p-2)}
\]
A similarly, for \(p > 4\):
\[\begin{align*}
\mathcal{I}_{(p)}&=\frac{x^p}{k}\sin kx\,\Biggr|_0^{\theta}+\frac{px^{p-1}}{k^2}\cos kx\,\Biggr|_0^{\theta}- \frac{p(p-1)x^{p-2}}{k^3}\sin kx\,\Biggr|_0^{\theta}\\
&~~~-\frac{p(p-1)(p-2)x^{p-3}}{k^4}\cos kx\,\Biggr|_0^{\theta}+\frac{p(p-1)(p-2)(p-3)}{k^4}\mathcal{I}_{(p-4)}
\end{align*}\]
Using the iteration process above, and creating two sums - one for sine terms and the other for cosine terms - we get the desired result:
\[\mathcal{I}_{(p)}=p!\,\left[\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^j\frac{x^{p-2j}\sin kx}{k^{2j+1}(p-2j)!}\Biggr|_0^{\theta}+ \sum_{j=0}^{\lfloor {(p-1)/2} \rfloor}(-1)^j\frac{x^{p-2j-1}\cos kx}{k^{2j+2}(p-2j-1)!)}\Biggr|_0^{\theta} \right]
\]
Inserting this into our moment integral gives:
\[\begin{align*}
&\mathbf{Cl}_{(1, p)}(\theta)=\int_0^{\theta}x^p\,\text{Cl}_{1}(x)\,\mathrm{d}x= -\int_0^{\theta}x^p\ln\left(2\sin\frac{x}{2}\right)\,\mathrm{d}x=-\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{\theta}x^p\cos kx\,\mathrm{d}x\\
&=-p!\,\sum_{k=1}^{\infty}\frac{1}{k}\,\left[\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^j\frac{x^{p-2j}\sin kx}{k^{2j+1}(p-2j)!}\Biggr|_0^{\theta}+ \sum_{j=0}^{\lfloor {(p-1)/2} \rfloor}(-1)^j\frac{x^{p-2j-1}\cos kx}{k^{2j+2}(p-2j-1)!)}\Biggr|_0^{\theta} \right]
\end{align*}\]
For the lower bound - \(x=0\) - the terms in the (finite!) sine series vanish. The same is true for all the lower bound term in the (finite) cosine series, except for the final term - containing \(x_0\), which is present only when \(p=2m+1\) is odd, in which case this final cosine term is \(\displaystyle (-1)^{\lfloor (p-1)/2 \rfloor +1} / k^{p+1}\,\). To account for this term, we introduce the function
\[\frac{[1+(-1)^{p+1}]}{2} = \begin{cases} 0, & \text{if }p\text{ is even} \\ 1, & \text{if }p\text{ is odd} \end{cases}
\]
Finally, expressing our sum in terms of Clausen functions, we arrive at:
\(\Large\mathbf{\color{Purple}{Result ~10:}}\)
\[\begin{align*}
&\int_0^{\theta}x^p\,\text{Cl}_{1}(x)\mathrm{d}x\\
&=-p!\,\left[ \sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^j\frac{{\theta}^{\,p-2j}}{(p-2j)!}\text{Cl}_{2j+2}(\theta)+ \sum_{j=0}^{\lfloor {(p-1)/2} \rfloor}(-1)^j\frac{{\theta}^{\,p-2j-1}}{(p-2j-1)!)}\text{Cl}_{2j+3}(\theta)\right]\\
&~~~+p! (-1)^{\lfloor (p-1)/2 \rfloor} \frac{[1+(-1)^{p+1}]}{2}\zeta(p+2)
\end{align*}\]
A number of simple - but nonetheless important - trigonometric integrals follow immediately from the previous evaluation. For example
\[\begin{align*}
&\int_0^{\theta}x^p\,\text{Cl}_{1}(x)\,\mathrm{d}x= -\int_0^{\theta}x^p\ln\left(2\sin\frac{x}{2}\right)\,\mathrm{d}x\\
&=-\frac{\theta^{\,p+1}}{p+1}\ln\left(2\sin\frac{\theta}{2}\right)+\frac{1}{2(p+1)}\,\int_0^{\theta}x^{p+1}\cot\frac{x}{2}\,\mathrm{d}x
\end{align*}\]
Which can be re-written in the more convenient form:
\(\Large\mathbf{\color{Purple}{Result ~11:}}\)
\[\begin{align*}
&\int_0^{\phi}x^{p+1}\cot x\,\mathrm{d}x\\
&=(p+1)! \frac{(-1)^{\lfloor (p-1)/2 \rfloor}[1+(-1)^{p+1}]}{2^{p+2}}\zeta(p+2)+\phi^{p+1}\ln(2\sin\phi)\\
&~~~-(p+1)!\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^j\frac{{\phi}^{\,p-2j}}{2^{2j+1}(p-2j)!}\text{Cl}_{2j+2}(\phi)\\
&~~~-(p+1)!\,\sum_{j=0}^{\lfloor {(p-1)/2} \rfloor}(-1)^j\frac{{\phi}^{\,p-2j-1}}{2^{2j+2}(p-2j-1)!}\text{Cl}_{2j+3}(\phi)
\end{align*}\]
\(\mathbf{\color{DarkOrange}{Logcosine ~moments - ~part ~1:}}\)
Following on from the logsine moments above, we find that the equivalent logcosine moments are slightly trickier - which was to be expected - although they are far richer, since the complex parts also yield useful information.
By analogy, we start off with:
\[\begin{align*}
&\int_0^{\theta}x^m\ln\left(2\cos\frac{x}{2}\right)\,\mathrm{d}x=\int_0^{\theta}x^m\ln\left(\frac{1+e^{-ix}}{e^{-ix/2}}\right)\,\mathrm{d}x\\
&=\int_0^{\theta}x^m\ln(1+e^{-ix})\,\mathrm{d}x+i\,\frac{\theta^{m+2}}{2(m+2)}
\end{align*}\]
We'll ignore that final complex function of \(\theta\) for now, and continue with the evaluation of the complex logarithmic integral part:
\[\begin{align*}
&\int_0^{\theta}x^m\ln(1+e^{-ix})\,\mathrm{d}x=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\int_0^{\theta}x^m(\cos kx -i\,\sin kx)\,\mathrm{d}x\\
&=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\int_0^{\theta}x^m\cos kx\,\mathrm{d}x-i\,\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\int_0^{\theta}x^m\sin kx\,\mathrm{d}x\\
&=\int_0^{\theta}x^m\left[\text{Cl}_1(x)-\frac{1}{2}\text{Cl}_1(2x)\right]\,\mathrm{d}x-i\,\int_0^{\theta}x^m\left[\text{Sl}_1(x)-\frac{1}{2}\text{Sl}_1(2x)\right]\,\mathrm{d}x\\
&\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{\theta}x^m\left(\cos kx-\frac{\cos 2kx}{2}\right)\,\mathrm{d}x-i\,\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{\theta}x^m\left(\sin kx-\frac{\sin 2kx}{2}\right)\,\mathrm{d}x
\end{align*}\]
We already have the closed form for the two leftmost CL-type integrals, so it remains to find the remaining two SL-types. [The real logcosine moments are omitted below, since they are easily deduced from the previous result]. Let
\[\mathcal{I}_{(p)}=\int_0^{\theta}x^p\sin kx\,\mathrm{d}x=-\frac{1}{k}\cos kx\Biggr|_0^{\theta}+\frac{px^{p-1}}{k^2}\sin kx\Biggr|_0^{\theta}-\frac{p(p-1)}{k^2}\mathcal{I}_{(p-2)}=\,\cdots
\]
As before, a similar iteration process leads to the general result:
\[\begin{align*}
&\int_0^{\theta}x^p\sin kx\,\mathrm{d}x= \\
&p!\,\left[\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{k^{2j+1}(p-2j)!}\cos kx \,\Biggr|_0^{\theta}+\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{k^{2j+2}(p-2j-1)!}\sin kx\,\Biggr|_0^{\theta}\right]
\end{align*}\]
So the imaginary part of \(\displaystyle \int_0^{\theta}x^p\ln\left(2\cos\frac{x}{2}\right)\,\mathrm{d}x\) yields
\[\begin{align*}
&p!\,\sum_{k=0}^{\infty}\frac{1}{k}\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{k^{2j+1}(p-2j)!}\cos kx \,\Biggr|_0^{\theta}\\
&+p!\,\sum_{k=0}^{\infty}\frac{1}{k}\,\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{k^{2j+2}(p-2j-1)!}\sin kx\,\Biggr|_0^{\theta}\\
&+\frac{p!}{2}\,\sum_{k=0}^{\infty}\frac{1}{k}\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{(2k)^{2j+1}(p-2j)!}\cos 2kx \,\Biggr|_0^{\theta}\\
&-\frac{p!}{2}\,\sum_{k=0}^{\infty}\frac{1}{k}\,\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{(2k)^{2j+2}(p-2j-1)!}\sin 2kx\,\Biggr|_0^{\theta}+\frac{\theta^{m+2}}{2(m+2)}=0
\end{align*}\]
\[\begin{align*}
&p!\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{(p-2j)!}\text{Sl}_{2j+2}(x)\,\Biggr|_0^{\theta}+p!\,\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{(p-2j-1)!}\text{Sl}_{2j+3}(x)\,\Biggr|_0^{\theta}\\
&-\frac{p!}{2}\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{2^{2j+1}(p-2j)!}\text{Sl}_{2j+2}(2x)\,\Biggr|_0^{\theta}\\
&-\frac{p!}{2}\,\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{2^{2j+2}(p-2j-1)!}\text{Sl}_{2j+3}(2x)\,\Biggr|_0^{\theta}+\frac{\theta^{m+2}}{2(m+2)}=0
\end{align*}\]
As promised before - on other threads, and indeed other forums - I'll start to find closed form expressions for various polygamma functions, at the rational arguments $1/2, 1/3, 2/3, 1/4 \(,\) 3/4, 1/6$, and \(5/6\). This might take a while, and be posted in stages.
To start with, let's consider the following particular Clausen function of (arbitrary) odd order:
\[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos (\pi k/3)}{k^{2m+1}}
\]
We want to split this into six sums, where the first sum contains the first of every six terms, the second contains the second of every six terms, and so on. We also change summation index so our new series start at \(k=0\), rather than \(k=1\) above.
\[\begin{align*}
&\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos (\pi k/3)}{k^{2m+1}}=\\
&\sum_{k=0}^{\infty}\frac{\cos\dfrac{\pi}{3}(6k+1)}{(6k+1)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{\cos\dfrac{\pi}{3}(6k+2)}{(6k+2)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{\cos\dfrac{\pi}{3}(6k+3)}{(6k+3)^{2m+1}}+\\
&\sum_{k=0}^{\infty}\frac{\cos\dfrac{\pi}{3}(6k+4)}{(6k+4)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{\cos\dfrac{\pi}{3}(6k+5)}{(6k+5)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{\cos\dfrac{\pi}{3}(6k+6)}{(6k+6)^{2m+1}}
\end{align*}\]
Simplify the trig term in each series:
\[\begin{align*}
\cos \frac{\pi}{3}(6k+n)&=\cos\left(2\pi k+\frac{\pi n}{3}\right)\\
&=\cos 2\pi k\cos\frac{\pi n}{3}-\sin 2\pi k\sin\frac{\pi n}{3}\equiv \cos\frac{\pi n}{3}
\end{align*}\]
Our new sextet of series is thus
\[\begin{align*}
&\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)=\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+1)^{2m+1}}+ \cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^{2m+1}}\\
&+\cos\left(\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+3)^{2m+1}}+ \cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^{2m+1}}\\
&+\cos\left(\frac{5\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+5)^{2m+1}}+ \cos\left(2\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+6)^{2m+1}}
\end{align*}\]
Multiply both sides by \(6^{2m+1}\), and then subtract the third and sixth series on the RHS from the Clausen term on the LHS (using \(\displaystyle \cos\pi = -1\,\) and \(\displaystyle \cos 2\pi=1\,\) ) to obtain:
\[\begin{align*}
&6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\sum_{k=0}^{\infty}\frac{1}{(k+1/2)^{2m+1}}-\sum_{k=0}^{\infty}\frac{1}{(k+1)^{2m+1}}\\
&=\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+1/6)^{2m+1}}+ \cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+1/3)^{2m+1}}\\
&~~~+\cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+2/3)^{2m+1}}+\cos\left(\frac{5\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+5/6)^{2m+1}}
\end{align*}\]
Express the cosine terms on the RHS in real/rational form to make the RHS
\[\begin{align*}
&\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+1/6)^{2m+1}}- \frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+1/3)^{2m+1}}\\
-&\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+2/3)^{2m+1}}+\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+5/6)^{2m+1}}
\end{align*}\]
Now use
\[\psi_{n\ge 1}(x)=(-1)^{n+1}n!\sum_{k=0}^{\infty}\frac{1}{(k+x)^{n+1}}
\]
to re-write the RHS as:
\[\begin{align*}
&\frac{1}{2}\left(\frac{(-1)^{2m}}{(2m)!}\right)\Bigg\{ \psi_{2m}\left( \frac{1}{6} \right) -\psi_{2m}\left( \frac{1}{3} \right) -\psi_{2m}\left( \frac{2}{3} \right) +\psi_{2m}\left( \frac{5}{6} \right) \Bigg\}\\
=&\frac{1}{2\,(2m)!}\, \Bigg\{ \psi_{2m}\left( \frac{1}{6} \right) -\psi_{2m}\left( \frac{1}{3} \right) -\psi_{2m}\left( \frac{2}{3} \right) +\psi_{2m}\left( \frac{5}{6} \right) \Bigg\}
\end{align*}\]
Next, apply the same process to the two series on the LHS (with the Clausen term):
\[\begin{align*}
&6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\sum_{k=0}^{\infty}\frac{1}{(k+1/2)^{2m+1}}-\sum_{k=0}^{\infty}\frac{1}{(k+1)^{2m+1}}\\
=&6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\frac{1}{(2m)!}\Bigg\{\psi_{2m}\left(\frac{1}{2}\right)-\psi_{2m}(1)\Bigg\}
\end{align*}\]
Multiplying both sides by \(2(2m)!\) then gives the identity
\[\begin{align*}
&2\, (2m)! \,6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+2\psi_{2m}\left(\frac{1}{2}\right)-2\psi_{2m}(1)\\
=& \psi_{2m}\left( \frac{1}{6} \right) -\psi_{2m}\left( \frac{1}{3} \right) -\psi_{2m}\left( \frac{2}{3} \right) +\psi_{2m}\left( \frac{5}{6} \right)
\end{align*}\]
Next, repeat all of the above, but this time in terms of the Clasuen function with argument \(2\pi /3\)
\[\begin{align*}
&\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos (2\pi k/3)}{k^{2m+1}}\\
&=\cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+1)^{2m+1}}+ \cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^{2m+1}}+\\
&~~~\cos\left(2\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+3)^{2m+1}}+ \cos\left(\frac{8\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^{2m+1}}+\\
&~~~\cos\left(\frac{10\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+5)^{2m+1}}+ \cos\left(4\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+6)^{2m+1}}\\
&=-\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+1)^{2m+1}} -\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^{2m+1}}+\sum_{k=0}^{\infty}\frac{1}{(6k+3)^{2m+1}}\\
&~~~- \frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^{2m+1}}-\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+5)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{1}{(6k+6)^{2m+1}}
\end{align*}\]
Continue exactly as before, and you get the second relation
\[\begin{align*}
&2\, (2m)! \,6^{2m+1}\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)-2\psi_{2m}\left(\frac{1}{2}\right)-2\psi_{2m}(1)\\
&=-\psi_{2m}\left( \frac{1}{6} \right) -\psi_{2m}\left( \frac{1}{3} \right) -\psi_{2m}\left( \frac{2}{3} \right) -\psi_{2m}\left( \frac{5}{6} \right)
\end{align*}\]
Relative to the arguments \(1/3, 2/3, 1/6\), and \(5/6\), the arguments \(1\) and \(1/2\) are pretty straightforward, so I'll simply state them now and prove them later.
\[\psi_{2m}(1)=-(2m)!\zeta(2m+1)~,~\psi_{2m}\left(\frac{1}{2}\right)=-(2m)!\,(2^{2m+1}-1)\zeta(2m+1)
\]
Now, if you add the final forms or relation \(1\) and relation \(2\) you get:
\[\begin{align*}
&2\, (2m)! \,6^{2m+1}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]-4\psi_{2m}(1)\\
=&2\, (2m)! \,6^{2m+1}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+4\,(2m)!\zeta(2m+1)\\
=&2\Bigg\{ \psi_{2m}\left( \frac{1}{3} \right) +\psi_{2m}\left( \frac{2}{3} \right)\Bigg\}
\end{align*}\]
Or
\[\begin{align*}
&(2m)! \,6^{2m+1}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+2\,(2m)!\zeta(2m+1)\\
=&\psi_{2m}\left( \frac{1}{3} \right) +\psi_{2m}\left( \frac{2}{3} \right)
\end{align*}\]
On the other hand, the reflection formula for the polygamma function gives:
\[\begin{align*}
&\psi_{2m}(x)-\psi_{2m}(1-x)=\pi\frac{\mathrm{d}^{2m}}{\mathrm{d}x^{2m}}\cot\pi x\\
\Rightarrow &\psi_{2m}\left( \frac{1}{3} \right) -\psi_{2m}\left( \frac{2}{3} \right)=\pi\frac{\mathrm{d}^{2m}}{\mathrm{d}x^{2m}}\cot\pi x\,\Biggr|_{x=1/3}
\end{align*}\]
So
\[\begin{align*}
&\psi_{2m}\left( \frac{1}{3} \right)\\
=&\displaystyle \frac{(2m)! \,6^{2m+1}}{2}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+\,(2m)!\zeta(2m+1)+\pi\frac{\mathrm{d}^{2m}}{\mathrm{d}x^{2m}}\cot\pi x\,\Biggr|_{x=1/3}
\end{align*}\]
and
\[\begin{align*}
&\psi_{2m}\left( \frac{2}{3} \right)\\
=&\frac{(2m)! \,6^{2m+1}}{2}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+\,(2m)!\zeta(2m+1)-\pi\frac{\mathrm{d}^{2m}}{\mathrm{d}x^{2m}}\cot\pi x\,\Biggr|_{x=1/3}
\end{align*}\]
