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Clausen functions (and related series, functions, integrals)



Posts: 138
Location: North Londinium, UK
A number of simple - but nonetheless important - trigonometric integrals follow immediately from the previous evaluation. For example

\(\displaystyle \int_0^{\theta}x^p\,\text{Cl}_{1}(x)\,dx=
-\int_0^{\theta}x^p\log\left(2\sin\frac{x}{2}\right)\,dx=\)

\(\displaystyle -\frac{\theta^{\,p+1}}{p+1}\log\left(2\sin\frac{\theta}{2}\right)+\frac{1}{2(p+1)}\,\int_0^{\theta}x^{p+1}\cot\frac{x}{2}\,dx\)


Which can be re-written in the more convenient form


\(\displaystyle \int_0^{\phi}x^{p+1}\cot x\,dx=\frac{(p+1)}{2^{p+1}}\int_0^{2\phi}x^p\,\text{Cl}_{1}(x)\,dx+\phi^{p+1}\log(2\sin\phi)\)


Or...



Result 11:



\(\displaystyle \int_0^{\phi}x^{p+1}\cot x\,dx=\)


\(\displaystyle (p+1)! \frac{(-1)^{\lfloor (p-1)/2 \rfloor}[1+(-1)^{p+1}]}{2^{p+2}}\zeta(p+2)+\phi^{p+1}\log(2\sin\phi)\)

\(\displaystyle -(p+1)!\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^j\frac{{\phi}^{\,p-2j}}{2^{2j+1}(p-2j)!}\text{Cl}_{2j+2}(\phi)\)

\(\displaystyle -(p+1)!\,\sum_{j=0}^{\lfloor {(p-1)/2} \rfloor}(-1)^j\frac{{\phi}^{\,p-2j-1}}{2^{2j+2}(p-2j-1)!}\text{Cl}_{2j+3}(\phi)\)


Posts: 138
Location: North Londinium, UK
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:


\(\displaystyle \int_0^{\theta}x^m\log\left(2\cos\frac{x}{2}\right)\,dx=\int_0^{\theta}x^m\log\left(\frac{1+e^{-ix}}{e^{-ix/2}}\right)\,dx=\)

\(\displaystyle \int_0^{\theta}x^m\log(1+e^{-ix})\,dx+i\,\frac{\theta^{m+2}}{2(m+2)}\)

We'll ignore that final complex function of \(\displaystyle \theta\,\) for now, and continue with the evaluation of the complex logarithmic integral part:

\(\displaystyle \int_0^{\theta}x^m\log(1+e^{-ix})\,dx=\)

\(\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\int_0^{\theta}x^m(\cos kx -i\,\sin kx)\,dx=\)

\(\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\int_0^{\theta}x^m\cos kx\,dx-i\,\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\int_0^{\theta}x^m\sin kx\,dx=\)

\(\displaystyle \int_0^{\theta}x^m\left[\text{Cl}_1(x)-\frac{1}{2}\text{Cl}_1(2x)\right]\,dx-i\,\int_0^{\theta}x^m\left[\text{Sl}_1(x)-\frac{1}{2}\text{Sl}_1(2x)\right]\,dx=\)

\(\displaystyle \sum_{k=1}^{\infty}\frac{1}{k}\int_0^{\theta}x^m\left(\cos kx-\frac{\cos 2kx}{2}\right)\,dx-i\,\sum_{k=1}^{\infty}\frac{1}{k}\int_0^{\theta}x^m\left(\sin kx-\frac{\sin 2kx}{2}\right)\,dx\)


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

\(\displaystyle \mathcal{I}_{(p)}=\int_0^{\theta}x^p\sin kx\,dx=\)

\(\displaystyle -\frac{1}{k}\cos kx\Biggr|_0^{\theta}+\frac{px^{p-1}}{k^2}\sin kx\Biggr|_0^{\theta}-\frac{p(p-1)}{k^2}\mathcal{I}_{(p-2)}=\,\cdots\)


As before, a similar iteration process leads to the general result:

\(\displaystyle \int_0^{\theta}x^p\sin kx\,dx=\)

\(\displaystyle p!\,\left[\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{k^{2j+1}(p-2j)!}\cos kx \,\Biggr|_0^{\theta}+\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{k^{2j+2}(p-2j-1)!}\sin kx\,\Biggr|_0^{\theta}\right]\)


So the imaginary part of \(\displaystyle \int_0^{\theta}x^p\log\left(2\cos\frac{x}{2}\right)\,dx\,\) yields


\(\displaystyle p!\,\sum_{k=0}^{\infty}\frac{1}{k}\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{k^{2j+1}(p-2j)!}\cos kx \,\Biggr|_0^{\theta}+\)

\(\displaystyle p!\,\sum_{k=0}^{\infty}\frac{1}{k}\,\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{k^{2j+2}(p-2j-1)!}\sin kx\,\Biggr|_0^{\theta}-\)

\(\displaystyle \frac{p!}{2}\,\sum_{k=0}^{\infty}\frac{1}{k}\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{(2k)^{2j+1}(p-2j)!}\cos 2kx \,\Biggr|_0^{\theta}-\)

\(\displaystyle \frac{p!}{2}\,\sum_{k=0}^{\infty}\frac{1}{k}\,\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{(2k)^{2j+2}(p-2j-1)!}\sin 2kx\,\Biggr|_0^{\theta}+\frac{\theta^{m+2}}{2(m+2)}=0=\)




\(\displaystyle p!\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{(p-2j)!}\text{Sl}_{2j+2}(x)\,\Biggr|_0^{\theta}+\)

\(\displaystyle p!\,\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{(p-2j-1)!}\text{Sl}_{2j+3}(x)\,\Biggr|_0^{\theta}-\)

\(\displaystyle \frac{p!}{2}\,\sum_{j=0}^{\lfloor {p/2} \rfloor}(-1)^{j+1}\frac{x^{p-2j}}{2^{2j+1}(p-2j)!}\text{Sl}_{2j+2}(2x)\,\Biggr|_0^{\theta}-\)

\(\displaystyle \frac{p!}{2}\,\sum_{j=0}^{\lfloor {(p-1)2} \rfloor}(-1)^{j+1}\frac{x^{p-2j-1}}{2^{2j+2}(p-2j-1)!}\text{Sl}_{2j+3}(2x)\,\Biggr|_0^{\theta}+\frac{\theta^{m+2}}{2(m+2)}=0\)


Posts: 138
Location: North Londinium, UK
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. So, here goes....


To start with, let's consider the following particular Clausen function of (arbitrary) odd order:


\(\displaystyle \text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos (\pi k/3)}{k^{2m+1}}\)


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.


\(\displaystyle \text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos (\pi k/3)}{k^{2m+1}}=\)


\(\displaystyle \sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+1)}{(6k+1)^{2m+1}}+
\sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+2)}{(6k+2)^{2m+1}}+
\sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+3)}{(6k+3)^{2m+1}}+\)

\(\displaystyle \sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+4)}{(6k+4)^{2m+1}}+
\sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+5)}{(6k+5)^{2m+1}}+
\sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+6)}{(6k+6)^{2m+1}}=\)


Simplify the trig term in each series:


\(\displaystyle \cos \frac{\pi}{3}(6k+n)=\cos\left(2\pi k+\frac{\pi n}{3}\right)=\)

\(\displaystyle \cos 2\pi k\cos\frac{\pi n}{3}-\sin 2\pi k\sin\frac{\pi n}{3}\equiv \cos\frac{\pi n}{3}\)


Our new sextet of series is thus

\(\displaystyle \text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)=\)

\(\displaystyle \cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+1)^{2m+1}}+
\cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^{2m+1}}+\)

\(\displaystyle \cos\left(\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+3)^{2m+1}}+
\cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^{2m+1}}+\)

\(\displaystyle \cos\left(\frac{5\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+5)^{2m+1}}+
\cos\left(2\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+6)^{2m+1}}\)


Multiply both sides by \(\displaystyle 6^{2m+1}\,\), 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:


\(\displaystyle 6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\sum_{k=0}^{\infty}\frac{1}{(k+1/2)^{2m+1}}-\sum_{k=0}^{\infty}\frac{1}{(k+1)^{2m+1}}=\)


\(\displaystyle \cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+1/6)^{2m+1}}+
\cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+1/3)^{2m+1}}+\)

\(\displaystyle \cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+2/3)^{2m+1}}+\cos\left(\frac{5\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+5/6)^{2m+1}}\)


Express the cosine terms on the RHS in real/rational form to make the RHS


\(\displaystyle \frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+1/6)^{2m+1}}-
\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+1/3)^{2m+1}}\)

\(\displaystyle -\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+2/3)^{2m+1}}+\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+5/6)^{2m+1}}\)



Now use


\(\displaystyle \psi_{n\ge 1}(x)=(-1)^{n+1}n!\sum_{k=0}^{\infty}\frac{1}{(k+x)^{n+1}}\)

to re-write the RHS as:

\(\displaystyle \frac{1}{2}\left(\frac{(-1)^{2m}}{(2m)!}\right)\Bigg\{
\psi_{2m}\left( \frac{1}{6} \right)
-\psi_{2m}\left( \frac{1}{3} \right)
-\psi_{2m}\left( \frac{2}{3} \right)
+\psi_{2m}\left( \frac{5}{6} \right)
\Bigg\}=\)

\(\displaystyle \frac{1}{2\,(2m)!}\, \Bigg\{
\psi_{2m}\left( \frac{1}{6} \right)
-\psi_{2m}\left( \frac{1}{3} \right)
-\psi_{2m}\left( \frac{2}{3} \right)
+\psi_{2m}\left( \frac{5}{6} \right)
\Bigg\}\)


Next, apply the same process to the two series on the LHS (with the Clausen term):


\(\displaystyle 6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\sum_{k=0}^{\infty}\frac{1}{(k+1/2)^{2m+1}}-\sum_{k=0}^{\infty}\frac{1}{(k+1)^{2m+1}}=\)

\(\displaystyle 6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\frac{1}{(2m)!}\Bigg\{\psi_{2m}\left(\frac{1}{2}\right)-\psi_{2m}(1)\Bigg\}\)


Multiplying BOTH sides by \(\displaystyle 2\, (2m)! \,\) then gives the identity


\(\displaystyle 2\, (2m)! \,6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+2\psi_{2m}\left(\frac{1}{2}\right)-2\psi_{2m}(1)=\)

\(\displaystyle \psi_{2m}\left( \frac{1}{6} \right)
-\psi_{2m}\left( \frac{1}{3} \right)
-\psi_{2m}\left( \frac{2}{3} \right)
+\psi_{2m}\left( \frac{5}{6} \right)\)


----------------------------------------------------


Next, repeat ALL of the above, but this time in terms of the Clasuen function with argument \(\displaystyle 2\pi/3\,\)


\(\displaystyle \text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos (2\pi k/3)}{k^{2m+1}}=\)


\(\displaystyle \cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+1)^{2m+1}}+
\cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^{2m+1}}+\)

\(\displaystyle \cos\left(2\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+3)^{2m+1}}+
\cos\left(\frac{8\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^{2m+1}}+\)

\(\displaystyle \cos\left(\frac{10\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+5)^{2m+1}}+
\cos\left(4\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+6)^{2m+1}}=\)


\(\displaystyle -\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+1)^{2m+1}}
-\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^{2m+1}}+\)

\(\displaystyle \sum_{k=0}^{\infty}\frac{1}{(6k+3)^{2m+1}}-
\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^{2m+1}}+\)

\(\displaystyle -\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+5)^{2m+1}}+
\sum_{k=0}^{\infty}\frac{1}{(6k+6)^{2m+1}}\)


Continue exactly as before, and you get the second relation


\(\displaystyle 2\, (2m)! \,6^{2m+1}\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)-2\psi_{2m}\left(\frac{1}{2}\right)-2\psi_{2m}(1)=\)

\(\displaystyle -\psi_{2m}\left( \frac{1}{6} \right)
-\psi_{2m}\left( \frac{1}{3} \right)
-\psi_{2m}\left( \frac{2}{3} \right)
-\psi_{2m}\left( \frac{5}{6} \right)\)



----------------------------------------------------


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.



\(\displaystyle \psi_{2m}(1)=-(2m)!\zeta(2m+1)\)

\(\displaystyle \psi_{2m}\left(\frac{1}{2}\right)=-(2m)!\,(2^{2m+1}-1)\zeta(2m+1)\)



Now, if you add the final forms or relation #1 and relation #2 you get:


\(\displaystyle 2\, (2m)! \,6^{2m+1}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]-4\psi_{2m}(1)=\)

\(\displaystyle 2\, (2m)! \,6^{2m+1}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+4\,(2m)!\zeta(2m+1)=\)

\(\displaystyle 2\Bigg\{ \psi_{2m}\left( \frac{1}{3} \right)
+\psi_{2m}\left( \frac{2}{3} \right)\Bigg\}\)


Or

\(\displaystyle (2m)! \,6^{2m+1}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+2\,(2m)!\zeta(2m+1)=\)


\(\displaystyle \psi_{2m}\left( \frac{1}{3} \right)
+\psi_{2m}\left( \frac{2}{3} \right)\)


On the other hand, the reflection formula for the polygamma function gives:


\(\displaystyle \psi_{2m}(x)-\psi_{2m}(1-x)=\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x\)


\(\displaystyle \Rightarrow\)


\(\displaystyle \psi_{2m}\left( \frac{1}{3} \right)
-\psi_{2m}\left( \frac{2}{3} \right)=\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x\,\Biggr|_{x=1/3}\)


So


\(\displaystyle \psi_{2m}\left( \frac{1}{3} \right)=\)

\(\displaystyle \frac{(2m)! \,6^{2m+1}}{2}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+\,(2m)!\zeta(2m+1)+\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x\,\Biggr|_{x=1/3}\)


and


\(\displaystyle \psi_{2m}\left( \frac{2}{3} \right)=\)

\(\displaystyle \frac{(2m)! \,6^{2m+1}}{2}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+\,(2m)!\zeta(2m+1)-\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x\,\Biggr|_{x=1/3}\)

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