Resources on Variants of the Clausen Functions I am interested in locating more information about the Clausen functions. Specifically I am looking for the closed forms of the Gl-type (or Sl-type as they are sometimes called) and the alternating analogues of the Cl and Gl-type Clausen functions.
In other words, I am looking for the closed forms of:
$$\sum_{n=1}^{\infty}{\frac{\cos(nx)}{n^m}}$$
$$\sum_{n=1}^{\infty}{(-1)^{n-1}\frac{\cos(nx)}{n^m}}$$
$$\sum_{n=1}^{\infty}{\frac{\sin(nx)}{n^m}}$$
$$\sum_{n=1}^{\infty}{(-1)^{n-1}\frac{\sin(nx)}{n^m}}$$
So far I have found a few examples, in places such as Mathworld, or Gradshteyn & Rhyzhik, or Abramowitz & Stegun. None of these sources provide anywhere near a comprehensive collection of the Clausen and Alternating Clausen Functions in closed form.  
I was hoping someone might know of a paper or book that could provide me with more information, especially on the Alternating Clausen Functions.  The help would be greatly appreciated!

Here's what I found:
let's pick the case where $m\quad =\quad 2k\quad /\quad k\in Z\\ $
As you know:
$\sum _{ n=1 }^{ \infty  }{ \frac { { t }^{ n } }{ { n }^{ m } }  } \quad ={ Li }_{ m }(t)$
Let's substitute $t={ e }^{ ix }$:
$\sum _{ n=1 }^{ \infty  }{ \frac { { { e }^{ inx } } }{ { n }^{ m } }  } \quad ={ Li }_{ m }({ e }^{ ix })$
We will use the following identities aswell:
$$
{ Li }_{ m }(z)\quad =\quad { (-1) }^{ m-1 }{ Li }_{ m }\left( \frac { 1 }{ z }  \right) -\frac { { (2\pi i) }^{ m } }{ m! } { B }_{ m }\left( \frac { ln(-z) }{ 2\pi i } +\frac { 1 }{ 2 }  \right) /z\notin (0,1) \quad (1)\\ \Re ({ Li }_{ m }(z))\quad =\quad \frac { 1 }{ 2 } \left( { Li }_{ m }(z)\quad +\quad { Li }_{ m }\left( \bar { z }  \right)  \right )\\
\Im \left( { Li }_{ m }(z) \right) \quad =\quad \frac { 1 }{ 2i } \left( { Li }_{ m }(z)-{ Li }_{ m }\left( \bar { z }  \right)  \right) 
$$
So:
$$
\sum _{ n=1 }^{ \infty  }{ \frac { \cos { (nx } ) }{ { n }^{ m } } \quad =\quad \Re \left( \sum _{ n=1 }^{ \infty  }{ \frac { { { e }^{ inx } } }{ { n }^{ m } }  }  \right)  } \\ \qquad \qquad \qquad \quad \quad \quad =\quad \Re \left( { Li }_{ m }\left( { e }^{ ix } \right)  \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 1 }{ 2 } \left( { Li }_{ m }\left( { e }^{ ix } \right) +{ Li }_{ m }\left( { e }^{ -ix } \right)  \right) 
$$
And using that m is even as we supposed we get from $(1)$:
$$
\frac { 1 }{ 2 } \left( { Li }_{ m }\left( { e }^{ ix } \right) +{ Li }_{ m }\left( { e }^{ -ix } \right)  \right) \quad =\quad -\frac { 1 }{ 2 } \frac { { (2\pi i) }^{ m } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { ln(-{ e }^{ ix }) }{ 2\pi i }  \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\frac { 1 }{ 2 } \frac { { (-1) }^{ \frac { m }{ 2 }  }{ (2\pi ) }^{ m } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { i\left( x- \pi  \right)  }{ 2\pi i }  \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\frac { 1 }{ 2 } \frac { { (-1) }^{ \frac { m }{ 2 }  }{ (2\pi ) }^{ m } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi  }  \right) 
$$
So we get the following:
$$
\ \sum _{ n=1 }^{ \infty  }{ \frac { \cos { \left( nx \right)  }  }{ { n }^{ m } }  } =\quad -\frac { 1 }{ 2 } \frac { { (-1) }^{ \frac { m }{ 2 }  }{ (2\pi ) }^{ m } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi  }  \right) \quad /\quad m\quad =\quad 2k
$$
For the alternating sum:
$$ 
\sum _{ n=1 }^{ \infty  }{ \frac { { (-1) }^{ n-1 }\cos { (nx) }  }{ { n }^{ m } }  } \quad =\quad -\sum _{ n=1 }^{ \infty  }{ \frac { { (-1) }^{ n }\cos { (nx) }  }{ { n }^{ m } }  } \\ \qquad \qquad \qquad \qquad \qquad \quad \quad =\quad -\Re \left( \sum _{ n=1 }^{ \infty  }{ \frac { { (-1) }^{ n }{ e }^{ inx } }{ { n }^{ m } }  }  \right) \\ \quad \qquad \qquad \qquad \qquad \qquad \quad =\quad -\Re \left( \sum _{ n=1 }^{ \infty  }{ \frac { { \left( { -e }^{ ix } \right)  }^{ n } }{ { n }^{ m } }  }  \right) \\ \qquad \qquad \qquad \qquad \qquad \quad \quad =-\Re \left( { Li }_{ m }\left( -{ e }^{ ix } \right)  \right) 
$$
We will calculate the latter just as we did:
$$ 
-\Re \left( { Li }_{ m }\left( -{ e }^{ ix } \right)  \right) \quad =\quad -\frac { 1 }{ 2 } \left( { Li }_{ m }\left( { -e }^{ ix } \right) \quad +\quad { Li }_{ m }\left( { -e }^{ -ix } \right)  \right) \\ \qquad \qquad \qquad \qquad =\quad \frac { -1 }{ 2 } \left(- \frac { { (2\pi i) }^{ m } }{ m! } { B }_{ m }\left( \frac { ln\left( { e }^{ ix } \right)  }{ 2\pi i } +\frac { 1 }{ 2 }  \right)  \right) \\ \qquad \qquad \qquad \qquad =\quad \frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m }{ 2 }  } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { x }{ 2\pi  }  \right)  \right) 
$$
We conclude that:
$$
\sum _{ n=1 }^{ \infty  }{ \frac { { (-1) }^{ n-1 }\cos { (nx) }  }{ { n }^{ m } }  } =\quad \frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m }{ 2 }  } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { x }{ 2\pi  }  \right)  \right) / m = 2k
$$
Now let's calculate $\sum _{ n=1 }^{ \infty  }{ \frac { \sin { (nx) }  }{ { n }^{ m } }  }$ but for odd values of m using the same identities:
$m = 2k+1$
$$
\sum _{ n=1 }^{ \infty  }{ \frac { \sin { (nx) }  }{ { n }^{ m } }  } =\Im \left( \sum _{ n=1 }^{ \infty  }{ \frac { { e }^{ inx } }{ { n }^{ m } }  }  \right) \\ \qquad \qquad \qquad \quad \quad =\quad \Im \left( { Li }_{ m }\left( { e }^{ ix } \right)  \right) \\ \qquad \qquad \qquad \quad \quad =\quad \frac { 1 }{ 2i } \left( { Li }_{ m }\left( { e }^{ ix } \right) -{ Li }_{ m }\left( { e }^{ -ix } \right)  \right) 
$$
So using $(1)$, we have:
$$
\frac { 1 }{ 2i } \left( { Li }_{ m }\left( { e }^{ ix } \right) -{ Li }_{ m }\left( { e }^{ -ix } \right)  \right) \quad =\quad \frac { 1 }{ 2i } \left( \frac { -{ (2\pi i) }^{ m } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { ln(-{ e }^{ ix }) }{ 2\pi i }  \right)  \right) \\ \qquad \qquad \qquad \qquad \qquad \qquad =\quad -\frac { 1 }{ 2i } \left( \frac { { (2\pi ) }^{ m }{ (i) }^{ m } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { i(x-\pi ) }{ 2\pi i }  \right)  \right) \\ \qquad \qquad \qquad \qquad \qquad \qquad =\quad -\frac { 1 }{ 2i } \left( \frac { { (2\pi ) }^{ m }{ (i) }^{ 2k+1 } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi  }  \right)  \right) \\ \qquad \qquad \qquad \qquad \qquad \qquad =\quad -\frac { 1 }{ 2i } \left( \frac { { (2\pi ) }^{ m }i{ (-1) }^{ k } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi  }  \right)  \right) \\ \qquad \qquad \qquad \qquad \qquad \qquad =\quad -\frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m-1 }{ 2 }  } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi  }  \right)  \right) 
$$
So our sum is:
$$
\sum _{ n=1 }^{ \infty  }{ \frac { \sin { (nx) }  }{ { n }^{ m } }  }  = \quad -\frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m-1 }{ 2 }  } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi  }  \right)  \right) \quad m = 2k+1
$$
And for the alternating sum it's the same technique:
$$
\sum _{ n=1 }^{ \infty  }{ \frac { { (-1) }^{ n-1 }\sin { (nx) }  }{ { n }^{ m } }  } =\quad \frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m-1 }{ 2 }  } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { x }{ 2\pi  }  \right)  \right) 
$$
This is all I can do for the time being, I'll try to find the sum for other values as well.
Reference:


*

*On a property of polylogarithm

*http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/17/ShowAll.html
