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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!

FofX
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1 Answers1

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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:

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Oussama Boussif
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  • Oussama, thank you for the help, I too was thinking that the polylogarithm would be the way to go, I was just hoping I wouldn't have to derive them myself ;). On another note, it just occurred to me that if you know one Clausen function you can integrate it to obtain the values for it's sister function. For instance since: $C_2= \sum_{n=1}^{\infty}{cos(nx)/n^2} = \pi^2/6 - \pi/2x +1/4x^2$ Integrating $C_2$ gives: $S_3= \sum_{n=1}^{\infty}{sin(nx)/n^3} = \pi^2/6x - \pi/4x^2 +1/12x^3$ – FofX Aug 05 '15 at 02:17
  • By repeating this process let's you calculate the values for as many orders as you'd like. I feel a little silly that I didn't see this earlier... Although obviously this only works if you known the value of one of the functions to begin with. But as I said, there are a few places that give the values of some of the lower-order functions. – FofX Aug 05 '15 at 02:24
  • Yes, but as you see I could only derive ${ C }{ m }$ for even values of$m$ and derive${ S }{ m }$ for odd values. But I'm going to work on it ;) – Oussama Boussif Aug 05 '15 at 10:47
  • Yep exactly! I'd definitely like to see what you find. Aklthough I did just run across this: $\sum_{n=1}^{\infty}{(-1)^{n-1}\frac{cos(nx)}{n}} = \log(\cos(x/2)) + \log(2)$, so this will gives us the ability to find alternating $C_m$ for odd values and alternating $S_m$ for even values. The only issue is knowing how to evaluate those log-cosine integrals! :) – FofX Aug 05 '15 at 16:16
  • Well that's the main issue. That's kind of why clausen functions are here ^^ – Oussama Boussif Aug 05 '15 at 17:57