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I would like to obtain an expression for the function $$g(s, \theta) = \sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n^{s}} \qquad (\#).$$

Here is what I've tried so far: we know from the definition of the Dirichlet convolution that $ (E*I_{1})(n) = \sigma_{1}(n). $ Here, $E(n) = 1$ for all $n$, $I_{a}(n) = n^{a}$ and $\sigma_{1}$ is the sum-of-divisors function. Now, instead of filling in numbers in these functions, I proceed with using functions as arguments. In particular, the Chebyshev polynomials of the first kind are relevant. Let $T_{n}(s)$ be the $n$'th Chebyshev polynomial of the first kind.

Then, I think: $(E*I_{1})(T_{n}(s)) = \sigma(T_{n}(s))$, where $E(\cdot)$ sends all polynomials to $1$, $I_{1}(\cdot)$ is the identity function for polynomials, and $\sigma_{1}(\cdot)$ is the sum of all polynomial divisors of its argument.

Question 1: is the equality $(E*I_{1})(T_{n}(s)) = \sigma(T_{n}(s))$ correct?

If we suppose it is, then I'd go on as follows: \begin{align*}L_{(E*I_{1})} (T_{n}(\cos(\theta))) &= L_{E} (T_{n}(\cos(\theta))) \cdot L_{I_{1}} (T_{n}(\cos(\theta))) \\ &= \zeta(s) \cdot g(s,\theta) = L_{\sigma}(\cos(n \theta)) . \end{align*}

Then I though of using an infinite product expression for $\cos(n \theta)$, in order to compute the sum of its polynomial divisors. An article like this one could come in handy, perhaps.

Question 2 Do you think this approach for finding $g(\cdot)$ as mentioned in $(\#)$ could work?

Question 3 Is there another method to obtain an expression for $g(\cdot)$ ? For instance, I think this paper could be very useful, but I'm not entirely sure how to apply their $\bigodot$ operator, and how it relates to the regular dirichlet convolution.

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For $\theta=a/q$ rational it is a sum of Dirichlet L-functions modulo $q$, otherwise there is no Dirichlet series simplification.

For $\Re(s) > 1, \theta\in (0,2\pi)$ with $C$ a contour enclosing $[0,\infty)$ clockwise

$$(e^{2i \pi s}-1)\Gamma(s) \sum_{n=1}^{\infty} \frac{\exp(in \theta)}{n^{s}} = \int_C \frac{x^{s-1}}{e^{x-i\theta}-1}dx$$ The RHS is entire and for $\Re(s) < 0$ with the residue theorem, being careful with the branch of $x^{s-1}$ it is

$$= 2i\pi \sum Res(\frac{x^{s-1}}{e^{x-i\theta}-1})=2i\pi \sum_k (2i\pi k+i\theta)^{s-1} $$ $$= (2i\pi)^s (\zeta(1-s,\frac{\theta}{2\pi})+(-1)^{s-1}\zeta(1-s,1-\frac{\theta}{2\pi}))$$ (the Hurwitz zeta function)

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  • $\begingroup$ Thank you, @reuns. Does this mean the value of the expression is not defined at $s=1$ ? Because then you get $e^{2i\pi} -1 = 1-1=0$ on the LHS of the first equation, right? (And on the right-hand side at the bottom, one obtains an imaginary number, it seems) $\endgroup$ – Max Muller Oct 26 '19 at 17:00
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    $\begingroup$ The contour integral on the RHS is zero at $s=1$, you should look there for the details of that contour integral representation $\endgroup$ – reuns Oct 26 '19 at 17:03
  • $\begingroup$ Thank you for the link. Now I also see the requirement that Re(s) > 1 above and Re(s) < 0 below, sorry, should've looked better $\endgroup$ – Max Muller Oct 26 '19 at 17:07
  • $\begingroup$ What do you think of this paper by the way (link: academia.edu/19563528/…) ? Do you think it formulas therein could be used to derive other identities for $g(s,\theta)$ ? $\endgroup$ – Max Muller Oct 26 '19 at 17:09
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I am not sure that this could help.

Using $$\sum_{n=1}^\infty\frac {e^{i n \theta}} {n^s}=\text{Li}_s\left(e^{i \theta}\right)$$ we have $$g(s, \theta) = \sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n^{s}}=\frac{1}{2} \left(\text{Li}_s\left(e^{-i \theta }\right)+\text{Li}_s\left(e^{i \theta }\right)\right)$$ where appears the polylogarithm function.

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You can get the value for even $s$, we start with the infinite sum
$\sum_{n=1}^{\infty}\cos(n\theta)$ and then integrate s times to get the disired result.

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  • $\begingroup$ The infinite sum $\sum_{n=1}^{\infty}\cos(n\theta)$ ?? Integrating $\sum_{n=1}^{\infty}z^n (e^{in\theta}+e^{-in\theta})$ gives the value of $\sum_{n=1}^{\infty} \cos(n\theta)n^{-k}$ only for $k=1$, for $k=2$ and $\cos(\theta)\ne \pm 1$ there is no known closed-form $\endgroup$ – reuns Oct 26 '19 at 18:46

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