# Infinite SUM Cosine and Sine times Zeta Function

Can anyone come up with the function that describes these infinite series?:

$$\sum_{n=1}^\infty \frac{\cos(n)}{n^s}$$

or

$$\sum_{n=1}^\infty \frac{\sin(n)}{n^s}.$$

It’s basically the zeta function, except the 1 in the numerator is replaced by the $\cos(n)$ and $\sin(n)$. The cosine function starts at $-0.5$ for $s=0$, then crosses the $x$-axis at $s \simeq .898635523$, and has an asymptote for $f(x)=\cos(1)$.

Thank you for your help and interest! Please let me know if I can clarify anything.

#### Solutions Collecting From Web of "Infinite SUM Cosine and Sine times Zeta Function"

For $|z| = 1$ and $Re(s) > 1$, the polylogarithm has the series representation :
$$Li_s(z) = \sum_{n=1}^\infty n^{-s} z^{n}$$
Since $n^{-s}\Gamma(s) = \int_0^\infty x^{s-1} e^{-nx}dx$ you get
$$\Gamma(s) Li_s(z) = \sum_{n=1}^\infty z^n\int_0^\infty x^{s-1} e^{-nx}dx = \int_0^\infty x^{s-1} \sum_{n=1}^\infty z^n e^{-nx}dx = \int_0^\infty \frac{x^{s-1}}{z^{-1}e^{x}-1}dx$$
For $z \ne 1, |z| = 1$ it converges for $Re(s) > 0$ and it has a nice analytic continuation, meromorphic on the whole complex plane with poles of order $1$ at the negative integers, since $\frac{1}{z^{-1}e^{x}-1} = \sum_{k=0}^\infty c_z(k) x^k$ is analytic at $x=0$, and
$$Li_s(z)\Gamma(s)-\sum_{k=0}^K \frac{c_z(k)}{s+k} = \int_0^\infty x^{s-1}\left(\frac{1}{z^{-1}e^{x}-1}-\sum_{k=0}^K c_z(k) x^k 1_{x < 1}\right)dx$$
that converges and is analytic for $Re(s) > -K-1$.

Hence $\displaystyle G(s)= \sum_{n=1}^\infty \frac{\cos(n)}{n^s} = \frac{Li_s(e^i)+Li_s(e^{-i})}{2}$ is entire, and you have $G(s) \Gamma(s) \sim G(-k)\frac{(-1)^k}{k!(s+k)} \sim \frac{c_i(k)+c_{-i}(k)}{2(s+k)}$ as $s \to -k$ and
$$G(-k) = \frac{(-1)^k k!}{2}\left.\frac{d^k}{dx^k}\left[\frac{1}{e^{x+i}-1}+\frac{1}{e^{x-i}-1}\right] \ \right|_{x= 0}$$

If you want a faster series for evaluating $Li_s(z) = \sum_{n=1}^\infty z^{n} n^{-s}, |z| = 1, z \ne 1$

then adapt what I wrote there about $\eta(s) = -Li_s(-1)$ to get
$$\forall s \in \mathbb{C}, \qquad \qquad Li_s(z) = \sum_{n=0}^\infty \frac{1}{2^{n+1}} \sum_{k=0}^n {n \choose k} \frac {z^{k+1}}{(k+1)^s}$$