Clausen zeta function For $0 < \theta < 2\pi$, define
$$\kappa(x,\theta) = \frac{1}{\zeta(x)}\sum_{n=1}^\infty \frac{e^{ in\theta}}{n^x}$$ 
for $\Re(x) > 1$. It is easy to see that 
$$\kappa(x,\theta) = \frac{1}{\zeta(x)\Gamma(x)}\int_0^\infty \frac{t^{x-1}}{e^{t - i\theta}-1}dt $$
My question is what the analytic continuation of this function $\kappa(x,\theta)$ looks like for $x=0$ and negative integer $x$. We can assume $\theta$ is fixed.  
 A: The Clausen function $\,\operatorname{Cl}_x(\theta)\,$ is often considered with $x$ a positive integer but allow analytic continuation for all complex values of $x$ (except $\,x=1$ and $\,\theta\equiv 0\pmod {2\pi}$).
Concerning non integer values of $x$ rewrite them as polylogarithms may be useful :
The polylogarithm function verifies indeed $\:\operatorname{Li}_x\left(e^{i\theta}\right)=\displaystyle \sum_{n=1}^\infty \frac{e^{ in\theta}}{n^x}\;$ so that 
$$\kappa(x,\theta) = \dfrac{\operatorname{Li}_x\left(e^{i\theta}\right)}{\zeta(x)}$$
and you may obtain values and representations for fixed $\theta$ using W. Alpha (here $\theta=1$) :

Concerning analytic extensions as $x=0$ and $\,x$ a negative integer they derive directly from $$\operatorname{Li}_0(z)=\sum_{k=1}^\infty \frac {z^k}{k^0}=\frac z{1-z}$$
at each step compute $\;\operatorname{Li}_{-n}(z)=z\,\dfrac d{dz}\operatorname{Li}_{-n+1}(z)$ to get :
\begin{align}
\operatorname{Li}_{\;0}(z)&= \frac z{1-z}\\
\operatorname{Li}_{-1}(z) &= \frac z{(1-z)^2}\\
\operatorname{Li}_{-2}(z) &= \frac {z\,(1+z)}{(1-z)^3}\\
\operatorname{Li}_{-3}(z) &= \frac {z\,(1+4z+z^2)}{(1-z)^4}\\
\end{align}
while $\,\operatorname{Li}_{\,1}(z)=-\log(1-z)\,$ will be fine for $z\neq 1$.
Hoping this helped even if very late (I saw this only recently sorry...).
A: You can perhaps use the following (if I am not wrong):
We have:
$$\frac{t^{x-1}}{\exp(t-i \theta)-1}-\frac{t^{x-1}}{\exp(t)(\exp(-i\theta)-1)}=-\frac{\exp(t)-1}{t}\frac{t^x}{\exp(t)(\exp(t-i\theta)-1)(\exp(-i\theta)-1)}$$
Hence
$$\int_0^{+\infty}\frac{t^{x-1}}{\exp(t-i \theta)-1}=\frac{\Gamma(x)}{\exp(-i\theta)-1}-g(x)$$
and $g(x)$ is well defined (and analytic ) for ${\rm Re}(x)>-1$.
