Does an analytic continuation for a particular Leibniz series exist? Define a Leibniz series as follows,
\begin{eqnarray*}
L(x) & = & \sum_{k=1}^{\infty}(-1)^{k}e^{-kx}\ln k,\ \ x>0
\end{eqnarray*}
I have two questions: (I) Is there an explicit formula for $L(x)$?
(II) Does the analytic continuation for $L(x)$ from region $x>0$
to region $x<0$ exist?
 A: For $0 < |z| < 1$ and $\Re(s) > 1$ 
$$F(s,z) = \Gamma(s)\sum_{n=1}^\infty z^n n^{-s} =\int_0^\infty x^{s-1} \sum_{n=1}^\infty e^{-nx} z^n dx= \int_0^\infty \frac{x^{s-1}}{e^x/z-1}dx$$ 
$$= \frac{ x^s/s}{1/z-1}+\int_0^\infty x^{s-1} \left(\frac{1}{e^x/z-1}- \frac{1_{x < 1}}{1/z-1}\right)dx $$
And this integral is well-defined and analytic for  $z \in \mathbb{C} \setminus [1,\infty),\Re(s) > -1$ as well as $\frac{F(s,z)}{\Gamma(s)}$
Finally 
$$G(y) = \sum_{n=1}^\infty (-1)^n e^{-ny} \log n=\frac{\partial}{\partial s}\frac{-F(s,-e^{-y})}{\Gamma(s)}|_{s=0}$$
is analytic for $-e^{-y} \not \in [1,\infty)$.
If you want to move the branch cut, try a change of contour $$F(s,z) =  \int_0^\infty  \frac{(\gamma(t))^{s-1}}{e^{\gamma(t)}/z-1}\gamma'(t)dt$$
where $\gamma$ is a curve from $0 \to \infty$, it will move the branch cut  to $e^{-\gamma(.)}$.
A: Mathematica evaluates the series as
$$-\text{PolyLog}^{(1,0)}\left(0,-e^{-x}\right).$$
A: Igor Rivin already gave Mathematica's answer to the first question.  As to the second: yes, $\text{PolyLog}(p,z)$ has a logarithmic branch point at $z=1$ (corresponding to $x=n\pi i$ for odd $n$); you can analytically continue around these to the rest of the complex plane.  Of course the continuation is not unique: if you want a single-valued version, there will be branch cuts.  And similarly you can analytically continue the derivative with respect to $p$.
