Picking the start of convergence $c$ for an integral expression of the Dirichlet $\eta$-function. What happens when $c \in \mathbb{C}$?

Question

Using the following expression for the Dirichlet $\eta(s)$-function:

$$\normalsize \eta(s,c) = -\frac{1}{\Gamma(s-c)} \int_0^{\infty} x^{s-c-1}\,\text{Li}_c(-\text{e} ^{-x}) \mathop{dx}$$

one could pick the start of its convergence $c$ for any $c \in \mathbb{R}$. Note that $\text{Li}$ is the polylog-function.

Question (maybe trivial):

What actually happens to the convergence of the function, when $c \in \mathbb{C}$? Will convergence then start at its absolute value?

Answer 1

(a)

$$f_c(x)=\sum_{n\ge 1} (-1)^{n+1}n^{-c} e^{-nx}$$ is continuous on $x\ge 0$ with $f_c(0)=\eta(c)$.

(b) whence for $\eta(c)\ne 0$, as $f_c(x)$ has exponential decay as $x\to \infty$, $$\int_0^\infty x^{s-c-1} f_c(x)dx$$ converges and is analytic for $\Re(s)>\Re(c)$.

(a) follows from that $(1+e^x)^k f_c(x)=\sum_{n\ge 1} (-1)^{n+1}n^{-c-k} g_{c,k}(1/n)e^{-nx}$ where $g_{c,k}(t)$ is analytic at $0$ by induction on $k$.