Series around $s=1$ for $F(s)=\int_{1}^{\infty}\text{Li}(x)\,x^{-s-1}\,dx$ 
Consider the function $$F(s)=\int_{1}^{\infty}\frac{\text{Li}(x)}{x^{s+1}}dx$$ where $\text{Li}(x)=\int_2^x \frac{1}{\log t}dt$ is the logarithmic integral.  What is the series expansion around $s=1$?

It has a logarithmic singularity at $s=1$, and I am fairly certain (although I cannot prove it) that it should expand as something of the form $$\log (1-s)+\sum_{n=0}^\infty a_n (s-1)^n.$$  (An expansion of the above form is what I am looking for)  I also have a guess that the constant term is $\pm \gamma$ where $\gamma$ is Euler's constant.  Does anyone know a concrete way to work out such an expansion?
Thanks!

Note that the integral $F(s)$ diverges at infinity for $s\leqslant1$ and redefine $F(s)$ for every $s\gt1$ as
$$
F(s)=\int_2^{+\infty}\frac{\text{Li}(x)}{x^{s+1}}\mathrm dx.
$$
An integration by parts yields
$$
sF(s)=\int_2^{+\infty}\frac{\mathrm dx}{x^s\log x},
$$
and the change of variable $x^{s-1}=\mathrm e^t$ yields $$sF(s)=\mathrm{E}_1(u\log2),\qquad u=s-1,$$ where the exponential integral function $\mathrm{E}_1$ is defined, for every complex $z$ not a nonpositive real number, by 
$$
\mathrm{E}_1(z)=\int_z^{+\infty}\mathrm e^{-t}\frac{\mathrm dt}t.
$$
One knows that, for every such $z$, 
$$
\mathrm{E}_1(z) = -\gamma-\log z-\sum\limits_{k=1}^\infty \frac{(-z)^{k}}{k\,k!}.
$$
On the other hand, $$\frac1s=\frac1{1+u}=\sum_{n\geqslant0}(-1)^nu^n,$$ hence $$F(s)=\frac1s\mathrm{E}_1(u\log2)=\sum_{n\geqslant0}(-1)^nu^n\cdot\left(-\gamma-\log\log2-\log u-\sum\limits_{k=1}^\infty \frac{(-1)^{k}(\log2)^k}{k\,k!}u^k\right).
$$
One sees that $F(1+u)$ coincides with a series in $u^n$ and $u^n\log u$ for nonnegative $n$, and that $G(u)=F(1+u)+\log u$ is such that $$G(0)=-\gamma-\log\log2.$$
Finally, 
$$
F(s)=-\gamma-\log\log2-\log(s-1)-\sum\limits_{n=1}^{+\infty}(-1)^{n}(s-1)^n\log(s-1)+\sum\limits_{n=1}^{+\infty}c_n(s-1)^n,
$$
for some coefficients $(c_n)_{n\geqslant1}$. Due to the logarithmic terms, this is a slightly more complicated expansion than the one suggested in the question, in particular $s\mapsto G(s-1)=F(s)+\log(s-1)$ is not analytic around $s=1$.
