How to solve this series with a log on the denominator?

So I stumbled upon a series I can't quite solve:

$$S=\sum_{n=2}^\infty\frac{(-1)^n}{n\ln(n)}\approx0.526418$$

Notice that if we generalize this:

$$S(x)=\sum_{n=2}^\infty\frac{(-1)^n}{n^x\ln(n)}$$

and differentiate with respect to $x$, we get

$$S'(x)=\sum_{n=2}^\infty\frac{(-1)^{n+1}}{n^x}=\eta(x)-1$$

where $\eta$ is the Dirichlet eta function. It then seems natural to try and integrate backwards:

$$S(x)=S(x_0)+x_0-x+\int_{x_0}^x\eta(x)\ dx$$

But I'm not sure how to compute this into some sort of closed form.

Also accepting any proofs that this series doesn't have a closed form.

By closed form, I mean something involving well-known constants such as $\pi,e,\gamma$, the Gamma function, the Reimann zeta function, and solutions to algebraic differential equations with algebraic initial conditions.

Solution 1:

Clearly this series has an integral representation: \begin{equation} S = \int\limits_0^\infty \left[1+ Li_{1+\theta}(-1)\right] d\theta = 0.526412246533310410930696501411\dots \end{equation} Now using the integral representation of the poly-logarithm: \begin{equation} Li_{1+\theta}(-1) = -\int\limits_0^\infty \frac{t^\theta}{\theta!(1+\exp(t))} dt \end{equation} and changing order of integration might be of some help.

On the other hand the original series converges quite slowly. After taking the first ten thousand terms only the first five digits are correct.