Integration of a polylogarithm: Is this function known?

Question

I would like to integrate a polylogarithm of a given order $$\int dx \mbox{Li}_{n-1}(x)$$ suppose that the order is $n\le 0$ and $x\in(-\infty,0]$, so the function is bounded. I know that it can be integrated by using the general rule for the derivative $\frac{d}{dx}\mbox{Li}_n(x)=\frac{1}{x}\mbox{Li}_{n-1}(x)$. So, we can write the integral as $$\int dx \frac{\mbox{Li}_{n-1}(x)}{x}x=\mbox{Li}_n(x)-\int dx\mbox{Li}_n(x),$$ integrating by parts. It can be integrated again by parts, finding in general $$\int dx \mbox{Li}_{n-1}(x)dx=\sum_{k=0}^{\infty}(-1)^k\mbox{Li}_{n+k}(x),$$

Expansion in series: Finally, the polylogarithms can be expanded in series (http://mathworld.wolfram.com/Polylogarithm.html) as $$\mbox{Li}_n(x)=\sum_{j=1}^{\infty}\frac{z^j}{j^n}$$ Substituing in the expression for the integral $$\int dx \mbox{Li}_{n-1}(x)=\sum_{j=1}^{\infty}\sum_{k=0}^{\infty}(-1)^k\frac{z^j}{j^{n+k}}$$ Can these series be summed up? Are they known? Thanks for any help!

Answer 1

$\int\text{Li}_{n-1}(x)~dx$

$=\int\sum\limits_{k=1}^{\infty}\dfrac{x^k}{k^{n-1}}~dx$

$=\sum\limits_{k=1}^{\infty}\dfrac{x^{k+1}}{k^{n-1}(k+1)}+C$

Even e.g. http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/21/01/01 can only get this form, unless when $n=3$ .