I have come across the following integral while going over this list (Problem $35$)
$$\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi \alpha)}~~~~s>0, \alpha\in(0,1)$$ where $\operatorname{Li}_s(x)$ denotes the Polylogarithm Function.
Using the series expansion of $\operatorname{Li}_s(-x)$ yields
$$\begin{align} \int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx&=\int_0^{\infty}\frac1{x^{\alpha+1}}\left[\sum_{n=1}^{\infty}\frac{(-x)^n}{n^s}\right]\mathrm dx\\ &=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^s}\int_0^{\infty}\frac{x^n}{x^{\alpha+1}}\mathrm dx\\ &=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^s}\int_0^{\infty}x^{n-\alpha-1}\mathrm dx \end{align}$$
One can easily see the problems concerning the convergence of the last integral. Also, I am not even sure whether it is possible to change the order of summation and integration in this case or not.
Another approach is based on an integral representation of $\operatorname{Li}_s(-x)$ so that the given integral becomes
$$\begin{align} \int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx&=\int_0^{\infty}\frac1{x^{\alpha+1}}\left[\frac1{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}}{e^t/(-x)-1}\mathrm dt\right]\mathrm dx\\ &=-\frac1{\Gamma(s)}\int_0^{\infty}\int_0^{\infty}\frac{t^{s-1}}{x^{\alpha}(e^t+x)}\mathrm dx\mathrm dt\\ \end{align}$$
From hereon, I do not know how to proceed. Since the solution reminds me of Euler's Reflection Formula it might be possible to reshape the integral in terms of the Gamma Function somehow.
I am asking for a whole evaluation of the given integral. I did not found anything closely related to this question but correct me if I am wrong.
Thanks in advance!