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!
Best Answer
You're just one step away: for $b>0$, we have $$\int_0^\infty {\frac{1}{{{x^\alpha }(b + x)}}dx} = {b^{ - \alpha }}\int_0^\infty {\frac{{{x^{ - \alpha }}}}{{1 + x}}dx} = \frac{{{b^{ - \alpha }}\pi }}{{\sin \alpha \pi }}$$ so $$\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}dx=-\frac{1}{\Gamma(s)}\int_0^{\infty}\int_0^{\infty}\frac{t^{s-1}}{x^{\alpha}(e^t+x)}dxdt = - \frac{\pi }{{\Gamma (s)\sin \alpha \pi }}\int_0^\infty {{t^{s - 1}}{e^{ - \alpha t}}dt} $$
The series expansion of $\text{Li}_s(-x)$ only converges for $|x| < 1$, so your first method is illegitimate.