Show that $\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi \alpha)}$ 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!
 A: 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.
A: I have finally figured out how to use Ramanujans Master Theorem in this case.

Ramanujans Master Theorem
Let $f(x)$ be an analytic function with a MacLaurin Expansion of the form
$$f(x)=\sum_{k=0}^{\infty}\frac{\phi(k)}{k!}(-x)^k$$then the Mellin Transform of this function is given by
$$\int_0^{\infty}x^{s-1}f(x)\mathrm dx=\Gamma(s)\phi(-s)$$

Lets get back to the given integral. The series representation of  $\operatorname{Li}_s(-x)$ is given by $\displaystyle\sum_{k=1}^{\infty}\frac{(-x)^k}{k^s}$ and therefore, by using the Gamma Function as extension of the factorial, we obtain $\displaystyle\phi(k)=\frac{\Gamma(k+1)}{k^s}$. Now applying the Mellin Transform with $s=-\alpha$ yields to
$$\int_0^{\infty}x^{-\alpha-1}\operatorname{Li}_s(-x)\mathrm dx=\Gamma(-\alpha)\phi(\alpha)=\Gamma(-\alpha)\frac{\Gamma(\alpha+1)}{\alpha^s}$$
This term can be simplified by using Eulers Reflection Formula with $z=\alpha+1$ which finally leads to
$$\frac1{\alpha^s}\Gamma(\alpha+1)\Gamma(-\alpha)=\frac1{\alpha^s}\frac{\pi}{\sin(\pi(\alpha+1))}=-\frac1{\alpha^s}\frac{\pi}{\sin(\pi\alpha)}$$
A: A contour integration approach:
For $s >0$,  $\operatorname{Li}_{s}(-z)$ has a branch cut on $(-\infty, -1]$.
On the upper side of the branch cut, the imaginary part of $\operatorname{Li}_{s}(-x) $ is $-\frac{\pi}{\Gamma(s)}\log^{s-1}(-x)$, while on the lower side of the branch cut, the imaginary part of $\operatorname{Li}_{s}(-x) $ is $\frac{\pi }{\Gamma(s)}\log^{s-1}(-x) $.
(See here and here.)
For the function $\frac{1}{z^{\alpha+1}}$, let's place the branch cut on the positive real axis, and define $\frac{1}{z^{\alpha+1}}$ to be real-valued on the upper side of the branch cut.
Let's then integrate the function $$ f(z) = \frac{\operatorname{Li}_{s}(-z)}{z^{\alpha +1}}, \quad ( s>0, \, 0 < \alpha <1),$$  around a double keyhole contour that is deformed around the two branch cuts.
The contour looks like the contour used here.
There are no singularities inside the contour, and the branch points at $z=0$ and $z=-1$ are integrable.
Furthermore, $f(z) \sim  - \frac{1}{\Gamma(s+1)} \frac{\ln^{s}(z)}{z^{\alpha +1}} $ as $ |z| \to \infty$ .  (See here.)
Therefore, letting the radii of the two big semicircles go to infinity (and the radii of the two small circles go to zero), we get
$$\begin{align} \oint f(z) \, \mathrm dz &=\int_{0}^{\infty} \frac{\operatorname{Li}_{s}(-x)}{x^{\alpha+1}} \, \mathrm dx + \int_{\infty}^{1} \frac{\Re \left(\operatorname{Li}_{s}(t) \right) - \frac{\pi i}{\Gamma(s)}\ln^{s-1} (t)}{(te^{\pi i})^{\alpha +1}} \,  (- \mathrm dt) \\ &+ \int_{1}^{\infty} \frac{\Re \left(\operatorname{Li}_{s}(t) \right) + \frac{\pi i}{\Gamma(s)} \ln^{s-1} (t)}{(te^{\pi i})^{\alpha +1}} (-\mathrm dt) \ +\int_{\infty}^{0} \frac{\operatorname{Li}_{s}(-t)}{(te^{2 \pi i})^{\alpha+1}} \, \mathrm dt =0 , \end{align}$$
from which it follows that
$$ \begin{align} \int_{0}^{\infty}  \frac{\operatorname{Li}_{s}(-x)}{x^{\alpha+1}} \, \mathrm dx  &= \frac{1 }{\Gamma(s)}\frac{ 2 \pi i \, e^{- \pi i (\alpha +1) } }{1-e^{-2 \pi i(\alpha+1)}} \int_{1}^{\infty}  \frac{\ln^{s-1}(t)}{t^{\alpha+1}} \, \mathrm dt \\ &=\frac{1}{\Gamma(s)}\frac{ \pi}{\sin \left(\pi(\alpha +1)\right)} \int_{0}^{\infty} u^{s-1}e^{- \alpha u} \, \mathrm du \\ &= -\frac{1}{\Gamma(s)}\frac{ \pi}{\sin(\pi \alpha)} \frac{1}{\alpha^s}\int_{0}^{\infty} v^{s-1} e^{-v} \, \mathrm d v \\ &= -\frac{ \pi}{\sin(\pi \alpha)} \frac{1}{\alpha^s}. \end{align}$$
A: $$\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{a+1}}\mathrm dx=\int_0^{\infty}\frac{1}{x^{a}}\left(\frac{\operatorname{Li}_s(-x)}{x}\right)\mathrm dx$$
$$=\int_0^{\infty}\frac{1}{x^{a}}\left(\frac{(-1)^{s}}{(s-1)!}\int_0^1\frac{\ln^{s-1}(y)}{1+yx}\mathrm  dy\right)\mathrm dx$$
$$=\frac{(-1)^{s}}{(s-1)!}\int_0^1 \ln^{s-1}(y)\left(\int_0^\infty \frac{x^{-a}}{1+yx}\mathrm dx\right)\mathrm dy$$
$$=\frac{(-1)^{s}}{(s-1)!}\int_0^1 \ln^{s-1}(y)\left(y^{a-1}\int_0^\infty \frac{t^{-a}}{1+t}\mathrm dt\right)\mathrm dy$$
$$=\frac{(-1)^{s}}{(s-1)!}\int_0^1 \ln^{s-1}(y)\left(y^{a-1}\pi \csc(\pi(1-a))\right)\mathrm dy$$
$$=\frac{(-1)^{s}}{(s-1)!}\pi\csc(a\pi)\int_0^1 \ln^{s-1}(y)\, y^{a-1}\mathrm dy$$
$$=\frac{(-1)^{s}}{(s-1)!}\pi\csc(a\pi)\frac{(-1)^{s-1}(s-1)!}{a^s}$$
$$=-\pi\, a^{-s}\csc(a\pi).$$

Using this approach, we also find:
$$\int_0^\infty\frac{\operatorname{Li}_a(-x)}{1+x^2}\mathrm{d}x=(2^{-2a}-2^{-a-1})\pi\zeta(a)-a\beta(a+1);$$
$$\int_0^1\frac{\operatorname{Li}_{2a+1}(-x)}{1+x^2}\mathrm{d}x=\frac{4^{-2a}-4^{-a}}{8}\pi\zeta(2a+1)-\frac{2a+1}{2}\beta(2a+2)$$
$$+\sum_{n=0}^a \eta(2a-2n)\beta(2n+2);$$
$$\int_0^\infty\frac{\operatorname{Li}_a(-x^2)}{1+x^2}\mathrm{d}x=(1-2^{a-1})\pi\zeta(a);$$
$$\int_0^1\frac{\operatorname{Li}_{2a+1}(-x^2)}{1+x^2}\mathrm{d}x=\frac{1-4^a}{2}\pi\zeta(2a+1)+2\sum_{n=0}^a 4^{n}\eta(2a-2n)\beta(2n+2).$$
A: With $\text{Li}’_s(-x)=\frac{1}x \text{Li}_{s-1}(-x) $, integrate by parts $s$ times
$$\int_0^{\infty}\frac{\operatorname{Li}_s(-x)}{x^{\alpha+1}}\mathrm dx \overset{ibp}=- \frac1{\alpha^s }\int_0^\infty \frac{x^{-\alpha}}{1+x}dx =-\frac1{\alpha^s}\frac{\pi}{\sin\pi \alpha}
$$
