Evaluating $\int_0^{\infty} \frac{\xi x^{\alpha}}{ e^{x}-\xi} \:\mathrm{d}x$ I am supposed to integrate for $\alpha \ge 0$
$$\int_0^{\infty} \frac{x^{\alpha}}{ \xi^{-1}e^x - 1} \:\mathrm{d}x,$$
where $\xi e^{-x} < 1$ which means, I want to express this in terms of simple functions (like the Gamma Function).
By using the geometric series I ended up with 
$$\Gamma( \alpha+1) \sum_{n=1}^{\infty} \frac{\xi^n}{n^{\alpha + 1}},$$
but I am not sure if this is correct.
I.e.  I need an explicit way to write down this integral for $\alpha =3$, but I don't know how to simplify this.
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You did it right because

\begin{align}&\color{#66f}{\large%
\int_0^{\infty}{x^{\alpha} \over \xi^{-1}\expo{x} - 1}\,\dd x}
=\xi\int_0^{\infty}{x^{\alpha}\expo{-x} \over 1 - \xi\expo{-x}}\,\dd x
=\xi\sum_{n\ =\ 0}^{\infty}\xi^{n}\int_0^{\infty}x^{\alpha}\expo{-\pars{n + 1}x} \,\dd x
\\[5mm]&=\xi\sum_{n\ =\ 0}^{\infty}
{\xi^{n} \over \pars{n + 1}^{\alpha + 1}}\int_0^{\infty}x^{\alpha}\expo{-x} \,\dd x
=\sum_{n\ =\ 1}^{\infty}{\xi^{n} \over n^{\alpha + 1}}\Gamma\pars{\alpha + 1}
=\Gamma\pars{\alpha + 1}\sum_{n\ =\ 1}^{\infty}{\xi^{n} \over n^{\alpha + 1}}
\\[5mm]&=\color{#66f}{\large\Gamma\pars{\alpha + 1}\Li{\alpha + 1}\pars{\xi}}\,,
\qquad\verts{\xi} < 1.
\end{align}

For $\ds{\alpha = 3}$, the result is
  $\ds{\Gamma\pars{3 + 1}\Li{3 + 1}\pars{\xi}=6\Li{4}\pars{\xi}}$. I guess that's all you can do about it.

