I have to calculate analytically this integral:
$$
{\rm J}\left(q\right)
=
\int\_{0}^{\infty}{{\rm d}x \over x^{q}\left({\rm e}^{kx}-1\right)}
$$
where $-1\le q\le N$
with:
$N\in\mathbb{N}$ and $q\in\mathbb{N}$, $k\le 5\times10^{-5}$
I didn't find anything on the Gradshteyn Ryzhik and Mathematica isn't able to integrate it. Is it possible to make some approximation because the little value of $k$? Thanks in advance.



Definite integral of a function containing an exponential
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\begin{align}
\color{#0000ff}{\large{\rm J}\left(q\right)}
&=
\int\_{0}^{\infty}{{\rm d}x \over x^{q}\left({\rm e}^{kx}-1\right)}
=
k^{q - 1}\int\_{0}^{\infty}{\expo{-x} \over x^{q}\pars{1 - \expo{-x}}}\,\dd x
=
k^{q - 1}\int\_{0}^{\infty}{\expo{-x} \over x^{q}}
\sum\_{\ell = 0}^{\infty}\expo{-\ell x}\,\dd x
\\[3mm]&=k^{q - 1}\sum\_{\ell = 0}^{\infty}\int\_{0}^{\infty}x^{-q}\expo{-\pars{\ell + 1}x}
\,\dd x
=k^{q - 1}\sum\_{\ell = 0}^{\infty}{1 \over \pars{\ell + 1}^{1 - q}}
\int\_{0}^{\infty}x^{-q}\expo{-x}\,\dd x
\\[3mm]&=
\color{#0000ff}{\large k^{q - 1}\zeta\pars{1 - q}\Gamma\pars{1 - q}}
\end{align}
The series converge when $1 - q > 1\quad\imp\quad \color{#0000ff}{\large q < 0 }$


