Integral $\int_0^\infty \frac{x^n}{(x^2+\alpha^2)^2(e^x-1)^2}dx$ Hey I am trying to integrate
$$
I_n:=\int_0^\infty \frac{x^n}{(x^2+\alpha^2)^2(e^x-1)^2}dx,\quad \alpha,n \geq 1.
$$
Thanks.
This integral is old. I am also looking for literature on these integrals as I have seen many for small values of n, and variations of this. Thanks. Maybe we can use residues. How can I make a contour with the $x^n$ piece involved with the possible quadrupole pole? Thanks
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$\ds{I_{n}\equiv\int_{0}^{\infty}
    {x^{n}\,\dd x \over \pars{x^{2} + \alpha^{2}}^{2}\pars{\expo{x} - 1}^{2}}\,,
     \qquad \alpha\,, n\ \geq\ 1}$.

\begin{align}
I_{n}&=-\,{1 \over 2\alpha}\,\partiald{}{\alpha}\int_{0}^{\infty}
{x^{n}\,\dd x \over \pars{x^{2} + \alpha^{2}}\pars{\expo{x} - 1}^{2}}
\\[3mm]&=-\,{1 \over 2\alpha}\,\partiald{}{\alpha}\int_{0}^{\infty}
{1 \over 2\alpha\ic}\pars{{1 \over x - \alpha\ic} - {1 \over x + \alpha\ic}}{x^{n}\,\dd x \over \pars{\expo{x} - 1}^{2}}
\\[3mm]&={1 \over 2\alpha}\,\partiald{}{\alpha}\bracks{{1 \over \alpha}\,
\Im\int_{0}^{\infty}
{1 \over x + \alpha\ic}{x^{n}\,\dd x \over \pars{\expo{x} - 1}^{2}}}
\\[3mm]&=-\,{1 \over 2\alpha}\sum_{\ell = 1}^{\infty}\ell\,\partiald{}{\alpha}\bracks{{1 \over \alpha^{2}}\,\Re\int_{0}^{\infty}
\pars{1 + {x \over \alpha\ic}}^{-1}x^{n}\expo{-\pars{\ell + 2}x}\,\dd x}
\\[3mm]&=-\,{1 \over 2\alpha}\sum_{\ell = 1}^{\infty}
{\ell \over \pars{\ell + 2}^{n + 1}}\,\partiald{}{\alpha}\bracks{%
{1 \over \alpha^{2}}\,\Re\int_{0}^{\infty}
\bracks{1 + {x \over \pars{\ell + 2}\alpha\ic}}^{-1}x^{n}\expo{-x}\,\dd x}
\end{align}

\begin{align}\color{#44f}{\large%
I_{n}}&=\color{#44f}{\large%
-\,{1 \over 2\alpha}\,\Gamma\pars{n + {3 \over 2}}\sum_{\ell = 1}^{\infty}
{\ell \over \pars{\ell + 2}^{3n/2 + 2}}\times}
\\[3mm]&\color{#44f}{\large\partiald{}{\alpha}\pars{%
{1 \over \alpha^{2}}\,\Re\braces{\pars{\alpha\ic}^{-n/2 - 1}\expo{\pars{\ell + 2}\alpha\ic}{\rm W}_{-n/2 - 1,n/2}\pars{\bracks{\ell + 2}\alpha\ic}}}}
\end{align}
with
$$
n \geq -\,{3 \over 2}\qquad\mbox{and}\qquad
\pars{-n - {3 \over 2}} \not\in {\mathbb Z}
$$

$\ds{{\rm W}_{k,m}\pars{z}}$ is the Whittaker Function ( see definition $\pars{5}$ in that link ). $\ds{\Gamma\pars{z}}$ is the
  Gamma Function ${\bf\mbox{6.1.1}}$.

