# 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

• In order to converge, $\large n > 1$. – Felix Marin Apr 10 '14 at 3:50
• @FelixMarin Yes, thanks. I made it clearer – Jeff Faraci Apr 10 '14 at 3:57
• Special case, $n=2$. Is there any chance of using contour integration techniques to evaluate $$\int_0^\infty \frac{x}{(\beta x^2+\alpha^2)(e^x-1)^2}dx,$$ and differentiating the result w.r.t $\beta$? Setting $\beta=1$ would give $2I_2$, where $I_n$ is the integral you are interested in. – Bennett Gardiner May 6 '14 at 2:30
• @BennettGardiner When $\large x \sim 0$, the integrand behaves as $\large x^{n - 2}$ $\large\left(~\mbox{with}\ \alpha \not= 0~\right)$. Then, $\large n - 2 > -1$ and, finally, $\large n > 1$. Similarly, when $\large\alpha = 0$, $\large n > 5$. – Felix Marin May 6 '14 at 4:24
• I see, so more generally $$I_n:=\int_0^\infty \frac{x^n}{(x^2+\alpha^2)^p(e^x-1)^m}dx$$ converges only for $n>m-1$? – Bennett Gardiner May 6 '14 at 4:29

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I_{n}\equiv\int_{0}^{\infty} {x^{n}\,\dd x \over \pars{x^{2} + \alpha^{2}}^{2}\pars{\expo{x} - 1}^{2}}\,, \qquad \alpha\,, n\ \geq\ 1}$.
\begin{align}\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 ${\bf\mbox{6.1.1}}$.