TITLE: Find the value of $\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$
QUESTION [40 upvotes]: I need to find a closed-form for the following integral. Please give me some ideas how to approach it:
$$\int_0^{\infty}\frac{x^3}{(x^4+1)(e^x-1)}\mathrm dx$$

REPLY [8 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\int_{0}^{\infty}{x^{3} \over \pars{x^{4} + 1}\pars{\expo{x} - 1}}\,\dd x:
     \ {\large ?}}$

\begin{align}&\color{#c00000}{%
\int_{0}^{\infty}{x^{3} \over \pars{x^{4} + 1}\pars{\expo{x} - 1}}\,\dd x}
=\half\int_{0}^{\infty}\pars{{1 \over x^{2} + \ic} + {1 \over x^{2} - \ic}}\,
{x \over \expo{x} - 1}\,\dd x
\\[3mm]&=\Re\int_{0}^{\infty}{x\,\dd x \over \pars{x^{2} + \ic}\pars{\expo{x} - 1}}
=\Re\int_{0}^{\infty}{x\,\dd x
\over \bracks{x^{2} + \ic\,/\pars{4\pi^{2}}}\pars{\expo{2\pi x} - 1}}\tag{1}
\end{align}

The last integral in $\pars{1}$ is related to the Digamma function $\ds{\Psi\pars{z}}$ by means of the identity ${\bf\mbox{6.3.21}}$:
$$
\Psi\pars{z} = \ln\pars{z} - {1 \over 2z}
-2\int_{0}^{\infty}
{t\,\dd t \over \pars{t^{2} + z^{2}}\pars{\expo{2\pi t} - 1}}\,,\qquad
\verts{{\rm arg}\pars{z}} < {\pi \over 2}
$$

Then, $\pars{1}$ is reduced to
  $\ds{\pars{~\mbox{with}\ \root{\ic} = \expo{\ic\pi/4}
={1 \over \root{2}}\pars{1 + \ic}~}}$:
  \begin{align}
&\color{#44f}{\large%
\!\!\!\!\!\int_{0}^{\infty}{x^{3} \over \pars{x^{4} + 1}\pars{\expo{x} - 1}}\,\dd x}
=\Re\bracks{-\,\half\,\Psi\pars{\root{\ic} \over 2\pi}
+\half\,\ln\pars{\root{\ic} \over 2\pi} - {1 \over 4}\,{2\pi \over \root{\ic}}}
\\[3mm]&=\color{#44f}{\large%
-\,\half\,\Re\Psi\pars{\root{\ic} \over 2\pi} -\half\,\ln\pars{2\pi}
-{\root{2} \over 4}\,\pi} \approx 0.3891
\end{align}