Calculate $\int_0^\infty\frac{\sin^3{x}}{e^x-1}\mathrm dx$ 
Calculate $$\int_0^\infty\frac{\sin^3{x}}{e^x-1}\,\mathrm dx.$$

It seems that the integral cannot be solved in terms of elementary functions, so I try to use the Cauchy (residue) theorem to evaluate it. However, I couldn't find a complex function $$f(z) = \frac{?}{e^z-1}$$ to evaluate this real integral. If the numerator were $\sin{x}$, we can consider $e^{iz}$ since $$ \sin{z} = \frac{e^{iz}-e^{-iz}}{2i}.$$
Is there any hint or method to solve this problem?
 A: Note that $\sin^3 x = \frac34 \sin x- \frac14 \sin(3x)$, and for any $a\in\mathbb{R}$,
$$\begin{align}
\int_{0}^{\infty} \frac{\sin(ax)}{e^x - 1} dx
&= \int_{0}^{\infty} \frac{\sin (ax) e^{-x}}{1 - e^{-x}}  dx 
= \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} \sin(ax) \, e^{-nx} \right)  dx \\
&=\sum_{n=1}^{\infty} \int_{0}^{\infty} \sin (ax) \, e^{-nx} \; dx 
= \sum_{n=1}^{\infty} \frac{a}{n^2+a^2}.
\end{align}.$$
Hence
$$\int_0^\infty\frac{\sin^3 x}{e^x-1}dx=\frac{3}{4}\sum_{n=1}^{\infty} \left(\frac{1}{n^2+1}-\frac{1}{n^2+9}\right).$$
In order to find a closed formula see How to sum $\sum_{n=1}^{\infty} \frac{1}{n^2 + a^2}$?
A: $$\sin^3 x=\dfrac34\sin x-\dfrac14\sin3x$$
with zeta function definition $\displaystyle\int_0^\infty\dfrac{u^{x-1}}{e^u-1}du=\Gamma(x)\zeta(x)$, and $\sin$ expansion one may write
\begin{align}
\int_0^\infty\dfrac{\sin^3x}{e^x-1}dx
&= \dfrac34\int_0^\infty\dfrac{\sin x}{e^x-1}dx-\dfrac14\int_0^\infty\dfrac{\sin3x}{e^x-1}dx \\
&= \sum_{n=0}^\infty\dfrac34\dfrac{(-1)^n}{(2n+1)!}\int_0^\infty\dfrac{x^{2n+1}}{e^x-1}dx-\sum_{n=0}^\infty\dfrac14\dfrac{(-1)^n3^{2n+1}}{(2n+1)!}\int_0^\infty\dfrac{x^{2n+1}}{e^x-1}dx \\
&= -\dfrac{3}{4}\sum_{n=1}^\infty i^{2n}\zeta(2n)+\dfrac{1}{12}\sum_{n=1}^\infty (3i)^{2n}\zeta(2n) \\
&= -\dfrac{3}{4}\frac12\left(1-\pi i\cot\pi i\right) + \dfrac{1}{12}\frac12\left(1-3\pi i\cot3\pi i\right)\\
&= \color{blue}{-\dfrac13 + \dfrac{3}{8}\pi \coth\pi - \dfrac{1}{8}\pi \coth3\pi}
\end{align}
which we used $\displaystyle\sum_{n=1}^\infty x^{2n}\zeta(2n)=\dfrac12\left(1-\pi x\cot\pi x\right)$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Note that
  $\ds{\int_{0}^{\infty}{\sin^{3}\pars{x} \over \expo{x} - 1}\,\dd x =
{3 \over 4}\int_{0}^{\infty}{\sin\pars{x} \over \expo{x} - 1}\,\dd x -
{1 \over 4}\int_{0}^{\infty}{\sin\pars{3x} \over \expo{x} - 1}\,\dd x}$

Lets apply the Abel-Plana Formula to the sum $\ds{\sum_{n = 0}^{\infty}\expo{-2\pi an}}$ where $\ds{a > 0}$.
Note that $\ds{\expo{-2\pi a\,\Re\pars{z} - 2\pi a\,\Im\pars{z}\ic}\expo{-2\pi\verts{a}\verts{\Im\pars{z}}} \to 0}$ as
$\ds{\Im\pars{z} \to \pm\infty}$.
\begin{align}
\left.\sum_{n = 0}^{\infty}\expo{-2\pi an}\right\vert_{\ a\ >\ 0} & =
\int_{0}^{\infty}\expo{-2\pi ax}\dd x +
\left.{1 \over 2}\expo{-2\pi an}\right\vert_{\ n\ =\ 0} -
2\int_{0}^{\infty}{\Im\pars{\expo{-2\pi a\pars{\ic x}}} \over \expo{2\pi x} - 1}\,\dd x
\\[5mm]
\implies{1 \over 1 - \expo{-2\pi a}} & =
{1 \over 2\pi a} + {1 \over 2} +
2\int_{0}^{\infty}{\sin\pars{2\pi ax} \over \expo{2\pi x} - 1}\,\dd x
\\[5mm]
\implies{1 \over 1 - \expo{-2\pi a}} & =
{1 \over 2\pi a} + {1 \over 2} +
{1 \over \pi}\int_{0}^{\infty}{\sin\pars{ax} \over \expo{x} - 1}\,\dd x
\\[5mm]
\implies &
\bbx{\int_{0}^{\infty}{\sin\pars{ax} \over \expo{x} - 1}\,\dd x = 
{\pi a\coth\pars{\pi a} - 1 \over 2a}}
\end{align}

which leads to

$$
\bbx{\int_{0}^{\infty}{\sin^{3}\pars{x} \over \expo{x} - 1}\,\dd x =
{3 \over 8}\,\pi\coth\pars{\pi} - {1 \over 8}\,\pi\coth\pars{3\pi} - {1 \over 3}} \approx 0.4565
$$
A: $$
\sin^3 x = \frac 14\left(3\sin x-\sin(3x)\right)\\
\frac{1}{e^x-1} = e^{-x}\sum_{k=0}^{\infty}e^{-kx}\;\;\mbox{with }\;\; x > 0
$$
then
$$
\int_0^\infty\frac{\sin^3{x}}{e^x-1}\,\mathrm dx = \frac 14\int_0^{\infty}\left(\left(3\sin x-\sin(3x)\right)e^{-x}\sum_{k=0}^{\infty}e^{-kx}\right) \mathrm dx
$$
now adding 
$$
\frac 14\int_0^{\infty}\left(\left(3\cos x-\cos(3x)\right)e^{-x}\sum_{k=0}^{\infty}e^{-kx}\right) \mathrm dx+ i\left(\frac 14\int_0^{\infty}\left(\left(3\sin x-\sin(3x)\right)e^{-x}\sum_{k=0}^{\infty}e^{-kx}\right) \mathrm dx\right)
$$
we have
$$
I = \frac 14\int_0^{\infty}\left(3e^{ix}-e^{e^{i 3x}}\right) e^{-x}\sum_{k=0}^{\infty}e^{-kx} \mathrm dx
$$
or
$$
I = \frac 14\left(\int_0^{\infty}3\sum_{k=0}^{\infty}e^{-(k+1-i)x}\right)\mathrm dx - \frac 14\left(\int_0^{\infty}\sum_{k=0}^{\infty}e^{-(k+1-3i)x}\right)\mathrm dx 
$$
hence
$$
I = \frac 14\left(\sum_{k=0}^{\infty}\frac{3}{k+1-i}-\frac{1}{k+1-3i}\right) = \frac 14\sum_{k=0}^{\infty}\left(\frac{3(k+1)}{(k+1)^2+1}-\frac{k+1}{(k+1)^2+3^2}+i\left(\frac{24}{\left((k+1)^2+1\right) \left((k+1)^2+3^2\right)}\right)\right)
$$
and finally
$$
\int_0^\infty\frac{\sin^3{x}}{e^x-1}\,\mathrm dx =\frac 14\sum_{k=0}^{\infty}\left(\frac{24}{\left((k+1)^2+1\right) \left((k+1)^2+3^2\right)}\right) = \pi  \cosh ^3(\pi ) \text{csch}(3 \pi )-\frac{19}{30}
$$
