Explicit calculation of $\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x$ Is it possible to confirm the value of this integral using the methods of complex analysis or similar? 
$$
\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x=\frac{\pi^2-9}{12}
$$
Of course, one can reduce it to the definition of the polylogarithm and a $\zeta$-function, but I was looking for an explicit calculation. 
 A: This will require some calculations but no special functions. Making the change of variables $x = -x$ on $\mathbb R^-$, we obtain
$$\int_{\mathbb R} \frac x {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx =
 \underbrace {\int_{\mathbb R^+} \frac {2 x} {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx}_
  {= I} +
 \int_{\mathbb R^+} \frac x {(x^2 + 1)^2} dx, \\
I = 2 \pi i \sum_{k \geq 1}
 \operatorname*{Res}_{x = i k} \frac x {(x^2 + 1)^2 (e^{2 \pi x} - 1)} +
 \frac 1 {2 (x^2 + 1)} \bigg\rvert_{x = 0}^\infty = \\
2 \pi i \frac {4 \pi^2 + 3} {96 \pi i} -
 2 \pi i \sum_{k \geq 2} \frac k {2 \pi i (k^2 - 1)^2} -
 \frac 1 2 = \\
\frac {4 \pi^3 - 21} {48} - \frac 1 4 \sum_{k \geq 2}
 \left( \frac 1 {(k - 1)^2} - \frac 1 {(k + 1)^2} \right) = \\
\frac {4 \pi^3 - 21} {48} - \frac 1 4 \left( 1 + \frac 1 4 \right) =
\frac {\pi^2 - 9} {12}.$$
A: This approach is unnecessarily complicated, but quite fun:
As explained in the comments and the other answers, we may write
$$ I \equiv \int \limits_0^\infty \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)} \, \mathrm{d}x = \int \limits_0^\infty \frac{2 x}{(1+x^2)(\mathrm{e}^{2\pi x}-1)} \, \mathrm{d} x \, . $$
We can then use the geometric series, integration by parts and this Laplace transform to obtain
\begin{align}
I &= \sum \limits_{n=1}^\infty \int \limits_0^\infty \frac{2 x}{(1+x^2)^2} \mathrm{e}^{-2 \pi n x} \, \mathrm{d} x = \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \int \limits_0^\infty \frac{\mathrm{e}^{-2 \pi n x}}{1+x^2} \, \mathrm{d} x \right] \\
&= \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \left(\frac{\pi}{2} - \operatorname{Si}(2 \pi n)\right)\right] \, ,
\end{align}
where $\operatorname{Si}$ is the sine integral . Plugging in its definition and integrating by parts twice more we arrive at
$$ I = \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \int \limits_1^\infty \frac{\sin(2 \pi n t)}{t} \, \mathrm{d} t \right] = 2 \sum \limits_{n=1}^\infty \frac{1}{2 \pi n} \int \limits_1^\infty \frac{\sin(2 \pi n t)}{t^3} \, \mathrm{d} t \, .$$
Now the Fourier series
$$ 1 - 2 \{t\} = 4 \sum \limits_{n=1}^\infty \frac{\sin(2 \pi n t)}{2 \pi n} \, , \, t \in \mathbb{R} \, , $$
and the fractional part integral
$$ \int \limits_0^1 x \left\{\frac{1}{x}\right\} \, \mathrm{d} t = \int \limits_1^\infty \frac{\{t\}}{t^3} \, \mathrm{d} t = 1 - \frac{\zeta(2)}{2} = 1 - \frac{\pi^2}{12} $$
are sufficient to derive the final result
$$ I = \frac{1}{2} \int \limits_1^\infty \frac{1 - 2\{t\}}{t^3} \, \mathrm{d} t = \frac{\pi^2}{12} - \frac{3}{4} = \frac{\pi^2 - 9}{12} $$
and thus prove $\pi > 3$ in a rather convoluted manner.
A: Not an answer but I believe that the form below is better for contour integration $$\int_0^\infty \frac{\sin(2\arctan(x))}{(1+x^2)(e^{2\pi x} -1)}dx=2\int_0^\infty \frac{\sin(\arctan(x))\cos(\arctan(x))}{(1+x^2)(e^{2 \pi x} -1)}dx=2\int_0^\infty \frac{x}{(1+x^2)^2 (e^{2 \pi x} -1)}dx$$ Now all that remains is to find a suitable contour that includes $[0,\infty)$, and apply residue theorem.
A: Here is an approach that first converts the integral to a double integral. 
Let
$$I = \int_0^\infty \frac{\sin (2 \tan^{-1} x)}{(1 + x^2) (e^{2 \pi x} - 1)} \, dx = 2 \int_0^\infty \frac{x}{(1 + x^2)^2 (e^{2 \pi x} - 1)} \, dx.$$
Observe that
$$\frac{1}{2} \int_0^\infty y e^{-y} \sin (xy) \, dy = \frac{x}{(1 + x^2)^2}.$$
Thus the integral can be rewritten as
$$I = \int_0^\infty y e^{-y} \int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx \, dy, \tag1$$
after a change of order has been made.
For the inner $x$-integral 
\begin{align}
\int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx &= \int_0^\infty \frac{e^{-2 \pi x} \sin (xy)}{1 - e^{-2 \pi x}} \, dx\\
&= \int_0^\infty \sum_{n = 0}^\infty e^{-2 \pi x} \sin (xy) \cdot e^{-2\pi n x} \, dx\\
&= \sum_{n = 1}^\infty \int_0^\infty e^{-2\pi n x} \sin (xy) \, dx,
\end{align}
where a shift in the index of $n \mapsto n - 1$ has been made. Now integrating by parts twice leads to
$$\int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx = \sum_{n = 1}^\infty \frac{y}{4 \pi^2 n^2 + y^2} = \frac{1}{2 \pi} \sum_{n = 1}^\infty \frac{\frac{y}{2\pi}}{\left (\frac{y}{2 \pi} \right )^2 + n^2}. \tag2$$
Using the well-known result of
$$\pi \coth (\pi z) = \frac{1}{z} + 2 \sum_{n = 1}^\infty \frac{z}{z^2 + n^2} , \quad z \neq 0,$$
the sum in (2) can be expressed as
$$\int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx = \frac{1}{4} \coth \left (\frac{y}{2} \right ) - \frac{1}{2y}.$$
On returning to our integral in (1), we have
\begin{align}
I &= -\frac{1}{2} \int_0^\infty e^{-y} + \frac{1}{4} \int_0^\infty y e^{-y} \coth \left (\frac{y}{2} \right ) \, dy = -\frac{1}{2} + \frac{1}{4} J,
\end{align}
where
$$J = \int_0^\infty y e^{-y} \coth \left (\frac{y}{2} \right ) \, dy.$$
Finding $J$ we have
\begin{align}
J &= \int_0^\infty y e^{-y} \frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}} \, dy\\
&= \int_0^\infty y e^{-y} \frac{1 + e^{-y}}{1 - e^{-y}} \, dy\\
&= \int_0^\infty y e^{-y} \left (1 + \frac{2 e^{-y}}{1 - e^{-y}} \right ) \, dy\\
&= \int_0^\infty y e^{-y} \, dy + 2 \int_0^\infty \frac{y e^{-2y}}{1 - e^{-y}} \, dy\\
&= 1 + 2 \sum_{n = 0}^\infty \int_0^\infty y e^{-y(n + 2)} \, dy\\
&= 1 + 2 \sum_{n = 0}^\infty \frac{1}{(n + 2)^2} \qquad \text{(by parts)}\\
&= 1 + 2 \sum_{n = 2}^\infty \frac{1}{n^2} \\
&= 1 + 2 \sum_{n = 1}^\infty \frac{1}{n^2} - 2\\
&= -1 + 2 \cdot \frac{\pi^2}{6}\\
&= -1 + \frac{\pi^2}{3}.
\end{align}
So finally we have
$$I = -\frac{1}{2} + \frac{1}{4} \left (-1 + \frac{\pi^2}{3} \right ),$$
or
$$\int_0^\infty \frac{\sin (2 \tan^{-1} x)}{(1 + x^2) (e^{2 \pi x} - 1)} \, dx = \frac{\pi^2 - 9}{12},$$
as expected. 
