Compute $I=\int_0^{+\infty}\frac{\arctan(t)}{e^{\pi t}-1}dt$ 
I would like to compute $\displaystyle I=\int_0^{+\infty}\frac{\arctan(t)}{e^{\pi t}-1}dt$

Let $D=(0,+\infty)$, I have $\frac{1}{e^{-\pi t}-1}=\frac{e^{-\pi t}}{1-e^{-\pi t}}$
So $$\frac{\arctan(t)}{e^{\pi t}-1}=\sum_{k=1}^{+\infty}\arctan(t)e^{-k \pi t}$$
Now, I can use integration term by term theorem, And finally I have,
$$
I=\sum_{k=1}^{+\infty}\int_0^{+\infty}\arctan(t)e^{-k \pi t}dt
$$
By integration by parts I get,
$$
\int_0^{X}\arctan(t)e^{k \pi t}dt\rightarrow \frac{1}{k\pi}\int_0^{+\infty}\frac{e^{-k \pi t}}{1+t^2}dt
$$
Therefore,
$$
I=\sum_{k=1}^{+\infty}\frac{1}{k\pi}\int_0^{+\infty}\frac{e^{-k \pi t}}{1+t^2}dt
$$
Now, I am stuck.
I would like to find a closed form,
Thank you in advance for your time.

Let $ \displaystyle I(z) = \int_{0}^{\infty}\frac{\arctan \frac{x}{z}}{e^{\pi x}-1} \ dx$.
Then
$$\begin{align}  I(z) &= \int_{0}^{\infty} \int_{0}^{\infty}\frac{1}{e^{\pi x}-1} \frac{\sin (xt)}{t}e^{-zt} \ dt \ dx \tag{1} \\ &= \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \frac{\sin (tx)}{e^{\pi x}-1} \ dx \ dt \\&= \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \sin (tx) \sum_{n=1}^{\infty} e^{-\pi nx}\ dx \ dt \\ &=\int_{0}^{\infty} \frac{e^{-zt}}{t} \sum_{n=1}^{\infty} \int_{0}^{\infty} \sin(tx) \ e^{-\pi n x} \ dx \ dt \\ &= \int_{0}^{\infty} \frac{e^{-zt}}{t} \sum_{n=1}^{\infty} \frac{t}{t^{2} + \pi^{2}n^{2}} \\&= \frac{1}{2}\int_{0}^{\infty} \frac{e^{-zt}}{t} \Big(\coth t -\frac{1}{t}\Big) \ dt \tag{2} \end{align}$$
Now differentiate inside of the integral with respect to $z$.
$$ I'(z) = - \frac{1}{2} \int_{0}^{\infty}e^{-zt} \Big( \coth t - \frac{1}{t}\Big) \ dt $$
And then integrate by parts.
$$ \begin{align} I'(z) &= -\frac{1}{2} e^{-zt} \Big(\log(\sinh t) - \log t \Big) \Big|^{\infty}_{0} - \frac{z}{2} \int_{0}^{\infty} e^{-zt} \Big( \log (\sinh t) - \log(t) \Big) \ dt \\ &=- \frac{z}{2} \int_{0}^{\infty} e^{-zt} \Big( \log (\sinh t) - \log(t) \Big) \ dt \\ &= - \frac{z}{2} \int_{0}^{\infty} e^{-zt} \Big(t - \log (2) + \log(1-e^{-2t}) - \log(t) \Big) \ dt  \\  &= -\frac{z}{2} \Big( \frac{1}{z^{2}} - \frac{\log(2)}{z} + \frac{\log(z) + \gamma}{z} \Big) + \frac{z}{2} \int_{0}^{\infty} e^{-zt} \sum_{n=1}^{\infty} \frac{e^{-2tn}}{n} \ dt \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} - \frac{\gamma}{2} +\frac{z}{2} \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} e^{-(2n+z)t} \ dt  \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} - \frac{\gamma}{2} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{z/2}{n(n+z/2)} \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} - \frac{\gamma}{2} + \frac{1}{2}\psi\left( \frac{z}{2}+1 \right)  + \frac{\gamma}{2} \tag{3} \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} + \frac{1}{2} \psi \Big( \frac{z}{2}+1 \Big) \end{align}$$
Then integrating back,
$$I(z) = - \frac{\log z}{2} + \frac{z \log 2}{2} - \frac{z \log z}{2} + \frac{z}{2} +  \log \Gamma \left( \frac{z}{2} +1 \right) + C$$
where using Stirling's formula the constant of integration is found to be $ \displaystyle -\frac{\log(\pi)}{2}$.
Therefore,
$$ \begin{align} \int_{0}^{\infty} \frac{\arctan x}{e^{\pi x}-1} \ dx &= I(1) \\ &= \frac{\log 2}{2} + \frac{1}{2} + \log \Gamma \left(\frac{3}{2} \right)- \frac{\log \pi}{2} \\ &= \frac{\log 2}{2} + \frac{1}{2} +  \log \left( \frac{\sqrt{\pi}}{2} \right) -\frac{\log \pi}{2} \\ &=\frac{1}{2} - \frac{\log 2}{2} \end{align}$$
$ $
$(1)$ $\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$ Evaluate Integral
$(2)$ Series expansion of $\coth x$ using the Fourier transform
$(3)$ http://en.wikipedia.org/wiki/Digamma_function#Series_formula
