Evaluating $\int_0^\infty\frac{\ln(1+x^2)}{e^{\pi x}-1}dx$ I want to evaluate 

$$\int_0^\infty\frac{\ln(1+x^2)}{e^{\pi x}-1}\,\mathrm dx$$

I tried to let $$I(a)=\int_0^\infty\frac{\ln(1+a^2x^2)}{e^{\pi x}-1}\,\mathrm dx,$$ and then 
$$
\begin{align}
I'(a)&=\int_0^\infty\frac{2ax^2}{(1+a^2x^2)(e^{\pi x}-1)}\,\mathrm dx\\[25pt]
&=2a\int_0^\infty\mathscr L^{-1}\Big(\frac{x}{1+a^2x^2}\Big)\mathscr L\Big(\frac{x}{e^{\pi x}-1}\Big)\,\mathrm dx\\[25pt]
&=-2a^{-2}\pi^{-2}\int_0^\infty \sin\frac xa \psi^{(1)}\left(1+\frac x\pi\right)\,\mathrm dx
\end{align}
$$
I can't go further.
 A: This is not a full solution, but a reduction to an integral that looks a bit simpler (though I am not sure whether it admits a closed-form expression).
We integrate by parts and use the series expansion of the logarithm to obtain
$$I \equiv \int \limits_0^\infty \frac{\ln(1+x^2)}{\mathrm{e}^{\pi x}-1} \, \mathrm{d} x = \frac{2}{\pi} \int \limits_0^\infty \frac{-x \ln(1-\mathrm{e}^{-\pi x})}{1+x^2} \, \mathrm{d} x = \frac{2}{\pi} \sum \limits_{n=1}^\infty \frac{1}{n} \int \limits_0^\infty \frac{x \mathrm{e}^{-n \pi x}}{1+x^2} \, \mathrm{d} x \, .$$
Now we employ the Laplace transform identity
$$ \mathcal{L} \left(t \mapsto \frac{t}{1+t^2}\right) (p) = \sin(p) \left[\frac{\pi}{2} - \operatorname{Si}(p)\right] - \cos(p) \operatorname{Ci}(p) $$
and the representation
$$ \operatorname{Ci} (p) = \gamma + \ln(p) - \operatorname{Cin}(p)$$
of the cosine integral ($\gamma$ is the Euler-Mascheroni constant and the trigonometric integrals are defined here) to write
$$ I = \frac{2}{\pi} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{n} \operatorname{Ci}(n \pi) = \frac{2}{\pi} \left[[\gamma + \ln(\pi)] \ln(2) - \eta'(1) - \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{n} \operatorname{Cin}(n \pi)\right] \, .$$
$\eta$ is the Dirichlet eta function and since $\eta'(1) = \gamma \ln(2) - \ln^2 (2) /2$ , we find
$$ I = \frac{\ln(2) \ln(2 \pi^2)}{\pi} - J \, ,$$
where
$$ J \equiv \frac{2}{\pi}\sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{n} \operatorname{Cin}(n \pi) = \frac{2}{\pi} \int \limits_0^{\pi/2} \frac{1}{t} \sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{n} [1-\cos(2 n t)] \, \mathrm{d} t \, . $$
Finally, the Fourier series  of $\ln(\cos)$ yields
$$ J = \frac{2}{\pi} \int \limits_0^{\pi/2} \frac{- \ln(\cos(t))}{t} \, \mathrm{d} t = \int \limits_0^1 - \ln(s) \tan \left(\frac{\pi}{2}s\right) \, \mathrm{d} s \approx 0.59921 \, .$$
I have not found a nice expression for $J$ yet, but maybe someone else knows how to calculate this integral. 
Power series expansions of the integrands lead to the following representations in terms of the Dirichlet lambda function and the dilogarithm:
$$ J = \frac{1}{\pi} \sum \limits_{n=1}^\infty \frac{\lambda(2n)}{n^2} = \frac{1}{\pi} \sum \limits_{k=0}^\infty \operatorname{Li}_2 [(2k+1)^{-2}] \, . $$
A: The integral can be transformed to
$$I=\int_0^\infty \frac{\log{(1+x^2)}}{e^{\pi x} -1}dx = \frac{4}{\pi}\int_0^1 \frac{du}{1-u^2} \big(h(1/u)-h(u)\big)$$
where
$$ h(u) = u\Big( \log{\Gamma(u/2)}-\frac{1}{2}(u-1)\log{(u/2)} + u/2 - \frac{1}{2}\log{(2\pi)} \Big) $$
An integration by parts enables one to pull out known constants involving $\pi$ and $\log{2},$  but the remaining part is ugly, with integrands involving 
$\log(1-u^2)\psi(u/2)$, for instance.  I derived the formula from
$$\int_0^\infty \frac{ \tan^{-1}(x\,u)}{u(1-u^2)}du = \frac{\pi}{4} \log(1+x^2). $$
The integral is a principal value integral where care must be taken at $u=1.$  Insert into definition, switch $\int$, and use the Binet formula for the log of the Gamma function in the inner integral. (The equation within the large parenthesis of the penultimate formula comes from the  Binet formula.)   The form presented comes from splitting the remaining integral at $u=1$ and making the transformation $u \to 1/u$ in the part that had previously extended to $\infty.$  It has been checked numerically with the original definition.
