Integral $\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$ Please help me to evaluate this integral in a closed form:
$$I=\int_0^\infty\text{Li}_2\left(e^{-\pi x}\right)\arctan x\,dx$$
Using integration by parts I found that it could be expressed through integrals of elementary functions:
$$I_1=\int_0^\infty\log\left(1-e^{-\pi x}\right)\log\left(1+x^2\right)dx$$
$$I_2=\int_0^\infty x\log\left(1-e^{-\pi x}\right)\arctan x\,dx$$

1. Solution
Here is another solution: Notice that
\begin{align*}
&\int_{0}^{\infty} \operatorname{Li}_2(e^{-\pi x}) \arctan x \, dx \\
&\qquad = \overbrace{\left[ -\tfrac{1}{\pi} \operatorname{Li}_3(e^{-\pi x}) \arctan x \right]_{0}^{\infty}}^{=0} + \int_{0}^{\infty} \frac{\operatorname{Li}_3(e^{-\pi x})}{\pi(1 + x^2)} \, dx \\
&\qquad = \sum_{n=1}^{\infty} \frac{1}{n^3} \int_{0}^{\infty} \frac{e^{-n\pi x}}{\pi(1+x^2)} \, dx \\
&\qquad = \sum_{n=1}^{\infty} \frac{1}{n^2} \int_{0}^{\infty} \frac{e^{-x}}{(\pi n)^2 + x^2} \, dx \qquad (n\pi x \mapsto x) \\
&\qquad = \frac{\pi^2}{2} \int_{0}^{\infty} \left( \frac{1}{x^4} + \frac{1}{3x^2} - \frac{\coth x}{x^3} \right) e^{-x} \, dx
\end{align*}
In order to evaluate the last integral, we introduce the function $I(s)$ defined by
$$ I(s) = \frac{\pi^2}{2} \int_{0}^{\infty} \left( \frac{1}{x^4} + \frac{1}{3x^2} - \frac{\coth x}{x^3} \right) x^{s} e^{-x} \, dx. $$
Then $I(s)$ is analytic for $\Re(s) > -1$ and our integral can be written as $I(0)$. Now assume for a moment that $\Re(s) > 3$. Then
\begin{align*}
I(s)
&= \frac{\pi^2}{2} \int_{0}^{\infty} \left( \frac{1}{x^4} + \frac{1}{x^3} + \frac{1}{3x^2} - \frac{2}{x^3(1 - e^{-2x})} \right) x^{s} e^{-x} \, dx \\
&= \frac{\pi^2}{2} \left( \Gamma(s-3) + \Gamma(s-2) + \frac{1}{3}\Gamma(s-1) - 2 \sum_{n=0}^{\infty} \frac{\Gamma(s-2)}{(2n+1)^{s-2}} \right) \\
&= \frac{\pi^2}{2} \left( \Gamma(s-3) + \Gamma(s-2) + \frac{1}{3}\Gamma(s-1) - 2 \Gamma(s-2)(1 - 2^{2-s})\zeta(s-2) \right).
\end{align*}
By the principle of analytic continuation, this relation continues to hold on $\Re(s) > -1$. So we only need to take limit as $s \to 0$. To this end, we treat two parts separately:


*

*It is easy to check that $\Gamma(s-3) + \Gamma(s-2) + \frac{1}{3}\Gamma(s-1) \to \frac{1}{9}$ as $s \to 0$.

*Using the functional equation of $\zeta(s)$ and the reflection formula for $\Gamma(s)$, we have
$$ 2 \Gamma(s-2)(1 - 2^{2-s})\zeta(s-2) = \frac{(1 - 2^{s-2})\pi^{s-2}}{\cos(\pi s/2)} \zeta(3-s). $$
Taking $s \to 0$, this converges to $\frac{3}{4\pi^2}\zeta(3)$.
Combining these two observations, we have
$$ I(0) = \frac{\pi^2}{18} - \frac{3}{8}\zeta(3). $$
2. Generalization
We can generalize the problem by considering the family of integrals
$$ I_n = \int_{0}^{\infty} \operatorname{Li}_{n}(e^{-\pi x}) \arctan x \, dx. $$
When $n$ is even, we can use the same technique as in our solution to obtain

Claim. For $m = 0, 1, 2, \cdots$ we have
  $$ I_{2m} = \frac{(-1)^m \pi^{2m}}{2} \sum_{k=0}^{2m+1} \frac{2^k B_k H_{2m+1-k}}{k!(2m+1-k)!} - \frac{1}{2}\eta(2m+1), \tag{1} $$
  where $(B_k)$ are the Bernoulli numbers, $(H_k)$ are the harmonic numbers, and
  $$\eta(s) = (1-2^{1-s})\zeta(s)$$
  is the Dirichlet eta function.

Remark 1. Following @Marco Cantarini's calculation, we have
$$ I_n = \sum_{k=0}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!(2k+1)} \eta(n-2k) - \frac{1}{2}\eta(n+1). \tag{2} $$
This reduces to a finite summation when $n = 2m$ is even:
$$ I_{2m} = (-1)^m \pi^{2m} \sum_{k=0}^{2m} \frac{(1-2^{k-1})B_k}{(2m+1-k)!(2m+1-k)(k!)} - \frac{1}{2}\eta(2m+1). \tag{3} $$
So we have two different representations for $I_{2m}$.
Remark 2. Using (1), we find that for $|z| < 1$,
$$ \sum_{m=0}^{\infty} I_{2m}z^{2m} = \frac{1}{2}\left( \frac{\operatorname{Si}(\pi z)}{\sin (\pi z)} - \int_{0}^{\infty} \frac{\cosh(z t)}{e^t + 1} \, dt \right). \tag{4} $$
Thus in principle, we can find the values of $I_{2m}$ by differentiating (4) $2m$ times.
