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:
$$I_1=\int_0^\infty\log\left(1-e^{-\pi x}\right)\log\left(1+x^2\right)dx$$
$$I_2=\int_0^\infty x\log\left(1-e^{-\pi x}\right)\arctan x\,dx$$
 A: Some heuristics suggest $\displaystyle\int\limits_0^\infty\operatorname{Li}_2\left(e^{-\pi x}\right)\arctan(x)\operatorname{d}x=\frac{\pi^2}{18}-\frac{3\zeta\left(3\right)}{8}.$
A: We have $$\int_{0}^{\infty}\textrm{Li}_{2}\left(e^{-\pi x}\right)\arctan\left(x\right)dx=\sum_{k\geq1}\frac{1}{k^{2}}\int_{0}^{\infty}e^{-\pi kx}\arctan\left(x\right)dx=\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k^{3}}\int_{0}^{\infty}\frac{e^{-\pi kx}}{1+x^{2}}dx
 $$ and this is the Laplace transform of $\frac{1}{1+x^{2}}
 $ at $s=\pi k
 $. This can be calculated (see for example here for $s=1
 $) $$\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k^{3}}\left(\textrm{Ci}\left(\pi k\right)\sin\left(\pi k\right)+\frac{\pi\cos\left(\pi k\right)}{2}-\textrm{Si}\left(\pi k\right)\cos\left(\pi k\right)\right)=$$ $$=-\frac{3}{8}\zeta\left(3\right)-\frac{1}{\pi}\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\textrm{Si}\left(\pi k\right)
 $$ where $\textrm{Ci}\left(x\right)
 $ and $\textrm{Si}\left(x\right)
 $ are the cosine and the sine integral. Now note, using the power series of $\textrm{Si}\left(x\right)
 $ $$\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\textrm{Si}\left(\pi k\right)=\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k^{3}}\sum_{n\geq1}\left(-1\right)^{n-1}\frac{\left(\pi k\right)^{2n-1}}{\left(2n-1\right)\left(2n-1\right)!}=$$ $$=\sum_{n\geq1}\left(-1\right)^{n-1}\frac{\pi^{2n-1}\left(2^{2n-3}-1\right)\zeta\left(4-2n\right)}{\left(2n-1\right)\left(2n-1\right)!}
 =\pi\left(2^{-1}-1\right)\zeta\left(2\right)-\frac{\pi^{3}\zeta\left(0\right)}{18}=$$ $$=-\frac{\pi^{3}}{18}
 $$ because $\zeta\left(-2n\right)=0,\,\forall n\geq1
 $. So we have $$\int_{0}^{\infty}\textrm{Li}_{2}\left(e^{-\pi x}\right)\arctan\left(x\right)dx=-\frac{3}{8}\zeta\left(3\right)+\frac{\pi^{2}}{18}
 $$ as conjectured by dbanet.
A: 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.
A: The following is another way to show that $$\sum_{n=1}^{\infty} \frac{(-1)^{k} \operatorname{Si}(k \pi )}{k^{3}}= - \frac{\pi^{3}}{18}.$$
(I'm surprised that replacing $\operatorname{Si}(k \pi)$ with its Maclaurin series and then switching the order of summation leads to the correct result because Fubini's theorem is not satisfied and $\sum_{k=1}^{\infty} (-1)^{k}k^{2n-4} = -\eta(4-2n)$ only when $n < 2$.)
The Fourier series of the piecewise continous function $$f(x) =  \begin{cases} x  &\text{if}  - \pi < x< \pi\\ f(x+ 2\pi) \,  &\text{otherwise} \end{cases}$$ is $$f(x) = 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \sin(kx).$$
Term-by-term integration gives $$\frac{x^{2}}{2} - \frac{\pi^{2}}{6} = 2 \sum_{k=1}^{\infty} \frac{(-1)^{k} \cos(nx)}{k^{2}}, \quad - \pi \le x \le \pi.$$
And term-by-term integration a second time gives $$\frac{x^{3}}{6} - \frac{\pi^{2}x}{6} = 2 \sum_{k=1}^{\infty} \frac{(-1)^{k} \sin(kx)}{k^{3}}, \quad - \pi \le x \le \pi. $$
Therefore, for $- \pi \le x \le \pi$, we have $$ \begin{align} \sum_{k=1}^{\infty} \frac{(-1)^{k}\operatorname{Si}(kx)}{k^{3}} &= \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k^{3}}\int_{0}^{kx} \frac{\sin(t)}{t} \, \mathrm dt \\ &=  \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k^{3}}\int_{0}^{x} \frac{\sin(ku)}{u} \, \mathrm du \\ &= \int_{0}^{x} \frac{1}{u} \sum_{k=1}^{\infty} \frac{(-1)^{k} \sin(ku)}{k^{3}} \\ &= \int_{0}^{x} \frac{1}{u} \left(\frac{u^{3}}{12}- \frac{\pi^{2}u}{12} \right) \, \mathrm du \\ &= \frac{1}{12} \int_{0}^{x} \left(u^{2} - \pi^{2} \right) \, \mathrm du  \\ &=\frac{1}{12} \left( \frac{x^{3}}{3}- \pi^{2}x \right). \end{align} $$
Letting $x= \pi$, we get $$\sum_{k=1}^{\infty} \frac{(-1)^{k}\operatorname{Si}(k\pi)}{k^{3}} = \frac{1}{12} \left( \frac{\pi^{3}}{3}- \pi^{3} \right)= - \frac{\pi^{3}}{18}. $$
