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:
$$I_1=\int_0^\infty\log\left(1-e^{-\pi x}\right)\log\left(1+x^2\right)dx$$
$$I_2=\int_0^\infty x\log\left(1-e^{-\pi x}\right)\arctan x\,dx$$

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 $$\sum_{k=1}^{\infty} \frac{(-1)^{k}\operatorname{Si}(k\pi)}{k^{3}} = \frac{1}{12} \left( \frac{\pi^{3}}{3}- \pi^{3} \right)= - \frac{\pi^{3}}{18}. $$
