Integral with dilogarithm

Let ${\rm Li}_2$ denote the dilogarithm function. Evaluate the integral

$$\int_0^\infty \frac{{\rm Li}_2(-x)}{1+x^2}\, {\rm d}x$$

Solution

From the link we know that for the Spences's function it holds that:

$${\rm Li}_2(-x) + {\rm Li}_2 \left ( -\frac{1}{x} \right )= - \frac{\pi^2}{6} + \frac{1}{2} \ln^2 x$$

Thus:

\begin{align*}
\int_{0}^{\infty} \frac{{\rm Li}_2(-x)}{1+x^2} \, {\rm d} x &= \int_{0}^{\infty} \frac{{\rm Li}_2\left ( -\frac{1}{x} \right )}{1+x^2} \, {\rm d}x\\
 &= \frac{1}{2}\left ( \int_{0}^{\infty} \frac{{\rm Li}_2(-x)}{1+x^2} \, {\rm d}x + \int_{0}^{\infty} \frac{{\rm Li}_2 \left ( -\frac{1}{x} \right )}{1+x^2} \, {\rm d}x \right )\\
 &= \frac{1}{2}\int_{0}^{\infty}\left (-\frac{1}{2} \frac{\ln^2 x}{1+ x^2} - \frac{\zeta(2)}{x^2+1} \right ) \, {\rm d}x\\
 &= -\frac{1}{2} \cdot \frac{\pi}{2} \cdot \frac{\pi^2}{6} -\frac{1}{4}\int_{0}^{\infty} \frac{\ln^2 x}{1+x^2} \, {\rm d}x\\
 &=-\frac{\pi^3}{24} -\frac{1}{2}\int_{0}^{1} \frac{\ln^2 x}{1+x^2} \, {\rm d}x \\
 &= -\frac{\pi^3}{24} - \frac{1}{2} \beta(3) \\
 &= -\frac{7 \pi^3}{96}
\end{align*}

A possible generalization may be the following:

$$\int_{0}^{\infty} \frac{x^{s-1} {\rm Li}_2(-x)}{1+x^s} \, {\rm d}x = -\frac{\pi^3}{4} \left ( \frac{1}{3s} +\frac{1}{s^3} \right ) \; , \quad  \mathfrak{Re}(s) \geq 2$$

The problem may also be found at integralsandseries.com