# Evaluating $\int_0^\infty \frac{\arctan x\operatorname{Li}_2( -x )}{x^2} \, dx$

Evaluate$$\int_0^{\infty}{\frac{\arctan x\operatorname{Li}_2(-x)}{x^2}dx}$$ My attempt: By Integrating by parts I got $$I=0+\int_0^\infty \frac{\ln(x+1)}{x}\left( \ln x-\frac{1}{2}\ln \left( x^2+1 \right) \right) \, dx-\int_0^\infty \frac{\arctan x\ln \left( x+1 \right)}{x^2} \, dx=I_1-I_2$$ $I_2$ can be evaluated by integrating by parts again. $$I_2=0-\int_0^\infty \frac{1}{1+x^2}\left( \ln x-\ln \left( x+1 \right) -\frac{\ln(x+1)}{x} \right) \, dx \\ =-0+G+\frac{1}{4}\pi \ln 2+\frac{5}{48}\pi^2$$ But I have no idea to deal with $I_1$. I tried integrating by parts and substitute $x\to\frac1x$ but I got nothing.

• $$I_1=-\frac{\pi}{2}G+\frac{3}{16}\zeta(3)$$ Jun 24, 2018 at 10:14
• For $I_1$, you may want to try splitting the integral and using integration by parts multiple times? Jun 24, 2018 at 16:06
• @Crescendo: I can't find a way using your method I just get to $$I_1=\int_0^{\infty } \frac{\text{Li}_2(-x)}{x^3+x} \, dx$$ and then get stuck. Jun 24, 2018 at 19:59
• @Crescendo Splitting the integral is impossible because $\int_0^\infty \frac{\ln(x+1)\ln x}{x}$ diverges. Jun 25, 2018 at 8:53

You can write $$I_1 = \frac{1}{2} \int \limits_0^\infty \frac{\ln(1+x)}{x} \ln \left(\frac{x^2}{1+x^2}\right) \, \mathrm{d} x = - \frac{1}{2} \int \limits_0^\infty \frac{\ln(1+x) \ln(1+x^{-2})}{x} \, \mathrm{d} x$$ and then have a look at this question to obtain the answer suggested in the comments.