Good afternoon. I would like to evaluate :
$$I = \int_0^{\infty} \frac{\arctan^2(x)}{x^2} \hspace{1mm} \mathrm{d}x$$
Apparently $I = \pi \ln(2)$.
My attempt :
I tried integration by parts but it was a mess. It leads to :
$$I = \left[ -\frac{\arctan^2(x)}{x} \right]_{\mathbb{R}_+} + 2\left[ \frac{\arctan(x)\ln(x)}{x^2+1} \right]_{\mathbb{R}_+} \hspace{-3mm} - \int_{\mathbb{R}_+} \ln(x)\left(\frac{2}{(x^2+1)^2} - \frac{4x\arctan(x)}{(x^2+1)^2}\right) \mathrm{d}x$$
Let $x = \tan(\theta)$, so $\mathrm{d}x = \frac{\mathrm{d}\theta}{\cos^2(\theta)}$. We obtain :
$$I = \int_0^{\frac{\pi}{2}} \frac{\theta^2}{\tan^2(\theta)} \hspace{1mm} \frac{\mathrm{d}\theta}{\cos^2(\theta)} = \int_0^{\frac{\pi}{2}} \frac{\theta^2}{\sin^2(\theta)} \mathrm{d}\theta$$
By king's property :
$$2I = \int_0^{\frac{\pi}{2}} \left( \frac{\theta^2}{\sin^2(\theta)} + \frac{(\frac{\pi}{2}-\theta)^2}{\cos^2(\theta)} \right) \mathrm{d}\theta$$
I don't think it is the right path. Any hint ?
