# Closer form for $\int_0^\infty\frac{(\arctan{x})^2\log^2({1+x^2})}{x^2}dx$

I Would like to know the value of this integral. $$\int_0^\infty\frac{(\arctan{x})^2\log^2({1+x^2})}{x^2}dx$$ I think $$I=\frac{a}{b}(\pi^3\ln2)+\frac{c}{d}(\pi\ln^32)+\frac{e}{f}(\pi\ln^22)+\frac{g}{h}(\pi\ln2)+\frac{i}{j}(\pi^3)+\frac{k}{m}\zeta({3})??$$ Where a,b,c,d...are integers Thanks.

• Where did this horrible, integral-creature come from? Feb 17, 2015 at 22:42
• W|A does not give me a value. But I think I've seen something similar in the past. I would start to tackle it using the series expansion of $\ln (1+x)$. I also think that complex analysis will come in handy here. Especially contour integration. Feb 17, 2015 at 22:49

Let $\displaystyle\small\gamma=\lim_{R\to\infty}[-R,R]\cup Re^{i[0,\pi]}$. Observe that $$\small\oint_\gamma\frac{\ln^4(1-iz)}{z^2}dz=\frac{1}{8}\int^\infty_0\frac{\ln^4(1+x^2)}{x^2}dx-3\int^\infty_0\frac{\ln^2(1+x^2)\arctan^2{x}}{x^2}dx+2\int^\infty_0\frac{\arctan^4{x}}{x^2}dx=0$$ since
• The integral over the arc vanishes as $\small\mathcal{O}\left(\dfrac{\ln^4{R}}{R}\right)$.
• The singularity at $\small 0$ is removable.
It has a closed form a bit different from what you conjectured: $$I=\frac{4\pi}3\ln^32+\frac{2\pi^3}3\ln2+\frac\pi2\zeta(3)$$