Gazeta Matematica 10/2015, Problem 27137

Problem:
Evaluate $$\lim_{y \to 0} \int_{y}^1 \left(\arctan x \arctan \dfrac{1}{x}+\dfrac{\log(1+x^2)}{1+x^2}\right) \ dx.$$

Proposed by Alessandro Ventullo, Milan, Italy

 

Solution:

Observe that $$\arctan x + \arctan \dfrac{1}{x}=\dfrac{\pi}{2}$$ for all $x>0$. Setting $a=\arctan x, b=\arctan \dfrac{1}{x}, c=-\dfrac{\pi}{2}$, we have $a+b+c=0$, so $$a^3+b^3+c^3=3abc.$$ This means that
$$-3\arctan x \cdot \arctan \dfrac{1}{x} \cdot \dfrac{\pi}{2}=\arctan^3 x + \arctan^3 \dfrac{1}{x}-\dfrac{\pi^3}{8}.$$
Hence
$$\int_0^1 \arctan x \arctan \dfrac{1}{x} \ dx=-\dfrac{2}{3\pi}\int_0^1 \left( \arctan^3 x + \arctan^3 \dfrac{1}{x}-\dfrac{\pi^3}{8}\right) \ dx.                  (1)$$
Integrating by part, we obtain
$$\int \arctan^3 x \ dx=x \arctan^3 x - \dfrac{3}{2}\log(1+x^2)\arctan^2 x+3 \int \dfrac{\log (1+x^2) \arctan x}{1+x^2} \ dx.$$
and
$$\int \arctan^3 \dfrac{1}{x} \ dx=x \arctan^3 \dfrac{1}{x} + \dfrac{3}{2}\log(1+x^2)\arctan^2 \dfrac{1}{x}+3 \int \dfrac{\log (1+x^2) \arctan 1/x}{1+x^2} \ dx.$$
Summing up and integrating from $x=0$ to $x=1$ and using the fact that $$\arctan x+\arctan \dfrac{1}{x}=\dfrac{\pi}{2},$$ we get
$$\int_0^1 \left( \arctan^3 x + \arctan^3 \dfrac{1}{x} \right) \ dx=\dfrac{\pi^3}{32}+\dfrac{3\pi}{2} \int_0^1 \dfrac{\log (1+x^2)}{1+x^2} \ dx.                 (2)$$
Substituting (1) into (2), we get
$$\int_{0}^1 \left(\arctan x \arctan \dfrac{1}{x}+\dfrac{\log(1+x^2)}{1+x^2}\right)=\dfrac{\pi^2}{16}.$$