Prove: $\int_0^{\infty} \frac{\ln{(1+x)}\arctan{(\sqrt{x})}}{4+x^2} \, \mathrm{d}x = \frac{\pi}{2} \arctan{\left(\frac{1}{2}\right)} \ln{5}$

Question

Prove: $$\int_0^{\infty} \frac{\ln{(1+x)}\arctan{(\sqrt{x})}}{4+x^2} \, \mathrm{d}x = \frac{\pi}{2} \arctan{\left(\frac{1}{2}\right)} \ln{5}$$
This might be a repeat question (I couldnt find a question of this here). If im being honest I dont know the first step really… Maybe a clever integration by parts, substitution, differentiation under integral sign, power series, or contour? If someone could give advice.

Answer 1

For $a>0$, $$\begin{aligned}I = \int_0^\infty {\frac{{\log (1 + x)\arctan \sqrt x }}{{{a^4} + {x^2}}}dx} &= \int_{ - \infty }^\infty {\frac{x}{{{a^4} + {x^4}}}\log (1 + {x^2})\arctan xdx} \\ &= -\Im \int_{ - \infty }^\infty {\frac{x}{{{a^4} + {x^4}}}{{\log }^2}(1 - ix)dx} \end{aligned}$$ The integrand is holomorphic on upper half plane, integral around big semicircle tends to $0$, calculating residues at $a\zeta, a\zeta^3$ (with $\zeta = e^{\pi i /4}$) give $$ I= \frac{{ \pi }}{{2{a^2}}}\Im\left[ {{{\log }^2}(1 + a\zeta ) - {{\log }^2}(1 - a{\zeta ^3})} \right]$$ when $a=\sqrt{2}$, it becomes $\frac{1}{2}\pi\arctan(1/2)\log 5$.