 Solution

Use the same indented upper-semicircle contour and branch of the complex logarithm, now for $F(z)=z^{1/2}\mathrm{Log}z/(1+z^2)$. Write $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$. On the negative axis, $z^{1/2}=i|z{|}^{1/2}$ and $\mathrm{Log}z=\log |z|+i\pi$. Consequently the two straight contributions tend toThe outer arc is $O(R^{-1/2}\log R)$ and the small arc is $O({\epsilon }^{3/2}(|\log \epsilon |+1))$, so neither contributes in the limit. At the enclosed pole $i$,The residue theorem gives $(1+i)J-\pi I(1/2)=i{\pi }^2{e}^{i\pi /4}/2$. Part (a) gives $I(1/2)=\pi /\sqrt{2}$. Substituting and using $e^{i\pi /4}=(1+i)/\sqrt{2}$, the right side after moving the $I$ term is ${\pi }^2(1+i)/(2\sqrt{2})$. Hence