 Upper-half-plane contour for square-root logarithmic integrals

Set $I={\int }_0^{\infty }{x}^{1/2}/(1+x^2)dx$ and $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$. In an indented upper semicircle use the principal complex logarithm and $z^{1/2}=e^{\log z/2}$. The negative real boundary contributes $iJ-\pi I$, the positive boundary contributes $J$, and both circular arcs vanish. The pole at $i$ has residue $\pi e^{i\pi /4}/4$. Thus $(1+i)J-\pi I=({\pi }^2/(2\sqrt{2}))(-1+i)$; comparing real and imaginary parts yields the displayed integrals.