 Solution

Introduce the Mellin transform integral $J(s)={\int }_0^{\infty }{x}^{s-1}/(1+x^2)dx$, valid for $0<\mathrm{Re}s<2$. A keyhole contour for $z^{s-1}/(1+z^2)$ with $0<\arg z<2\pi$ has residues $i^{s-1}/(2i)$ and $(-i{)}^{s-1}/(-2i)$, with the latter argument $3\pi /2$. The two sides of the positive-axis cut give $(1-e^{2\pi i(s-1)})J(s)$. The small and large arcs vanish in this strip; simplifying the residue theorem identity yieldsDifferentiation under the integral is valid on compact substrips because the extra $\log x$ is dominated at both endpoints. ThereforeAt $s=3/2$, the requested values are