 Solution

Put $a=1/n$, so $0<a<1$, and write $I(a)={\int }_0^{\infty }{x}^a/(1+x^2)dx$. Choose the branch of the complex logarithm with $-\pi /2<\arg z<3\pi /2$, whose cut is the downward imaginary axis. Define $z^a=\exp (a\mathrm{Log}z)$ and integrate $z^a/(1+z^2)$ along the positively oriented upper semicircle, indented above zero. The straight pieces run from $-R$ to $-\epsilon$ and from $\epsilon$ to $R$; the small arc runs clockwise from argument $\pi$ to zero.

On the negative real axis $z^a=|z{|}^a{e}^{i\pi a}$, so the two straight pieces tend to $(1+e^{i\pi a})I(a)$. The outer arc is $O(R^{a-1})$ and the inner arc is $O({\epsilon }^{a+1})$, both tending to zero. The only enclosed pole is $i$, with residue $e^{i\pi a/2}/(2i)$. The residue theorem therefore givesUsing $1+e^{i\pi a}=2e^{i\pi a/2}\cos (\pi a/2)$ yieldsThe same bounds show convergence of the real integral at both zero and infinity.