= Solution
Use the branch in (i) and the positively oriented slit-annulus contour. Its upper ray runs from $r$ to $R$, its outer circle counterclockwise, its lower ray from $R$ back to $r$, and its inner circle clockwise. Initially keep their arguments at $\alpha$ and $2\pi-\alpha$ as in the figure, with $0<\alpha<\pi/2$; let $\alpha\downarrow0$ for fixed $0<r<1<R$. Both poles $i,-i$ lie inside.
Write $\sigma=\Re w\in(-1,1)$. On the outer circle the absolute integral is $O(R^{\sigma-1})$, since $|z^w|\le e^{2\pi|\Im w|}R^\sigma$ and $|z^2+1|\ge R^2-1$. On the inner circle it is $O(r^{\sigma+1})$. Both tend to zero. The real integral converges absolutely because its endpoint behaviors are $x^\sigma$ and $x^{\sigma-2}$.
The two residues, with the chosen arguments, are
$$
\frac{e^{i\pi w/2}}{2i},\qquad-\frac{e^{3i\pi w/2}}{2i}.
$$
The <residue theorem> and the two ray orientations therefore give
$$
(1-e^{2\pi iw})I(w)=\pi e^{i\pi w/2}(1-e^{i\pi w}),\qquad I(w)=\int_0^\infty\frac{x^w}{1+x^2}\,dx.
$$
For $w\ne0$ in this strip we may cancel $1-e^{i\pi w}$; its only possible zero here is $w=0$. Consequently
$$
\boxed{I(w)=\frac{\pi e^{i\pi w/2}}{1+e^{i\pi w}}=\frac\pi{2\cos(\pi w/2)}.}
$$
At $w=0$ the same result follows directly from $I(0)=\int_0^\infty(1+x^2)^{-1}dx=\pi/2$, or by <dominated convergence>. The denominator has no zero in the open strip.
Back to article page
