 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 $\alpha \downarrow 0$ 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, areThe residue theorem and the two ray orientations therefore giveFor $w\ne 0$ in this strip we may cancel $1-e^{i\pi w}$; its only possible zero here is $w=0$. ConsequentlyAt $w=0$ 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.