 Solution

Near zero the magnitude of the integrand is of order $t^{\mathrm{Re}z-1}$, and at infinity it is of order $t^{\mathrm{Re}z-3}$. Thus the improper integral converges exactly in the stripAt a boundary of this strip the logarithmic-variable oscillation has no limiting primitive, so conditional convergence does not enlarge it.

Use a keyhole contour about the positive real axis for $w^{z-1}/(1+w{)}^2$, taking $0<\arg w<2\pi$. The circular arcs vanish in the indicated strip. The upper and lower banks contribute $(1-e^{2\pi iz})F(z)$. The only enclosed pole is the double pole at $w=-1$, whose residue is $(z-1)e^{i\pi z}$. The residue theorem therefore givesAt $z=1$ the quotient has a removable singularity, with limit one, agreeing with ${\int }_0^{\infty }(1+t{)}^{-2}dt=1$.