= Solution
Near zero the magnitude of the integrand is of order $t^{\operatorname{Re}z-1}$, and at infinity it is of order $t^{\operatorname{Re}z-3}$. Thus the <improper integral> converges exactly in the strip
$$
\boxed{0<\operatorname{Re}z<2.}
$$
At 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 gives
$$
(1-e^{2\pi iz})F(z)=2\pi i(z-1)e^{i\pi z},\qquad
\boxed{F(z)=\frac{\pi(1-z)}{\sin\pi z}.}
$$
At $z=1$ the quotient has a <removable singularity>, with limit one, agreeing with $\int_0^\infty(1+t)^{-2}dt=1$.
Back to article page
