 Solution

Poles and residues. An isolated singularity $z_0$ is a pole of order $m$ if its Laurent series has finitely many negative powers, the lowest being a nonzero multiple of $(z-z_0{)}^{-m}$. The residue is the coefficient of $(z-z_0{)}^{-1}$.

If $h(z_0)=0$ and $h^{\prime }(z_0)\ne 0$, factor $h(z)=(z-z_0)H(z)$, where $H$ is holomorphic near $z_0$ and $H(z_0)=h^{\prime }(z_0)$. For $g(z_0)\ne 0$,The last quotient is holomorphic and nonzero at $z_0$, proving a simple pole with

The given function. For noninteger $a$, specify a branch of a multivalued function before using $z^a$. Choose $z^a=\exp (a\log z)$ with $-\pi /2<\arg z<3\pi /2$, a cut along the negative imaginary axis. This leaves all three roots of $1+z^3$ off the cut. They are $z_{+}=e^{i\pi /3}$, $z_{-}=e^{-i\pi /3}$ and $z_0=-1$, all simple. The numerator is nonzero at these points. Since the denominator has derivative $3z^2$, the residues areFor noninteger $a$, zero is a branch point, rather than an isolated pole. This remains true when $-1<a<0$ and the numerator diverges there. For nonnegative integer $a$ it is a regular point. Other logarithm branches change the phases of the residues consistently.

Integral and convergence. Near zero the integrand behaves as $x^a$, requiring $a>-1$. At infinity it behaves as $x^{a-3}$, requiring $a<2$. Therefore the exact convergence range isUse a counterclockwise sector contour with rays at angles $0$ and $2\pi /3$, inner radius $\epsilon$ and outer radius $R$. The same complex logarithm branch is valid throughout this sector. Only $z_{+}=e^{i\pi /3}$ lies inside. The small circular arc contributes $O({\epsilon }^{a+1})$ and the large one contributes $O(R^{a-2})$, so both vanish precisely in the stated range.

The outward positive-real ray contributes $I$. On the returning ray set $z=x{e}^{2\pi i/3}$: the denominator is still $1+x^3$, while $z^{a}dz=e^{2\pi i(a+1)/3}{x}^{a}dx$, with reversed limits. The residue theorem givesPut $\alpha =\pi (a+1)/3$. The right-hand phase is $-e^{i\alpha }$ and $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$. HenceThe sine is positive in this range, consistent with a positive real integral. This is a special case of the Mellin integral of a reciprocal binomial.