= Solution
\b[The ordinary <Fourier transform> does not exist.] There is a real <simple pole> at $x=-1$, with
$$
\frac1{1+x^3}=\frac1{3(x+1)}+O(1)\quad(x\to-1).
$$
The two separate improper integrals diverge logarithmically. Thus no ordinary large-$k$ expansion is defined under the stated integral convention. A <Cauchy principal value> or a contour prescription would be additional data, and must be stated rather than silently introduced.
For completeness, the symmetric <Cauchy principal value> has a precise answer. The real <pole> contributes the <principal-value Fourier transform of a real pole>
$$
\mathcal F\!\left(\operatorname{PV}\frac1{3(x+1)}\right)=-\frac{i\pi}{3}\operatorname{sgn}(k)e^{ik}.
$$
The other <poles> are $z_\pm=(1\pm i\sqrt3)/2$ with <residues> $r_\pm=1/(3z_\pm^2)=(-1\mp i\sqrt3)/6$. For $k>0$, close the <contour integration> in the lower half-plane, with clockwise orientation. The nonreal <pole> contributes $-2\pi i r_-e^{-ikz_-}$, while the real principal-value <pole> supplies the half-residue above. For $k<0$, the upper-half-plane contour gives the conjugate result. With $s=\operatorname{sgn}(k)$, the exact principal-value formula for $k\ne0$ is
$$
\boxed{\widehat f_{\rm PV}(k)=-\frac{i\pi}{3}s e^{ik}+\frac\pi3(\sqrt3+is)\exp\left(-\frac{\sqrt3}{2}|k|-\frac i2k\right)}.
$$
These are its two nonzero asymptotic contributions: a nondecaying oscillatory real-pole term and an exponentially small complex-pole term. There is no second nonzero inverse-power term. This formula is qualified by the principal-value choice; it is not the ordinary transform requested in the statement.
Other prescriptions change the leading term. For example, replacing the real <pole> by its upper or lower boundary value changes the distribution by $\mp i\pi\delta(x+1)/3$, and hence changes the transform by $\mp i\pi e^{ik}/3$. This demonstrates why a <pole> prescription is essential. The smooth <Taylor series> at the origin alone would miss the decisive real-pole contribution.
Back to article page
