Although I've found similar integrals, I haven't found this.
Determine the values $a \geq 0 $ such that the integral:
$\int_0^\infty\frac{x^a}{x^3+1}dx $ is convergent.
For these values calculate it.
I have managed to check that $ a \in [0, 2) $, checking the interval $ (0, 1] $, in which there is no issues, because the limit tends to zero.
Also in $ [1, \infty) $, I get $$ \int_1^\infty x^{(a-3)} dx $$ and get that $ a < 2 $ or else it diverges because of the p-test.
Now to calculate by residue I'm unsure how to proceed, since it does not state that $ a $ is natural.
Best Answer
Calculate the integral $$\oint \frac{z^a}{z^3+1} \, dz = \oint \frac{\exp[a \log z]}{z^3+1} \, dz$$ over a suitable contour.
The contour could be a wedge from the origin to the real point $R$, then an arc of a circle to the point $\beta$ and back to the origin.
The point $\beta$ could be $R\exp[\frac{2 \pi i}{3}]$ containing one pole inside. Other similarly constructed contours will also work.