<s>[INST] Solving the improper integral $\int_0^\infty \frac{x^{1/3}}{1+x^2} \mathrm dx$ I'm trying to solve:
$\displaystyle \int \limits_0^\infty \dfrac{x^{1/3}}{1+x^2} \mathrm dx$
I have tried contour integration with $C_R^+$ and the real line like this:
$\displaystyle \int \limits_T \dfrac{z^{1/3}}{1+z^2} \mathrm dz = \int \limits_{-\infty}^\infty \dfrac{z^{1/3}}{1+z^2}\mathrm  dz + \int \limits_{C_R^+} \frac{z^{1/3}}{1+z^2} \mathrm dz$
Where the last integral tends to $0$ as $R \longrightarrow \infty$
$\text{Res}(f(z);i) = \dfrac{i^{1/3}}{2i}$
and
$\displaystyle \int \limits_{-\infty}^\infty \frac{z^{1/3}}{1+z^2}\mathrm  dz = \int \limits_{0}^\infty \frac{z^{1/3}}{1+z^2} \mathrm dz + \int \limits_{-\infty}^0\frac{z^{1/3}}{1+z^2} \mathrm dz$
If i manipulate the last term by changing the limits and substitute $u=-t$ i get:
$\displaystyle  \int \limits_{-\infty}^0\dfrac{z^{1/3}}{1+z^2} \mathrm dz  = -\int \limits_{0}^{-\infty}\frac{z^{1/3}}{1+z^2} \mathrm dz$
If i now substitue $u=-z, u'=-1$
$\displaystyle \int \limits_{-\infty}^0\dfrac{z^{1/3}}{1+z^2} \mathrm dz  = \int \limits_{0}^{\infty}\dfrac{(-u)^{1/3}}{1+u^2}\mathrm  dz = (-1)^{1/3}\int \limits_{0}^{\infty}\dfrac{u^{1/3}}{1+u^2}\mathrm  dz$
$\displaystyle \int \limits_{-\infty}^\infty \dfrac{z^{1/3}}{1+z^2}\mathrm  dz = \left(1+e^\frac{i\pi}{3}\right) \int \limits_{0}^\infty \frac{z^{1/3}}{1+z^2} \mathrm dz $
So i end up with:
$\displaystyle \dfrac{2i\cdot\pi \cdot e^{i\cdot\pi /6}}{2i\cdot\left(1+e^\frac{i\pi}{3}\right)} = \dfrac{\pi\cdot e^{i\cdot \pi /6}}{\left(1+e^\frac{i\pi}{3}\right)}  = \int \limits_{0}^\infty \dfrac{z^{1/3}}{1+z^2} \mathrm dz$ which is wrong answer
 [/INST] Hint: using the substitution $x=u^{\frac{3}{2}}$ your integral will be equal to $$\frac{3}{2}\int\frac{u}{u^3+1}du$$ which can be evaluated by decomposing partial fractions. Additional note: if you divide the integral as an integral from $0$ to $1$ and an integral from $1$ to $\infty$, it is clearly that the first one is divergent.
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