<s>[INST] How to integrate $\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$? I'm trying to evaluate the integral $\displaystyle\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$.
My book explains that to evaluate integrals of the form $\displaystyle\int_0^\infty x^\alpha R(x)dx$, with real $\alpha\in(0,1)$ and $R(x)$ a rational function, one first starts with a substitution $x=t^2$, to transform the integral to
$$
2\int_0^\infty t^{2\alpha+1} R(t^2) dt.
$$
It then observes that
$$
\int_{-\infty}^\infty z^{2\alpha+1}R(z^2)dz=\int_0^\infty(z^{2\alpha+1}+(-z)^{2\alpha+1})R(z^2)dz.
$$
Since $(-z)^{2\alpha}=e^{2\pi i\alpha}z^{2\alpha}$, the integral then equals
$$
(1-e^{2\pi i\alpha})\int_0^\infty z^{2\alpha+1}R(z^2)dz.
$$
How are you suppose to apply the residue theorem to this integral if the integrand is not a rational function?
In my case, I have
$$
\int_0^\infty x^{1/3}R(x)dx=2\int_0^\infty t^{5/3}R(t^2)dt.
$$
Also
$$
\int_0^\infty z^{5/3} R(z^2)dz=\frac{1}{1-e^{(2\pi i)/3}}\int_{-\infty}^{\infty}\frac{z^{5/3}}{1+z^4}dz.
$$
I don't know what to do after that, since this last integrand still has a fractional power in the numerator. How does this work? Thanks.
(This is part (g) of #3 on page 161 of Ahlfors' Complex Analysis, part of some self-study.)
 [/INST] Here is the simplest I could get without introducing other variables or making strange substitutions. Everything is in complex on the main branch ($0 \le arg(z) < 2\pi$).
First we calculate a bound change : 
 $$\begin{eqnarray*}
\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz &=& \int_{-\infty}^0 \frac{z^{1/3}}{1+z^2} dz + \int_{0}^\infty \frac{z^{1/3}}{1+z^2} dz\\
&=& \int_{0}^\infty \left( \frac{z^{1/3}}{1+z^2} + \frac{(-z)^{1/3}}{1+(-z)^2}\right) dz \\
&=& \int_{0}^\infty \frac{z^{1/3} + (e^{i\pi}z)^{1/3}}{1+z^2} dz \\
&=& (1+e^{i\frac{\pi}{3}})\int_{0}^\infty \frac{z^{1/3}}{1+z^2} dz \\
\end{eqnarray*}$$
So we have (1)
$$\int_{0}^\infty \frac{z^{1/3}}{1+z^2} dz = \frac{1}{1+e^{i\frac{\pi}{3}}} \int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz
$$
Now we only have to find $\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz$. Because $1+z^2 = (z-i)(z+i)$, the only singularity in the upper half-plane is $i$ and by the Residue Theorem we immediatly get
$$
\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz = 2 \pi i \text{Res}_i \text{  with  } \text{Res}_i = \lim_{z \to i} (z-i) \frac{e^{log(z)/3}}{(z-i)(z+i)} = \frac{e^{i\frac{\pi}{6}}}{2i}
$$
So we have (2)
$$
\int_{-\infty}^\infty\frac{z^{1/3}}{1+z^2} dz = \pi e^{i\frac{\pi}{6}}
$$
Finally, by (1) and (2) we get 
$$
\int_{0}^\infty\frac{z^{1/3}}{1+z^2} dz = \frac{\pi e^{i\frac{\pi}{6}}}{1+e^{i\frac{\pi}{3}}} = \frac{\pi}{\sqrt{3}}
$$
</s>