How to integrate $\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$?

Continue reading the next few paragraphs where it is explained how to go on. In fact, that substitution isn't really necessary if you integrate over a "keyhole contour", which is also explained a bit further down the page.

You have found, correctly, that $$\begin{eqnarray*} I &\equiv& \int_0^\infty\frac{x^{1/3}dx}{1+x^2} \\ &=& \frac{2}{1-e^{2\pi i/3}} \int_{-\infty}^\infty d z\, \frac{z^{5/3}}{1+z^4}. \end{eqnarray*}$$ The second integral has a branch cut at $z=0$ and singularities at the roots of $1+z^4$. Close the contour in the upper half-plane. Let $\gamma$ denote the contour. We will pick up residues at $e^{i\pi/4}$ and $e^{i 3\pi/4}$.

We deform the contour slightly near $z=0$ to avoid the cut. Letting $z=\epsilon e^{i\theta}$ we see that the contribution to the integral here goes like $\epsilon^{1+5/3} = \epsilon^{8/3}$ and so vanishes in the limit.

Letting $z = R e^{i\theta}$, we find the integral over the semicircle at infinity goes like $R^{1+5/3-4} = 1/R^{4/3}$. Thus, the contribution from this part of the contour is also zero. Therefore, $\int_{-\infty}^\infty d z\, z^{5/3}/(1+z^4) = \int_\gamma d z\, z^{5/3}/(1+z^4)$.

We find $$\begin{eqnarray*} I &=& \frac{2}{1-e^{2\pi i/3}} \int_\gamma d z\, \frac{z^{5/3}}{1+z^4} \\ &=& \frac{2}{1-e^{2\pi i/3}} 2\pi i \sum\mathrm{Res} \, \frac{z^{5/3}}{1+z^4} \\ &=& \frac{\pi}{\sqrt{3}} \end{eqnarray*}$$ where the residues are to be taken from the upper half-plane.

The residue of $f(z)$ at $z=z_0$ is just the coefficient of $(z-z_0)^{-1}$ in the Laurent series and, of course, $z^{5/3}/(1+z^4)$ has a Laurent series about the zeros of $1+z^4$. (It does not, however, have a Laurent series about $z=0$.)

A simpler solution

Recognize that the integral $I$ is already in the form $$\int_0^\infty d t\, t^\beta f(t^2).$$ But $$\begin{eqnarray*} \int_0^\infty d t\, t^\beta f(t^2) &=& \frac{1}{1+e^{\beta \pi i}} \int_{-\infty}^\infty d t\, t^\beta f(t^2) \\ &=& \frac{1}{1+e^{\beta \pi i}} \int_\gamma d t\, t^\beta f(t^2) \\ &=& \frac{1}{1+e^{\beta \pi i}} 2\pi i \sum \mathrm{Res} \, t^\beta f(t^2) \end{eqnarray*}$$ assuming the contributions from the contour around the branch cut at $z=0$ and the contour at infinity vanish. (Again, we have closed the contour in the upper half-plane and pick up only the residues residing there.) It is also important that the singularities of $f$ do not lie on the real line.

The contribution near $z=0$ goes like $\epsilon^{1+1/3} = \epsilon^{4/3}$. On the semicircle at infinity the integral goes like $R^{1+1/3-2} = R^{-2/3}$. Thus, $\int_{-\infty}^\infty d t\, t^{1/3}/(1+t^2) = \int_\gamma d t\, t^{1/3}/(1+t^2)$.

There is one residue, at $t=i=e^{i\pi/2}$, so we have halved our work. We find $$\begin{eqnarray*} I &=& \frac{1}{1+e^{\pi i/3}} 2\pi i \frac{e^{i\pi/6}}{2i} \\ &=& \frac{\pi}{2\cos \pi/6} \\ &=& \frac{\pi}{\sqrt{3}}. \end{eqnarray*}$$