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.)
 A: 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.
A: 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
$$\begin{eqnarray*}
I &=& \frac{1}{1+e^{\pi i/3}} 2\pi i \frac{e^{i\pi/6}}{2i} \\
&=& \frac{\pi}{2\cos \pi/6} \\
&=& \frac{\pi}{\sqrt{3}}.
\end{eqnarray*}$$
A: 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 
$$
\int_{0}^\infty\frac{z^{1/3}}{1+z^2} dz = \frac{\pi e^{i\frac{\pi}{6}}}{1+e^{i\frac{\pi}{3}}} = \frac{\pi}{\sqrt{3}}
$$
A: Suppose $\alpha \in (-1, 1)$. Consider a keyhole contour $\Gamma$ of outer radius $R$ and inner radius $\varepsilon$ about the positive real axis. Let $\gamma_R$ denote the outer arc, $\gamma_\varepsilon$ denote the inner arc, and $\gamma_\pm$ denote the segment going away from and towards the origin. Note that this contour contains two singularities of $f(z) = \frac{z^\alpha}{z^2 + 1}$, namely $z = \pm i = \exp\left(\frac{\pi i}{2}\right ), \exp\left(\frac{3\pi i}{2}\right )$. We have
\begin{align*}
\int_\Gamma \frac{z^\alpha}{z^2 + 1} \ \mathrm{d}z &= 2\pi i \left (\text{Res}_f\left(\exp\left(\frac{\pi i}{2}\right ) \right ) + \text{Res}_f\left(\exp\left(\frac{3\pi i}{2}\right ) \right )\right )\\
&= \pi \left (\exp\left(\frac{\alpha \pi i}{2}\right ) - \exp\left(\frac{3\alpha \pi i}{2}\right )\right )
\end{align*}
On the other hand,
\begin{align*}
\int_{\gamma_R} \frac{z^\alpha}{z^2 + 1} \ \mathrm{d}z &= \int_0^1 \frac{R^\alpha \exp(2\alpha\pi i t)}{R^2 \exp(2\pi i t) + 1} 2\pi i R \exp(2\pi i t) \ \mathrm{d}t\\
&= 2\pi i R \int_0^1 \frac{R^\alpha \exp(2\alpha\pi i t)\exp(2\pi i t) }{R^2 \exp(2\pi i t) + 1} \ \mathrm{d}t\\
\end{align*}
using $z = R\exp(2\pi it)$. Bounding, we get
\begin{align*}
\left | 2\pi i R \int_0^1 \frac{R^\alpha \exp(2\alpha\pi i t)\exp(2\pi i t) }{R^2 \exp(2\pi i t) + 1} \ \mathrm{d}t\right | &\leq 2\pi R \int_0^1 \left | \frac{R^\alpha \exp(2\alpha\pi i t)\exp(2\pi i t) }{R^2 \exp(2\pi i t) + 1} \ \right |\mathrm{d}t\\
&\leq 2\pi R \int_0^1 \frac{R^\alpha}{R^2 - 1} \ \mathrm{d}t = \frac{2\pi R^{\alpha + 1}}{R^2 - 1}
\end{align*}
Since $\alpha \in (-1, 1)$, as $R \to \infty$, this tends to $0$. Further note that for $\gamma_\varepsilon$, this should be identical except $R$ becomes $\frac{1}{R}$, so we have bound $\frac{2\pi R^{-1-\alpha}}{1-R^{-2}} = \frac{2\pi R^{1-\alpha}}{R^2-1}$. But again, $\alpha \in (-1, 1)$ implies that this expression tends to $0$ as $R \to \infty$ (or $\varepsilon \to 0$), so this also disappears.
For $\gamma_+$, we have
\begin{align*}
\int_{\gamma_+} \frac{z^\alpha}{z^2 + 1} \ \mathrm{d}z &= \int_\varepsilon^R \frac{x^\alpha}{x^2 + 1} \ \mathrm{d}x
\end{align*}
For $\gamma_-$, we substitute $z = x \exp(2\pi i)$, so we have
\begin{align*}
\int_{\gamma_-} \frac{z^\alpha}{z^2 + 1} \ \mathrm{d}z &= \int_R^\varepsilon \frac{x^\alpha \exp(2\alpha \pi i)}{x^2 + 1} \ \mathrm{d}x
\end{align*}
and so by adding all the integrals together (and sending $R \to \infty$, $\varepsilon \to 0$), we get
\begin{align*}
\pi \left (\exp\left(\frac{\alpha \pi i}{2}\right ) - \exp\left(\frac{3\alpha \pi i}{2}\right )\right ) &= \left (1 - \exp(2\alpha\pi i) \right ) \int_0^\infty \frac{x^\alpha}{x^2 + 1} \ \mathrm{d}x
\end{align*}
Hence, if $\omega = \exp\left(\frac{\alpha \pi i}{2}\right )$, we get
\begin{align*}
\int_0^\infty \frac{x^\alpha}{x^2 + 1} \ \mathrm{d}x &= \frac{\pi \left(\omega - \omega^3 \right )}{1 - \omega^4}\\
&= \frac{\pi}{\omega + \omega^{-1}}\\
&= \frac{\pi}{2}\sec\left (\frac{\alpha \pi}{2} \right )
\end{align*}
Therefore,
\begin{align*}
\boxed{\int_0^\infty \frac{x^\alpha}{x^2 + 1} = \frac{\pi}{2}\sec\left (\frac{\alpha \pi}{2} \right )}
\end{align*}
Substituting $\alpha = \frac{1}{3}$ gives $\frac{\pi}{\sqrt{3}}$.
