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.)

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 $\frac{\pi}{\sqrt{3}}$.
