Evaluating $\int_{0}^{\infty} \frac{1}{1+x^{3/2}}\,\textrm{d}x$ I'm trying to evaluate this integral using contour integration (over a Riemann surface), but I'm stuck at the step where I need to calculate the residues. The roots of $1+z^{3/2}$ are $1$ and $e^{2\pi*i/3}$, but I don't see any way to extract poles or whatnot from this information, as $z^{1/2}$ seems inseparable from $1+z^{3/2}$.
Help is appreciated!
 A: First, sub $x=y^2$ to get that the integral is equal to
$$2 \int_0^{\infty} dy \frac{y}{1+y^3}$$
Now we can proceed straightforwardly.  Consider the contour integral
$$\oint_C dz \frac{z \, \log{z}}{1+z^3}$$
where $C$ is a keyhole contour about the positive real axis.  In this case, one may show that this contour integral is simply
$$-i 2 \pi \int_0^{\infty} dy \frac{y}{1+y^3}$$
and is also equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand, or $z=e^{i \pi/3}$, $z=e^{i \pi}$, and $z=e^{i 5 \pi/3}$.  This sum is equal to
$$\frac{e^{i \pi/3} (i \pi/3)}{3 e^{i 2 \pi/3}} + \frac{e^{i \pi} (i \pi)}{3 e^{i 2 \pi}}+\frac{e^{i 5 \pi/3} (i 5 \pi/3)}{3 e^{i 10 \pi/3}}$$
which, upon simplifying, is seen to be equal to 
$$i \frac{\pi}{9} (i 2 \sqrt{3}) = -\frac{2 \pi}{3 \sqrt{3}}$$
Therefore the orginal integral is
$$\int_0^{\infty} \frac{dx}{1+x^{3/2}} = \frac{4 \pi}{3 \sqrt{3}}$$
A: Here's another approach.  Consider the general integral
$$\int_0^{\infty} \frac{dx}{1+x^a}$$
where $a>1$.  Sub $y=x^a$ and transform the integral to
$$\frac{1}{a} \int_0^{\infty} dy \frac{y^{-(a-1)/a}}{1+y}$$
This integral may be evaluated by considering the contour integral
$$\oint_C dz \frac{z^{-(a-1)/a}}{1+z}$$
where $C$ is a keyhole contour about the positive real axis.  One may then show that this integral is equal to
$$\left ( 1-e^{-i 2 \pi (a-1)/a}\right ) \int_0^{\infty} dy \frac{y^{-(a-1)/a}}{1+y} $$
which in turn is equal to $i 2 \pi$ times the residue at $z=e^{i \pi}$, or 
$$i 2 \pi e^{-i \pi (a-1)/a}$$
Therefore the integral is
$$\int_0^{\infty} \frac{dx}{1+x^a} = \frac{\pi/a}{\sin{(\pi/a)}}$$
Plugging in $a=3/2$, I get $(2 \pi/3)/\sin(2\pi/3) = 4 \pi/(3 \sqrt{3})$.
A: As shown in this answer,
$$
\int_0^\infty\frac{x^n}{1+x^m}\mathrm{d}x=\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)
$$
Setting $n=0$ and $m=3/2$, we get
$$
\begin{align}
\int_0^\infty\frac{1}{1+x^{3/2}}\mathrm{d}x
&=\frac{2\pi}{3}\csc\left(\frac{2\pi}{3}\right)\\
&=\frac{4\pi}{3\sqrt3}
\end{align}
$$
