Need a hint for this integral I'm trying to evaluate the following integral
$$\int_0^{\infty} \frac{1}{x^{\frac{3}{2}}+1}\,dx.$$
This is an old complex analysis exam question, so I plan to use the residue theorem.  
How can I first deal with the square-root, cubed term?  I've been trying to find a clever substitution to reduce the problem to a simple one but so far I have not found a good one...
Any ideas are welcome.
Thanks,
 A: Hint: Every time I see integrand like this which integrate over $(0,\infty)$, I will transform it to a beta integral and follows my nose.
Spoiler 1

 Let $\displaystyle\;y = x^{3/2}\;$ and $\displaystyle\;z = \frac{y}{1+y}\;$, we have

Spoiler 2

 $$\begin{align} \int_0^\infty \frac{dx}{x^{3/2}+1}&= \int_0^\infty \frac{dy^{2/3}}{y+1} = \frac23 \int_0^\infty \frac{y^{2/3-1} dy}{y+1}\\ &= \frac23 \int_0^\infty \left(\frac{y}{1+y}\right)^{2/3-1}\left(\frac{1}{1+y}\right)^{1/3-1}\frac{dy}{(1+y)^2}\\ &= \frac23 \int_0^1 z^{2/3-1}(1-z)^{1/3-1}dz \\ &= \frac23\frac{\Gamma(2/3)\Gamma(1/3)}{\Gamma(2/3+1/3)} = \frac23 \frac{\pi}{\sin\frac{\pi}{3}}\\ & = \frac{4\pi}{3\sqrt{3}} \end{align} $$

Update (sorry, this part is too hard to setup as spoiler correctly)
If you really want to compute the integral using residue, you can


*

*change variable to $y = x^{3/2}$.  

*pick the branch cut of $y^{-1/3}$ in the resulting integrand along the positive real axis.  

*set up a integral over the contour:
$$C :=  +\infty - \epsilon i\quad\to\quad -\epsilon-\epsilon i\quad \to \quad -\epsilon + \epsilon i \quad\to\quad +\infty + \epsilon i$$   


If you fix the argument of $y^{-1/3}$ to be zero on the upper branch of $C$, you will have 
$$\left(1 - e^{-\frac{2\pi i}{3}}\right)\int_0^\infty \frac{y^{-1/3}dy}{y+1} = \int_C \frac{y^{-1/3}dy}{y+1} =^{\color{blue}{[1]}} 2\pi i\mathop{\text{Res}}_{y = -1}\left(\frac{y^{-1/3}}{y+1}\right) = 2\pi i e^{-\frac{\pi i}{3}}$$ 
This will give you
$$\int_0^\infty \frac{dx}{x^{3/2}+1} = \frac23 \int_0^\infty \frac{y^{-1/3} dy}{y+1} = \frac{4\pi i}{3}\left(\frac{e^{-\frac{\pi i}{3}}}{1 - e^{-\frac{2\pi i}{3}}}\right) = \frac{2\pi}{3\sin\frac{\pi}{3}} = \frac{4\pi}{3\sqrt{3}}$$
Notes


*

*$\color{blue}{[1]}$ Since the integrand $\frac{y^{-1/3}}{y+1}$ goes to $0$ faster than $\frac{1}{|y|}$ as $|y| \to \infty$, we can complete the contour $C$ by a circle of infinite radius and evaluate the integral over $C$ by taking residue at poles within the extended contour.

A: Let $t=\sqrt x$. Then $dt=\frac{1}{2}x^{-\frac{1}{2}}dx$, so $dx=2\sqrt x\ dt=2t\ dt$.
We have
$$\int\frac{2t}{t^3+1}dt$$
This integral is not going to get very simple as you can see via wolfram. However we can integrate by parts and use arctan.
$u=\frac{1}{t^3+1}$ and $dv=2t\ dt$
So $du=\frac{-3t^2}{(t^3+1)^2}dt$ and $v=t^2$
We have
$$\frac{t^2}{t^3+1}+\int \frac{3t^4}{(t^3+1)^2}$$
Now at least the term outside the integral is $0$, so we can focus on the integral. You should be able to use partial fractions at this point, and like I said there will be an arctan. I've checked this far with wolfram, by the way, but I'll leave the rest to you.
A: $$\int\frac{dx}{x^{3/2} + 1}$$
$$=2\int\frac{u}{u^3 + 1}du$$
Depending on how you want to solve the problem, you could start working from here  using residue theory or save some work and break down into partial fractions
$$= \frac{2}{3} \int \frac{u}{u^2 - u + 1}du + \frac{2}{3} \int \frac{1}{u^2 - u + 1}du - \frac{2}{3}\int \frac{1}{u+1}du$$
The first should fall to a $u$-sub, the second by completing the square (giving an answer in terms of $\arctan$) and the last is trivial. Of course, if you have to use residue theory than this is somewhat redundant :)
