Integration using a contour I am trying to evaluate $$I=\int_0^\infty\frac{1}{(\sqrt{x})^3+1}dx$$ I know that I can probably do a change of variable and then split it to partial fraction, which should be integrable without any complex analysis. However, I am struggling with solving this integral with a contour. For example, I tried using the semi-circle in the upper half-plane, but I cannot figure out $\int_{-\infty}^0\frac{1}{(\sqrt{x})^3+1}dx$ as a multiple of $I$. I changed it to $\int_\infty^0\frac{1}{(\sqrt{e^{i\pi}x})^3+1}dx$, and do not know how to proceed. Is partial fraction here inevitable?
 A: Hint
Replace $x=u^2$ to obtain
$$
I=\int_0^\infty \frac{2udu}{u^3+1}
$$
and perform contour integration on the above expression.
A: This is actually a standard integral. I don't really like partial fractions since this will involve an irreducible quadratic, and therefore many logs and arctans in the integration which is just ugly.
Suppose $\alpha>1$ is any real number. Then, $\int_0^{\infty}\frac{dx}{1+x^{\alpha}}=\frac{\pi/\alpha}{\sin(\pi/\alpha)}$ (I've seen this integral a few times so I just so happen to have commited this to memory). The way I prefer to prove this is to first make the substitution $t=x^{\alpha}$, which yields
\begin{align}
\int_0^{\infty}\frac{dx}{1+x^{\alpha}}&=\int_0^{\infty}\frac{\frac{1}{\alpha}t^{\frac{1}{\alpha}-1}}{1+t}\,dt=\frac{1}{\alpha}\int_0^1\frac{dt}{t^{1-\frac{1}{\alpha}}(1+t)}
\end{align}
and this is of the form $\int_0^{\infty}\frac{R(t)}{t^{\beta}}\,dt$ for some rational function $R(t)$ (namely $\frac{1}{1+t}$) and some $0<\beta<1$. Here, the answer can be obtained by integrating along a "pacman contour" (see the here for the picture, and an explanation of why the contour works, if you haven't already seen this). Anyway, by following the procedure there, you can prove that $\int_0^{\infty}\frac{dt}{t^{\beta}(1+t)}=\frac{\pi}{\sin(\pi \beta)}=\frac{\pi}{\sin(\pi/\alpha)}$, where the last equal sign is just the addition formula.
I'm pretty sure you can also evaluate the integral $\int_0^{\infty}\frac{dx}{1+x^{\alpha}}$ by integrating over the sector of angle $\frac{2\pi}{\alpha}$.
So, for the particular case of $\alpha=\frac{3}{2}$, this yields $\frac{2\pi/3}{\sin(2\pi/3)}=\frac{2\pi/3}{\sqrt{3}/2}=\frac{4\pi}{3\sqrt{3}}$.
