Another residue theory integral I need to evaluate the following real convergent improper integral using residue theory (vital that i use residue theory so other methods are not needed here)
I also need to use the following contour (specifically a keyhole contour to exclude the branch cut):

$$\int_0^\infty \frac{\sqrt{x}}{x^3+1}\ \mathrm dx$$
 A: Consider the integral of $\sqrt{z}/(z^3+1)$ around the given contour, using a branch of $\sqrt{z}$ with branch cut on the positive real axis.  This can be evaluated using residues.
Note that (in the appropriate limit) the integrals over $L_1$ and $L_2$ both approach $\int_0^\infty \frac{\sqrt{x}}{x^3+1}\ dx$ (for $L_2$ you're going from right to left, but also $\sqrt{z}$ approaches $-\sqrt{x}$ as $z$ approaches $x$ from below the branch cut).  The integrals over the circular arcs should both go to $0$.
A: Close format for this type of integrals:
$$ \int_0^{\infty} x^{\alpha-1}Q(x)dx =\frac{\pi}{\sin(\alpha \pi)} \sum_{i=1}^{n} \,\text{Res}_i\big((-z)^{\alpha-1}Q(z)\big) $$
$$ I=\int_0^\infty \frac{\sqrt{x}}{x^3+1} dx   \rightarrow \alpha-1=\frac{1}{2} \rightarrow \alpha=\frac{3}{2}$$
$$ g(z) =(-z)^{\alpha-1}Q(z) =\frac{(-z)^{\frac{1}{2}}}{z^3+1} =\frac{i \sqrt{z}}{z^3+1}$$
$$ z^3+1=0 \rightarrow  \hspace{8mm }z^3=-1=e^{i \pi} \rightarrow \hspace{8mm }z_k=e^{\frac{\pi+2k \pi}{3}}
$$
$$z_k= \begin{cases} k=0 & z_1=e^{i \frac{\pi}{3}}=\frac{1}{2}+i\frac{\sqrt{3}}{2}    \\ k=1 &
z_2=e^{i \pi}=-1 \\k=2 & z_3=e^{i \frac{5 \pi}{3}}=\frac{1}{2}-i\frac{\sqrt{3}}{2}  \end{cases}$$
$$R_1=\text{Residue}\big(g(z),z_1\big)=\frac{i \sqrt{z_1}}{(z_1-z_2)(z_1-z_3)}$$
$$R_2=\text{Residue}\big(g(z),z_2\big)=\frac{i \sqrt{z_2}}{(z_2-z_1)(z_2-z_3)}$$
$$R_3=\text{Residue}\big(g(z),z_3\big)=\frac{i \sqrt{z_3}}{(z_3-z_2)(z_3-z_1)}$$
$$ I=\frac{\pi}{\sin\left( \frac{3}{2} \pi\right)} (R_1+R_2+R_3) = \frac{\pi}{-1} \left(\frac{-1}{3}\right)=\frac{\pi}{3}$$
Matlab  Program
     syms x
     f=sqrt(x)/(x^3+1);
     int(f,0,inf) 
     ans =

     pi/3

Compute R1,R2,R3 with Malab
     z1=exp(i*pi/3);
     z2=exp(i*pi);
     z3=exp(5*i*pi/3);
     R1=i*sqrt(z1)/((z1-z2)*(z1-z3));
     R2=i*sqrt(z2)/((z2-z1)*(z2-z3));
     R3=i*sqrt(z3)/((z3-z2)*(z3-z1));
     I=(-pi)*(R1+R2+R3);

A: Since a solution involving contour integration has been given, I am providing an alternative method without contour integration.  Let $u:=\sqrt{x}$.  Then, the integral $I:=\displaystyle\int_0^\infty\,\frac{\sqrt{x}}{x^3+1}\,\text{d}x$ equals
$$I=2\,\int_0^\infty\,\frac{u^2}{u^6+1}\,\text{d}u=\int_{-\infty}^{+\infty}\,\frac{u^2}{u^6+1}\,\text{d}u\,.$$
Note that
$$\frac{u^2}{u^6+1}=\frac{1}{3}\,\left(\frac{u^2+1}{u^4-u^2+1}\right)-\frac13\,\left(\frac{1}{u^2+1}\right)\,.$$
Now, let $v:=u-\frac{1}{u}$.  Then,
$$\frac{u^2+1}{u^4-u^2+1}=\frac{1+\frac{1}{u^2}}{\left(u-\frac{1}{u}\right)^2+1}=\left(\frac{1}{v^2+1}\right)\,\frac{\text{d}v}{\text{d}u}\,.$$
Thus,
$$\begin{align}\int\,\frac{u^2+1}{u^4-u^2+1}\,\text{d}u&=\int\,\frac{1}{v^2+1}\,\text{d}v\\&=\text{arctan}(v)+C\\&=\text{arctan}\left(u-\frac{1}{u}\right)+C\,,\end{align}$$
where $C$ is a constant of integration.
Thus,
$$\int_0^{+\infty}\,\frac{u^2+1}{u^4-u^2+1}\,\text{d}u=\pi=\int_{-\infty}^0\,\frac{u^2+1}{u^4-u^2+1}\,\text{d}u\,.$$
On the other hand,
$$\int_{-\infty}^{+\infty}\,\frac{1}{u^2+1}\,\text{d}u=\Big.\big(\text{arctan}(u)\big)\Big|_{u=-\infty}^{u=+\infty}=\pi\,,$$
making
$$I=\frac{1}{3}\,(2\pi)-\frac{1}{3}\,\pi=\frac{\pi}{3}\,.$$
This result agrees with the computation made by Amir Alizadeh approximately six years ago.
