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):
$$\int_0^\infty \frac{\sqrt{x}}{x^3+1}\ \mathrm dx$$
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 $0$.
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.
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);
Here is another method without the use of contours or the arctan function. We take advantage of two Mellin transforms, namely, $$\begin{align*} \mathcal{M}\left\{\frac{1}{1+t};s\right\}&=\frac{\pi}{\sin(\pi s)} \\ \mathcal{M}\left\{f(t^a);s\right\}&=\frac{1}{a}\mathcal{M}\left\{f(t);s/a\right\}. \end{align*}$$ The first result is proved by first substituting $1+t=(1-x)^{-1},dt={(1-x)^{-2}}dx $, $$\begin{align*} \mathcal{M}\left\{\frac{1}{1+t};s\right\}&=\int_0^{+\infty}\frac{t^{s-1}}{1+t}\ dt& \\ &=\int_0^1x^{s-1}(1-x)^{-s}\ dx \\ &=B(s,1-s) \\ &=\Gamma(s)\Gamma(1-s) \\ &=\frac{\pi}{\sin(\pi s)}\end{align*}$$ by Euler's reflection formula, where $B(n,m)$ is the Beta function.
For the second result substitute $t\to t^a$, $$\mathcal{M}\{f(t^a);s\}=\int_0^{+\infty}f(t^a)t^{s-1}\ dt \stackrel{t\to t^a}{=}\frac{1}{a}\int_0^{+\infty}f(t)t^{(s/a)-1}\ dt =\frac{1}{a}\mathcal{M}(f(t);s/a).$$ Hence applying the second result on the first gives, $$\mathcal{M}\left\{\frac{1}{1+t^a};s\right\}=\frac{\pi}{a}\csc\left(\frac{\pi}{a}s\right)$$ taking $s=3/2,a=3$, $$\begin{align*} \mathcal{M}\left\{\frac{1}{1+t^3};3/2\right\}&=\int_0^{+\infty}\frac{t^{1/2}}{1+t^3}\ dt=\frac{\pi}{3}\csc\left(\frac{\pi}{2}\right)=\frac{\pi}{3}. \end{align*}$$
