I am trying to show $\displaystyle{\int_{0}^{\infty} \frac{dx}{x^3+1} = \frac{2\pi}{3\sqrt{3}}}.$
Any help? (I am having troubles using the half circle infinite contour)
Or more specifically, what is the residue $\text{res} \left(\frac{1}{z^3+1},z_0=e^\frac{\pi i}{3} \right )$
Thanks!
Hint
Substitute $u=1/x \ $ then $ \ \text{d}u=(-1/x^2)\text{d}x=-u^2\text{d}x$ $$ I=\int_{0}^{+\infty}\frac{\text{d}x}{1+x^3}=-\int_{+\infty}^{0}\frac{1}{1+(1/u)^3}\left(\frac{\text{d}u}{u^2}\right) $$ Then $$ \int_{0}^{+\infty}\frac{\text{d}x}{1+x^3}=\int_{0}^{+\infty}\frac{u}{1+u^3}\text{d}u$$ What is beautiful here is that you can sum the two different expressions
$$ 2I=\int_{0}^{+\infty}\frac{1+u}{1+u^3}\text{d}u=\int_{0}^{+\infty}\frac{\text{d}u}{1-u+u^2} $$
Then you can write that $$ 1-u+u^2=\left(u-\frac{1}{2}\right)^2+\frac{3}{4}=\frac{3}{4}\left(\frac{4}{3}\left(u-\frac{1}{2}\right)^2+1\right)$$ Then you can write $\displaystyle \frac{4}{3}$ as $\displaystyle \left(\frac{2}{\sqrt{3}}\right)^2$ So far we have $$ 2I=\frac{4}{3}\int_{0}^{+\infty}\frac{\text{d}u}{\displaystyle 1+\left(\frac{2}{\sqrt{3}}\left(u-\frac{1}{2}\right)\right)^2}=\frac{2}{\sqrt{3}}\int_{0}^{+\infty}\frac{2/\sqrt{3}}{\displaystyle 1+\left(\frac{2}{\sqrt{3}}\left(u-\frac{1}{2}\right)\right)^2}\text{d}u $$ There we have the arctangente and $$ 2I=\frac{2}{\sqrt{3}}\left[\text{arctan}\left(\frac{2}{\sqrt{3}}\left(u-\frac{1}{2}\right)\right)\right]^{+\infty}_{0}=\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}-\text{arctan}\left(-\frac{1}{\sqrt{3}}\right)\right) $$ Then with $\displaystyle \text{arctan}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$ we finally have $$ 2I=\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}+\frac{\pi}{6}\right)=\frac{2}{\sqrt{3}}\frac{4 \pi}{6}$$ Here we have the claimed
$$ I=\frac{2 \pi}{3\sqrt{3}} $$
Beautiful equality
use that $$\frac{1}{1+x^3}=1/3\, \left( x+1 \right) ^{-1}+1/3\,{\frac {-x+2}{{x}^{2}-x+1}}$$
There are already two answers showing how to find the integral using just calculus. It can also be done by the Residue Theorem:
It sounds like you're trying to apply RT to the closed curve defined by a straight line from $0$ to $A$ followed by a circular arc from $A$ back to $0$. That's not going to work, because there's no reason the intergal over the semicircle should tend to $0$ as $A\to\infty$.
How would you use RT to find $\int_0^\infty dt/(1+t^2)$? You'd start by noting that $$\int_0^\infty\frac{dt}{1+t^2}=\frac12\int_{-\infty}^\infty\frac{dt}{1+t^2},$$and apply RT to the second integral.
You can't do exactly that here, because the function $1/(1+t^3)$ is not even. But there's an analogous trick available.
Hint: Let $$f(z)=\frac1{1+z^3}.$$If $\omega=e^{2\pi i/3}$ then $$f(\omega z)=f(z).$$(Now you're going to apply RT to the boundary of a certain sector of opening $2\pi/3$... be careful about the "$dz"$...)
$$\int_{0}^{+\infty}\frac{dx}{1+x^3}=\int_{0}^{1}\frac{dx}{1+x^3}+\int_{1}^{+\infty}\frac{dx}{1+x^3}=\int_{0}^{1}\frac{1+x}{1+x^3}\,dx$$
clearly equals $\int_{0}^{1}\frac{dx}{1-x+x^2}$, which is an elementary integral. We may also state
$$\int_{0}^{1}\frac{dx}{1+x^3}=\int_{0}^{1}\frac{1+x-x^3-x^4}{1-x^6}\,dx = \sum_{n\geq 0}\left[\frac{1}{6n+1}+\frac{1}{6n+2}-\frac{1}{6n+4}-\frac{1}{6n+5}\right]$$
and by denoting as $\chi_3$ and $\chi_6$ the non-principal Dirichlet's characters $\!\!\pmod{3}$ and $\!\!\pmod{6}$ the RHS can be written as $L(\chi_6,1)+\frac{1}{2}\,L(\chi_3,1)$, which is related to $\sqrt{3}$ and $\pi$ via the class number formula.
Another approach is given by the reflection formula for the $\psi$ (digamma) function, since
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b},\qquad \psi(x)-\psi(1-x)=-\pi\cot(\pi x)$$
ensure
$$ \sum_{n\geq 0}\left[\frac{1}{6n+1}-\frac{1}{6n+5}\right]=\tfrac{\pi}{6}\cot\left(\tfrac{\pi}{6}\right),\qquad \sum_{n\geq 0}\left[\frac{1}{3n+1}-\frac{1}{3n+2}\right]=\tfrac{\pi}{3}\cot\left(\tfrac{\pi}{3}\right). $$
Here is an approach that makes use of Euler's Beta function. Enforcing a substitution of $x \mapsto \sqrt[3]{x}$ one has \begin{align*} \int_0^\infty \frac{dx}{x^3 + 1} &= \frac{1}{3} \int_0^\infty \frac{x^{-2/3}}{x + 1} \, dx\\ &= \frac{1}{3} \int_0^\infty \frac{x^{\frac{1}{3} - 1}}{(x + 1)^{\frac{1}{3} + \frac{2}{3}}}\\ &= \frac{1}{3} \text{B} \left (\frac{1}{3}, \frac{2}{3} \right ) \tag1\\ &= \frac{1}{3} \frac{\Gamma \left (\frac{1}{3} \right ) \Gamma \left (\frac{2}{3} \right )}{\Gamma (1)} \tag2\\ &= \frac{1}{3} \Gamma \left (1 - \frac{1}{3} \right ) \Gamma \left (\frac{1}{3} \right )\\ &= \frac{1}{3} \cdot \frac{\pi}{\sin (\pi/3)} \tag3\\ &= \frac{1}{3} \cdot \frac{2\pi}{\sqrt{3}}\\ &= \frac{2\pi}{3 \sqrt{3}}. \end{align*}
Explanation
(1) Since $\displaystyle{\text{B}(x,y) = \int_0^\infty \frac{t^{x - 1}}{(1 + t)^{x + y}} \, dt}$.
(2) Since $\text{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$.
(3) From Euler's reflection formula.