1. ## [SOLVED] Evaluating residues

I'm having troubles evaluating this improper integral
Show: $\displaystyle \int_0^{\infty} \frac{1}{x^3+1} dx = \frac{2\pi}{3\sqrt{3}}$ using residues and a closed contour.

So, i used a closed contour C that only contains the singularity $\displaystyle z= e^{\frac{i\pi}{3}}$. I let R be the radius of the contour and I let R tend to infinity

So, $\displaystyle C = C_{+} + C_{R} + C_{\frac{2\pi}{3}}$
where $\displaystyle C_{+} = \{z=r: 0<r<R, 1<R<\infty \}$
$\displaystyle C_{R} = \{ z = Re^{i\theta}: 0\leq\theta\leq\frac{2\pi}{3}, 1<R<\infty \}$
$\displaystyle C_{\frac{2\pi}{3}} = \{z = re^{\frac{i\pi}{3}}: 0<r<R, 1<R<\infty \}$

I let $\displaystyle I = \lim_{R\to\infty}{\int_0^{R} \frac{1}{z^3 +1}dz}, I_{R} = \int_{C_R}\frac{1}{z^3 +1}dz, I_{\frac{2\pi}{3}} = \int_{C_{\frac{2\pi}{3}}}\frac{1}{z^3 +1}dz$

By Integral theorems, I get $\displaystyle I_{R} = 0$ as $\displaystyle R\to\infty$ and $\displaystyle I_{\frac{2\pi}{3}} = - e^{\frac{i\pi}{3}}* I$

By the Cauchy Residue Theorem, I get
$\displaystyle (1 - e^{\frac{i\pi}{3}})* I = 2\pi i{Res}_{z=e^{\frac{i\pi}{3}}} \frac{1}{z^3+1}$$\displaystyle = 2\pi i\lim_{z\to e^{\frac{i\pi}{3}}} ((1 - e^{\frac{i\pi}{3}})\frac{1}{z^3+1})$

but I can't get the answer to evaluate. I just wanna know if there is something I'm missing in complex arithmetic, or is it my set-up that is wrong

2. $\displaystyle I_{\frac{2\pi}{3}} = - e^{\frac{i\pi}{3}}* I$

I am confused here . isn't it $\displaystyle - e^{\frac{2\pi i}{3} } I$ ?

and also consider the residue

$\displaystyle Res_{z = e^{\frac{\pi i}{3}} } \left (\frac{1}{z^3 + 1} \right )$

$\displaystyle = \lim_{z \to e^{\frac{\pi i}{3}} } ( z - e^{\frac{\pi i}{3}} )\cdot \left (\frac{1}{z^3 + 1} \right )$

Use L'hospital to finish it

Here is my attempt

$\displaystyle (1 - e^{\frac{2\pi i}{3} })I = 2\pi i \frac{1}{3z^2} |_{z = e^{\frac{\pi i}{3}} } = \frac{2\pi i e^{\frac{\pi i}{3}}}{-3}$

$\displaystyle e^{\frac{\pi i}{3}} ( -2i \sin( \frac{\pi}{3} )) I = - \frac{2\pi i }{3} \cdot e^{\frac{\pi i}{3}}$

$\displaystyle \frac{\sqrt{3}}{2} I = \frac{\pi}{3}$

$\displaystyle I = \frac{2\pi}{3\sqrt{3}}$

3. That all looks right to me. So then, when you bring everything to one side and calculate the residue, I get:

$\displaystyle \int_0^{\infty} \frac{1}{1+x^3}dx=\frac{2\pi i}{(1+e^{\pi i/3})(e^{\pi i/3}-e^{-\pi i/3})(1-e^{2\pi i/3})}$

Not sure about the easiest way to evaluate that by hand, but I just reduced it to a+bi format and got as one step:

$\displaystyle \frac{2\pi i}{\sqrt{3}i(3/2+\sqrt{3}/2 i)(3/2-\sqrt{3}/2 i)}$

which eventually reduces down to $\displaystyle \frac{2\pi}{3\sqrt{3}}$.