# Trouble with use of the Residue Theorem

1. Sep 20, 2009

### jeffreydk

I have just learned the residue theorem and am attempting to apply it to this intergral.

$$\int_{0}^{\infty}\frac{dx}{x^3+a^3}=\frac{2\pi}{3\sqrt{3}a^2}$$

where $a$ is real and greater than 0. I want to take a ray going out at $\theta=0$ and another at $\theta=\frac{2\pi}{3}$ and connect them with an arc at infinity to evaluate the integral using the residue theorem; but so far I am having problems getting the correct result. I have shown

$$\left|\frac{1}{x^3+a^3}\right|=\frac{1}{|x|^3}\frac{1}{\left|1+\frac{a^3}{x^3}\right|}\leq \frac{1}{|x|^3}\frac{1}{\left|1-\frac{|a|^3}{|x|^3}\right|}$$

by the triangle inequality. If we can say that $\frac{|a|}{|x|}<\frac{1}{2^{1/3}}$ then it follows that

$$\left| \int_{\text{arc}}\frac{dx}{x^3+a^3}\right|\leq \left(\frac{2\pi R}{3}\right)\frac{2}{R^3}=\frac{4\pi}{3R^2}\longrightarrow 0$$

Thus the arc integral goes to 0 as the radius become infinite as needed. For the ray along $\theta=\frac{2\pi}{3}$ using a parametrization of $x=(R-t)e^{2\pi i/3}$ which gives $dx=-e^{2\pi i/3}dt$ the integral becomes

$$\int_{\text{ray}_2}\frac{dx}{x^3+a^3}=\int_{\text{ray}_2}\frac{-e^{2\pi i/3}dt}{(R-t)^3e^{-2\pi i}+a^3}\longrightarrow 0$$

again as needed. This would mean to me that the integral should just be the sum of residues lying inside my contour (times $2\pi i$ of course), which is just the one pole $x=ae^{i\pi/3}$. This would give

$$\text{Res}f(ae^{i\pi/3})=\lim_{x\rightarrow ae^{i\pi/3}}(x-ae^{i\pi/3})\left(\frac{1}{x^3+a^3}\right)=-\frac{e^{i\pi/3}}{3a^2}$$

and since it is a first order pole this is also the residue. Therefore the answer I am getting is

$$\int_{0}^{\infty}\frac{dx}{x^3+a^3}=-\frac{2\pi i e^{i\pi/3}}{3a^2}$$

I'm not sure where I went wrong--any help would be greatly appreciated.

2. Sep 20, 2009

### lurflurf

Your residue is right, but you should know that the integral along z/|z|=exp(2*pi*i/3) is not zero.
let I be the given integral
You will see
ray integral=C*I (for some costant C find it)
then
closed integral=given+arc+ray=(1-0-C)*I=2pi*i*res(f,z0)=-2pi*i*exp(2pi*i/3)/(3a^2)
recall
sin(z)=[exp(iz)-exp(-iz)]/(2i)

3. Sep 20, 2009

### jeffreydk

Ah!! Thank you very much for that insight, it hit me as soon as I read what you had to say. I haven't used that trick in awhile so I think I temporarily forgot about it.