Trouble with use of the Residue Theorem

In summary, the speaker is trying to apply the residue theorem to evaluate the integral \int_{0}^{\infty}\frac{dx}{x^3+a^3}=\frac{2\pi}{3\sqrt{3}a^2} where a is a real number greater than 0. They are attempting to use a ray and an arc at infinity to evaluate the integral, but they are having trouble getting the correct result. They have shown that the arc integral goes to 0 as the radius becomes infinite and the ray integral also goes to 0. However, they are not sure where they went wrong and are seeking help. The expert summarizer then provides a solution, stating that the integral along z/
  • #1
jeffreydk
135
0
I have just learned the residue theorem and am attempting to apply it to this intergral.

[tex]
\int_{0}^{\infty}\frac{dx}{x^3+a^3}=\frac{2\pi}{3\sqrt{3}a^2}
[/tex]

where [itex]a[/itex] is real and greater than 0. I want to take a ray going out at [itex]\theta=0[/itex] and another at [itex]\theta=\frac{2\pi}{3}[/itex] 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

[tex]\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|}[/tex]

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

[tex]\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[/tex]

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

[tex] \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 [/tex]

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

[tex]\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}[/tex]

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

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

I'm not sure where I went wrong--any help would be greatly appreciated.
 
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  • #2
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
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.
 

1. What is the Residue Theorem and how is it used?

The Residue Theorem is a mathematical tool used in complex analysis to evaluate integrals of complex functions. It states that the integral of a complex function around a closed contour is equal to the sum of the residues of the function at its singularities inside the contour. This theorem is often used to simplify and solve complex integrals.

2. What are the common challenges when using the Residue Theorem?

One of the main challenges when using the Residue Theorem is identifying the singularities of the complex function. This can be difficult, especially for functions with multiple poles or essential singularities. Another challenge is finding the correct contour to use for the integration, as it must enclose all singularities of the function.

3. Can the Residue Theorem be used for any type of integral?

No, the Residue Theorem is only applicable to integrals of complex functions. It cannot be used for real integrals or integrals with real variables.

4. Are there any restrictions or limitations when using the Residue Theorem?

Yes, there are some limitations when using the Residue Theorem. It can only be used for integrals with simple poles or essential singularities. Integrals with branch points or other types of singularities cannot be evaluated using this theorem.

5. How can I ensure that I am using the Residue Theorem correctly?

To ensure that you are using the Residue Theorem correctly, it is important to carefully identify and analyze the singularities of the complex function. Double check that the contour being used encloses all the singularities and that the residues are calculated correctly. It may also be helpful to compare your solution to known solutions or use a computer program to verify the result.

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