I have just learned the residue theorem and am attempting to apply it to this intergral.(adsbygoogle = window.adsbygoogle || []).push({});

[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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# Trouble with use of the Residue Theorem

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