- #1
jeffreydk
- 133
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I have just learned the residue theorem and am attempting to apply it to this intergral.
<br /> \int_{0}^{\infty}\frac{dx}{x^3+a^3}=\frac{2\pi}{3\sqrt{3}a^2}<br />
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.
<br /> \int_{0}^{\infty}\frac{dx}{x^3+a^3}=\frac{2\pi}{3\sqrt{3}a^2}<br />
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.