Sunday, 24 November 2013

complex analysis - Evaluatingintinftyinftyfracx6(4+x4)2dxusing residues



I need help to solve the next improper integral using complex analysis:



x6(4+x4)2dx




I have problems when I try to find residues for the function f=1(z4+4)2.



This is what I tried.



res(f,2ei(π4+kπ2))=limz2ei(π4+kπ2)((z2ei(π4+kπ2))2(z44)2)



with k\in\{0,1,2,3\}. What do you think about it?



I know there is a little more general problem involving this integral; for all a>0




\int_{-\infty}^{\infty} \frac{x^6}{(a^4+x^4)^2} dx= \frac{3\pi\sqrt{2}}{8a}



Edit.






I've had an idea: from the integration by parts
\int u dv = uv - \int vdu



and if we let

dv = \frac{4x^3 }{(4+x^4)^2}, \, u = \frac{x^3}{4}



with



dv = -\frac{d}{dx} \frac{1}{4+x^4} = \frac{4x^3 }{(4+x^4)^2}



we get finally



\int_{-\infty}^{\infty} \frac{x^6}{(4+x^4)^2} dx = 0 + \frac{3}{4} \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx




which I think is more easy to solve.



Anyway, if you know another idea or how to complete my first try will be welcome.


Answer



If you are right, then \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx can be easily done by a semi-circle contour, computing the residues at (1+i)/\sqrt{2} and (-1+i)/\sqrt{2}. It is easy to check that the integral of \frac{z^2}{1+z^4} on the arc of the semi-circle goes to 0.



\frac{z^2}{1+z^4}=\frac{z^2}{(z-e^{i\pi/4})(z-e^{i3\pi/4})(z-e^{i5\pi/4})(z-e^{i7\pi/4})} shows that it is a simple pole at those residues.



\text{ Res}(\frac{z^2}{1+z^4},e^{i\pi/4})=\lim_{z\to e^{i\pi/4}}\frac{z^2(z-e^{i\pi/4})}{1+z^4}=\lim_{z\to e^{i\pi/4}}\frac{3z^2-2ze^{i\pi/4}}{4z^3}=(\frac{3}{4}-\frac{1}{2})e^{-i\pi/4}.




\text{ Res}(\frac{z^2}{1+z^4},e^{i3\pi/4})=\lim_{z\to e^{i3\pi/4}}\frac{z^2(z-e^{i3\pi/4})}{1+z^4}=\lim_{z\to e^{i3\pi/4}}\frac{3z^2-2ze^{i3\pi/4}}{4z^3}=(\frac{3}{4}-\frac{1}{2})e^{-i3\pi/4}.



I used L'Hospital's rule above because it seems simpler.



So the answer is \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx=2\pi i\frac{1}{4}\left(\frac{1-i}{\sqrt{2}}+\frac{-1-i}{\sqrt{2}}\right)=\frac{\sqrt{2}}2\pi.



Edit: Something's wrong with the integration by parts; we do not seem to get the right answer. It is a 4 in the bottom and not 1. But we can do a further substitution to get it right.



\int_{-\infty}^{\infty} \frac{x^2}{4+x^4} dx=\frac{1}{\sqrt{2}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2}}\frac{(\frac{x}{\sqrt{2}})^2}{1+(\frac{x}{\sqrt{2}})^4} dx=\frac{1}{\sqrt{2}}\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx.




So the final answer is \int_{-\infty}^{\infty} \frac{x^6}{(4+x^4)^2} dx=\frac{3}{4}\int_{-\infty}^{\infty} \frac{x^2}{4+x^4} dx=\frac{3}{8}\pi. You also have to argue that you are limiting r\to\infty in \left[\frac{x^3}{4(4+x^4)}\right]_{-r}^r which is why it is 0.


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