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{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




x6(a4+x4)2dx=3π28a



Edit.






I've had an idea: from the integration by parts
udv=uvvdu



and if we let

dv=4x3(4+x4)2,u=x34



with



dv=ddx14+x4=4x3(4+x4)2



we get finally



x6(4+x4)2dx=0+34x21+x4dx




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 x21+x4dx can be easily done by a semi-circle contour, computing the residues at (1+i)/2 and (1+i)/2. It is easy to check that the integral of z21+z4 on the arc of the semi-circle goes to 0.



z21+z4=z2(zeiπ/4)(zei3π/4)(zei5π/4)(zei7π/4) shows that it is a simple pole at those residues.



 Res(z21+z4,eiπ/4)=limzeiπ/4z2(zeiπ/4)1+z4=limzeiπ/43z22zeiπ/44z3=(3412)eiπ/4.




 Res(z21+z4,ei3π/4)=limzei3π/4z2(zei3π/4)1+z4=limzei3π/43z22zei3π/44z3=(3412)ei3π/4.



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



So the answer is x21+x4dx=2πi14(1i2+1i2)=22π.



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.



x24+x4dx=1212(x2)21+(x2)4dx=12x21+x4dx.




So the final answer is x6(4+x4)2dx=34x24+x4dx=38π. You also have to argue that you are limiting r in [x34(4+x4)]rr which is why it is 0.


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