Friday, 16 November 2018

integration - How to show intinfty0fracdxx3+1=frac2pi3sqrt3




I am trying to show 0dxx3+1=2π33.




Any help?
(I am having troubles using the half circle infinite contour)



Or more specifically, what is the residue res(1z3+1,z0=eπi3)



Thanks!


Answer



There are already two answers showing how to find the integral using just calculus. It can also be done by the Residue Theorem:




It sounds like you're trying to apply RT to the closed curve defined by a straight line from 0 to A followed by a circular arc from A back to 0. That's not going to work, because there's no reason the intergal over the semicircle should tend to 0 as A.



How would you use RT to find 0dt/(1+t2)? You'd start by noting that 0dt1+t2=12dt1+t2,and apply RT to the second integral.



You can't do exactly that here, because the function 1/(1+t3) is not even. But there's an analogous trick available.



Hint: Let f(z)=11+z3.If ω=e2πi/3 then f(ωz)=f(z).(Now you're going to apply RT to the boundary of a certain sector of opening 2π/3... be careful about the "dz"...)


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