Saturday, 16 February 2013

contour integration - Example of Improper integral in complex analysis



I'm doing this example of Cauchy principle value



0dxx3+1=2π33



After some steps i got,




[0,R]+CRdzz3+1=2πi(B1) where B1=Resz=z01z3+1



also I got that |CRdzz3+1|0 as R



There is problem to to finding residue at z0=1+3i2



Here i am considering the following contour:



enter image description here




please help me.thanks in advance.


Answer



The contour is good. Two things though:



1) You have to consider the integral along the angled line of the wedge contour. The angle of the contour was chosen to preserve the integrand. 2) Write z=ei2π/3x and get that the contour integral is



(1ei2π/3)0dxx3+1=i2π13ei2π/3



The term on the right is the residue at the pole z=eiπ/3 times i2π. I used the fact that, if f(z)=a(z)/b(z), then the residue of a simple pole zk of f is a(zk)/b(zk).




Note that ei2π/3ei4π/3=i3. The result follows.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...