Compelx variable question, the Residue Thm and applications

[alphas]

Hello,

I am stuck with two complex variable questions. Among others, I can't find residues when I have the poles, and these two questions are getting me out of hope. Maybe one of the geniuses on this site knows how to drive those 2 problems to an answer:

1) Computing the following integral:
Intergral from -oo to +oo of: (x*x) / (x*x*x*x - 4x*x + 5 ) dx


2) Computing the following integral:
Integral from 0 to 2 pi of: d(theta) / [( 1 + B cos theta)*( 1 + B cos theta) ]


Formulas: residue theorem
http://en.wikipedia.org/wiki/Residue_theorem"


Attempts:
For the 1st one,
First I searched zeros for Q(z) = (z*z*z*z - 4z*z + 5 )
I searched for z*z first. I found
z*z = 2 - i
or
z*z = 2+i

Then using polar coord., I found:
5^(1/4) * e^(i theta/2)
- 5^(1/4) * e^(i theta/2)
5^(1/4) * e^(- i theta/2)
5^(1/4) * e^(i theta/2)
Are they right?

with theta = arcsin (5^(-1/2))

and then I would like to get to the solution but I am stuck.
Same for 2nd problem.
Thanks for your help

 

[Roberto Bramb]

The denominator D factorizes in D=x1*x2*x3*x4, where
x1=sqrt(2+i) (first quadrant)
x2=-sqrt(2-i) (second quadrant)
x3=-sqrt(2+i) (third quadrant)
x4=sqrt(2-i) (forth quadrant)

The integral is given by the residue theorem

int=(2*pi*i)[Res(x1)+Res(x2)]

and Res(x1)=-(i/4)*sqrt(2+1), Res(x2)=-(i/4)*sqrt(2-1), so

int=(pi/2)(sqrt(2-i)+sqrt(2+i))= pi*sqrt[2+sqrt(5)/2]

 

[HallsofIvy]

Roberto, those are the zeros of the denominator. What you meant to say was that D= (z- x1)(z- x2)(z-x3)(z-x4). And the "residue theorem does NOT say that- at least not directly.

alphas, the "residue theorem" says that the integral of a function of a complex variable on a closed path in the complex plane is the sum of the residues at each pole inside that path.

Your integral is on a straight line, not a closed path. You will need to decide on a closed path that includes that, at least in the limit.

I recommend: integrate from -R to R on the "real" axis, then around the upper half of a circle with radius R in the complex plane from R to -R. Take the limit as R goes to infinity. IF you can show that the integral around the semi-circle goes to 0, then the integral on the line goes to the sum of the residues.