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Problem with contour integration

  1. Apr 17, 2012 #1
    1. The problem statement, all variables and given/known data
    Evaluate [itex]\int[/itex]dx/(x2+2x+2) (limits from -∞ to ∞) by contour integration

    Ok so, this is what i did...:

    we have ---> [itex]\Gamma[/itex] : |z|=R

    [itex]\oint[/itex]dz/(z2+2z+2) =
    [itex]\int[/itex]dx/x2+2x+2 (limits -R to R) +[itex]\int[/itex]dz/(z2+2z+2) (lower limit [itex]\Gamma[/itex])--> (1)

    [where f(z)=1/(z2+2z+2)]

    we have, limz→∞ zf(z) = limz→∞ z/(z2+2z+2)
    = 0
    therefore,
    limR→∞ [itex]\int[/itex]dz/(z2+2z+2) [lower limit[itex]\Gamma[/itex]
    = 0

    Taking the limit R→∞ on both sides of (1)

    [itex]\oint[/itex]dz/(z2+2z+2)= [itex]\int[/itex]dx/x2+2x+2 [limits from -∞ to ∞) + 0


    where f(z)= 1/z2+2z+2 so for finding out the poles of f(z)....
    yeah this is where im stuck, cuz in the previous one we did f(z) was→1/z4+1

    so there we solved for z the way we solve for the roots of a complex number...
    all that k= n-1 and nth root of z = rei(θ+2k∏)1/n......

    so how do we solve in this case?

    Thanks alot in advance....
     
  2. jcsd
  3. Apr 17, 2012 #2

    Dick

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    To find the poles of 1/(z^2+2z+2) you need to find the roots of z^2+2z+2=0. That shouldn't be hard for you. It's a quadratic equation.
     
  4. Apr 17, 2012 #3
    okay...after solving quadratically, then we go for the residues, right? But i dont get the answer im supposed to...its supposed to be π...
     
  5. Apr 17, 2012 #4

    micromass

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    Tell us what you did!!
     
  6. Apr 17, 2012 #5

    Dick

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    Sure, it's pi. Show why you didn't get pi. What are the residues and what kind of contour are you picking? Can you show your work?
     
  7. Apr 17, 2012 #6
    I completed the square for that qudratic equation and its like this...→(z+1)2+1
    then what? we find the residue for z→-1 to the order of 2?....
     
  8. Apr 17, 2012 #7

    Dick

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    You find the roots where the poles are and pick a contour first. Then decide what residues you need to evaluate.
     
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