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Complex integration problem using residues .

  1. Dec 9, 2011 #1
    1. The problem statement, all variables and given/known data

    I =[itex]\int \frac{cosx}{x^{2}-2x+2}dx[/itex] the integral runs from -inf to inf

    evaluate the integral using the calculus of residues.


    2. Relevant equations

    shown in my attempt


    3. The attempt at a solution

    Re [itex]\oint\frac{e^{iz}}{z^{2}-2z+2}[/itex]

    with singularities at z=1+i and z-i they are both simple poles since we can turn

    [itex]z^{2}-2z+2[/itex]

    to

    [itex](z-2)z+2[/itex]

    now all we need to do is find the residues

    residue=(z-pole)f(z)

    the problem is i get an answer with imaginary numbers.

    my contour uses a semi circle above the real axis, so i only need to use the 1+i pole.
     
  2. jcsd
  3. Dec 9, 2011 #2
    [tex]\int_{-\infty}^{\infty}\frac{\cos(z)}{z^2-2z+2}=\text{Re}\left\{\int_{-\infty}^{\infty} \frac{e^{iz}}{z^2-2z-2}\right\}[/tex]

    so that when you compute the e^(iz) integral and obtain an answer in the form of a+bi, a is the value of the cosine integral. Also, need to write the integrand in factored form as

    [tex]\frac{e^{iz}}{(z-(1+i))(z-(1+i))}[/tex]

    then compute the residue.
     
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